Namma Kalvi - The No.1 Educational Website for 9th, 10th ......11th, 12th, TRB TET & TNPSC Materials...

Namma Kalvi - The No.1 Educational Website for 9th, 10th,11th, 12th, TRB TET & TNPSC Materials

www.nammakalvi.weebly.com

(ii)

CHAIRPERSON

Dr. R. MurthyAssociate Professor

Department of MathematicsPresidency College (Autonomous)

Govt. of Tamil Nadu, Chennai – 600 005

Dr. G.P. YouvarajAssociate Professor

Ramanujan Institue for Advanced Study in MathematicsUniversity of Madras, Chennai-5

M. MadhivananHeadmaster

Govt. High SchoolBommahalli, KarimangalamDharmapuri District-635 111

Dr. S.A. SettuAssociate Professor

Department of MathematicsD.G. Vaishnav College

Arumbakkam, Chennai-106

V. MunianHeadmaster

Chennai High SchoolJafferkhanpet, Chennai- 600 083

S. BabuGraduate Teacher

Govt. Girls Hr. Sec. SchoolVillupuram - 605 602

A. SenthilkumarGraduate Teacher

Govt. Hr. Sec. SchoolThiruthuraiyur, Cuddalore Dt - 607 205

S.M. HarikrishnanGraduate Teacher

General Cariappa Hr. Sec. SchoolSaligramam, Chennai - 600 093

M. KuzhandhaiveluGraduate Teacher

Govt. Girls Hr. Sec. SchoolVillupuram - 605 602

M. SellamuthuGraduate TeacherGovt. High School

Goonipalayam, Thiruvallur - 602 026

S. Panneer SelvamP.G.Assistant

Govt. Hr. Sec. SchoolM.G.R. Nagar, Chennai- 600 078

REVIEWERS

AUTHORS

Assisted by

© Tamilnadu Parent Teacher AssociationFirst Edition - 2011

This book has been brought out with the Guidance ofThe Directorate of School Education

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Price. `

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Namma Kalvi - The No.1 Educational Websitefor 9th, 10th, 11th, 12th,

TRB TET & TNPSC Materials

(v)

We, the writing team of X-standard Mathematics book, have great pleasure in bringing out the SCORE BOOK consisting of Solutions to the exercise problems, Creative questions based on the content of the textbook, Objective type questions with solutions, Rating (classification) of Questions and Evaluation of Model Question Paper for the public examination.

One should never forget that solution to a problem in Mathematics can be obtained in more than one way. We have tried our level best in giving the proofs in an elegant and simple way.

It is the moral duty of a teacher to encourage a student to adopt a new but

correct approach while solving mathematical problems.

This book does not contain an exhaustive list of creative questions. They are only sample questions. Similar types of questions may also be framed and included in the public examination.

This book is only a tool to develop mathematical skill among the student community. It is only through one’s own interest that mastery over mathematics can be achieved. Hence, this book is an attempt to make the students learn Mathematics with zest, zeal and enthusiasm.

We thank the Directorate of School Education and Tamilnadu Parent Teacher Association for rendering all assistance in the preparation of SCORE BOOK. Suggestions to improve the contents of this book are welcome and highly appreciated.

R. MurthyChairperson - Writing team

Prefacewww.nammakalvi.weebly.com

(vi)

Contents

Solutions

1. Sets and Functions ...... 1

2. Sequences and Series of Real Numbers ...... 22

3. Algebra ...... 54

4. Matrices ...... 115

5. Coordinate Geometry ...... 133

6. Geometry ...... 161

7. Trigonometry ...... 179

8. Mensuration ...... 199

9. Practical Geometry ...... 221

10. Graphs ...... 241

11. Statistics ...... 258

12. Probability ...... 269

Classification of Questions ...... 283

Creative Questions ...... 350

Model Question Papers ...... 371

Departmental Model Question Paper - Evaluation ...... 393


Solution - Sets and Functions 1

1. SETS AND FUNCTIONS

Exercise 1.1 1. If ,A B1 then show that A B B, = (use Venn diagram).

Solution: A B B, =

2. If ,A B1 then find A B+ and \A B (use Venn diagram).Solution: A B A+ = \A B z=

3. Let { , , }, { , , , } { , , , }P a b c Q g h x y R a e f sand= = = . Find the following:

(i) \P R (ii) Q R+ (iii) \R P Q+^ h

Solution: (i) Given, P = { , , }, { , , , }a b c Q g h x y= and { , , , }R a e f s=

Now, \P R = { ; }t P t Rg! , the set of all elements of P which are not in R. = { , }b c .

(ii) Q R+ . Now, Q R+ is the set of elements which belong to both the sets Q and R. Thus, ,Q R+ Q= the empty set.

(iii) \R P Q+^ h. Now, P Q+ is empty. Thus, \( ) { , , , } .R P Q R a e f s+ = =

4. If { , , , , }, { , , } { , , , , , }A B C4 6 7 8 9 2 4 6 1 2 3 4 5 6and= = = , then find (i) A B C, +^ h, (ii) A B C+ ,^ h, (iii) \ \A C B^ h.

Solution: (i) Now, B C+ = {2, 4, 6} + {1, 2, 3, 4, 5, 6} = {2, 4, 6}. Thus, ( )A B C, + = {4, 6, 7, 8, 9} , {2, 4, 6} = {2, 4, 6, 7, 8, 9}.(ii) Now, B C, = {2, 4, 6} , {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}. Thus, ( )A B C+ , = {4, 6, 7, 8, 9} + {1, 2, 3, 4, 5, 6} = {4, 6}(iii) Now, \C B = {1, 2, 3, 4, 5, 6} \ {2, 4, 6} = {1, 3, 5} Hence, \ ( \ )A C B = {4, 6, 7, 8, 9} \ {1, 3, 5} = {4, 6, 7, 8, 9}.

5. Given { , , , , }, { , , , , }A a x y r s B 1 3 5 7 10= = - , verify the commutative property of set union.Solution: Given { , , , , }, {1,3,5,7, 10}.A a x y r s B= = -

Let us verify that A B B A, ,= .

Sets and Functions 1www.nammakalvi.weebly.com

10th Std. Mathematics - SCORE book2

Now, A B, = { , , , , } {1,3,5,7, 10}.a x y r s , -

= { , , , , , , , , , }a x y r s 1 3 5 7 10- g (1) B A, = {1,3,5,7, 10} { , , , , }a x y r s,-

= { , , , , , , , , , }a x y r s 1 3 5 7 10- g (2)From (1) and (2), we get A B B A, ,= . That is, the set union is commutative.

6. Verify the commutative property of set intersection for

{ , , , , , , , } { , , , , , , , }A l m n o B m n o p2 3 4 7 2 5 3 2and= = - .Solution: Let us verify that A B B A+ += .Now, A B+ = { , , , , , , , } { , , , , , , , }l m n o m n o p2 3 4 7 2 5 3 2+ -

= { , , , , }m n o2 3 g (1) B A+ = { , , , , , , , } { , , , , , , , }m n o p l m n o2 5 3 2 2 3 4 7+-

= { , , , , }m n o2 3 g (2)From (1) and (2), we get A B B A+ += . That is, the set intersection is commutative.

7. For { }A x x 42is a prime factor of;= , { , }B x x x5 12 N1; # != and

{ , , , }C 1 4 5 6= , verify A B C A B C, , , ,=^ ^h h .Solution: {2,3,7}, {6,7,8,9,10,11,12, {1,4,5,6}A B C= = =

Let us verify that ( ) ( )A B C A B C, , , ,= . Now, B C, = {6, 7, 8, 9, 10, 11, 12} , {1, 4, 5, 6} = {1, 4, 5, 6, 7, 8, 9, 10, 11, 12} Thus, ( )A B C, , = {2, 3, 7} , {1, 4, 5, 6, 7, 8, 9, 10, 11, 12} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} g (1) Now, A B, = {2, 3, 7} , {6, 7, 8, 9, 10, 11, 12} = {2, 3, 6, 7, 8, 9, 10, 11, 12} Thus, ( )A B C, , = {2, 3, 6, 7, 8, 9, 10, 11, 12} , {1, 4, 5, 6} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} g (2)From (1) and (2), we get ( ) ( )A B C A B C, , , ,= . (Note: The set union is associative)

8. Given { , , , , }, { , , , , } { , , , }P a b c d e Q a e i o u R a c e gand= = = . Verify the associative property of set intersection.Solution: Given that { , , , , }, { , , , , }P a b c d e Q a e i o u= = and { , , , }R a c e g=

Let us verify the associative property of set intersection ( ) ( )P Q R P Q R+ + + +=

Now, P Q+ = { , , , , } { , , , , } { , }a b c d e a e i o u a e+ =

Thus, ( )P Q R+ + = { , } { , , , } { , }a e a c e g a e+ = g (1) Now, Q R+ = { , , , , } { , , , } { , }a e i o u a c e g a e+ =

Thus, ( )P Q R+ + = { , , , , } { , } { , }a b c d e a e a e+ = g (2)From (1) and (2), we get ( ) ( )P Q R P Q R+ + + += .

Thus, the set intersection is associative.


9. For { , , , }; { , , , , } { , , , , , }A B C5 10 15 20 6 10 12 18 24 7 10 12 14 21 28and= = = ,

verify whether \ \ \ \A B C A B C=^ ^h h . Justify your answer.Solution:Given that { , , , }, { , , , , }A B5 10 15 20 6 10 12 18 24= = and { , , , , , }C 7 10 12 14 21 28=

Now, \B C = { , , , , }\{ , , , , , }6 10 12 18 24 7 10 12 14 21 28

= {6, 18, 24}Thus, \ ( \ )A B C = {5, 10, 15, 20} \ {6, 18,24} = {5, 10, 15, 20} g (1) \A B = { , , , }\{ , , , , } { , , }5 10 15 20 6 10 12 18 24 5 15 20=

Thus, ( \ ) \A B C = {5, 15, 20}\{7,10,12,14,21,28} = {5, 15, 20} g (2)From (1) and (2), we get \ ( \ ) ( \ ) \A B C A B C! .

10. Let { , , , }, { , , }, { , , }A B C5 3 2 1 2 1 0 6 4 2and= - - - - = - - = - - - . Find

\ \ ( \ ) \A B C A B Cand^ h . What can we conclude about set difference operation?Solution: Given that { 5, 3, 2, 1}, { 2, 1,0}A B= - - - - = - - and { , , }C 6 4 2= - - - .Now, \B C = { , , }\{ , } { , }2 1 0 6 4 2 1 0- - - - - = -

Thus, \( \ )A B C = { , , , }\{ , } { , , }5 3 2 1 1 0 5 3 2- - - - - = - - - g (1) \A B = { , , , }\{ , , } { , }5 3 2 1 2 1 0 5 3- - - - - - = - -

Thus, ( \ )\A B C = { , }\{ , , } { , }5 3 6 4 2 5 3- - - - - = - - g (2)From (1) and (2) we get, \( \ ) ( \ )\A B C A B C! . Thus, the set difference is not associative.

11. For { , , , , , , }, { , , , , , }A B3 1 0 4 6 8 10 1 2 3 4 5 6= - - = - - and

{ , , , , , },C 1 2 3 4 5 7= - show that (i) A B C, +^ h= A B A C, + ,^ ^h h (ii) A B C+ ,^ h

= A B A C+ , +^ ^h h (iii) Verify A B C, +^ h= A B A C, + ,^ ^h h using Venn diagram (iv) Verify A B C+ ,^ h= A B A C+ , +^ ^h h using Venn diagram.Solution: Given that { , , , , , , }A 3 1 0 4 6 8 10= - - , { , , , , , }B 1 2 3 4 5 6= - - and { , , , , , }C 1 2 3 4 5 7= - .(i) B C+ = { 1, 2,3,4,5,6} { 1,2,3,4,5,7}+- - -

= { 1,3,4,5}- g (1) So, ( )A B C, + = { , , , , , , } { , , , }3 1 0 4 6 8 10 1 3 4 5+- - -

= { , , , , , , , , }3 1 0 3 4 5 6 8 10- -

A B, = { , , , , , , } { , , , , , }3 1 0 4 6 8 10 1 2 3 4 5 6,- - - -

= { , , , , , , , , , }3 2 1 0 3 4 5 6 8 10- - -

A C, = { , , , , , , } { , , , , , }3 1 0 4 6 8 10 1 2 3 4 5 7,- - -

= { , , , , , , , , , , }3 1 0 2 3 4 5 6 7 8 10- -

( ) ( )A B A C, + , ={ 3, 2, 1,0,3,4,5,6,8, } { 3, 1,0,2,3,4,5,6,7,8,10}10 +- - - - -

= { , , , , , , , , }3 1 0 3 4 5 6 8 10- - g (2)From (1) and (2), we get ( ) ( ) ( )A B C A B A C, + , + ,= .


(ii) B C, = { , , , , , } { , , , , , }1 2 3 4 5 6 1 2 3 4 5 7,- - -

= { 2, 1,2,3,4,5,6,7}- -

Thus, ( )A B C+ , = { , , , , , , } { , , , , , , , }3 1 0 4 6 8 10 2 1 2 3 4 5 6 7+- - - -

= { , , }1 4 6- g (1) A B+ = { , , , , , , } { , , , , , }3 1 0 4 6 8 10 1 2 3 4 5 6+- - - -

= { , , }1 4 6-

A C+ = { 3, 1,0,4,6,8,10} { 1,2,3,4,5, }7+- - -

= { , }1 4-

Thus, ( ) ( )A B A C+ , + = { , , } { , } { , , }1 4 6 1 4 1 4 6,- - = - g (2)From (1) and (2) we get, ( ) ( ) ( )A B C A B A C+ , + , += .(iii) Venn diagram of ( ) ( ) ( )A B C A B A C, + , + ,=

From (2) and (5), we get ( ) ( ) ( )A B C A B A C, + , + ,= .(iv) Venn diagram of ( ) ( ) ( )A B C A B A C+ , + , +=

From (2) and (5), we get ( ) ( ) ( )A B C A B A C+ , + , +=


Exercise 1.2

1. Represent the following using Venn diagrams (i) ,{ , , , , }, { , , , } { , , , , }U A B5 6 7 8 13 5 8 10 11 5 6 7 9 10andg= = =

(ii) { , , , , , , , }, { , , , } { , , , , }U a b c d e f g h M b d f g N a b d e gand= = =

Solution: (i) (ii)

2. Write a description of each shaded area. Use symbols U, , , , ,A B C , + , l and \ as

necessary. (Many answers are possible. One such description is given below)

3. Draw Venn diagram of three sets ,A B Cand illustrating the following: (i) A B C+ + (ii) A Band are disjoint but both are subsets of C (iii) \A B C+^ h

(iv) \B C A,^ h (v) A B C, +^ h (vi) \C B A+^ h (vii) C B A+ ,^ h.


Solution:

4. Use Venn diagram to verify \A B A B A+ , =^ ^h h .Solution:

From (1) and (2), we get ( ) ( \ )A B A B A+ , = .

5. Let {4,8,12,16,20,24,28}U = , {8,16,24}A = and {4,16,20,28}B = .

Find A B A Band, +l l^ ^h h .Solution:Given that { , , , , , , }, { , , }U A4 8 12 16 20 24 28 8 16 24= = and { , , , }B 4 16 20 28= . Now, A B, = { , , } { , , , } { , , , , , }8 16 24 4 16 20 28 4 8 16 20 24 28, =

Thus, ( )A B, l = \ ( ) {4,8,12,16,20,24,28} \ {4,8,16,20,24,28} {12}U A B, = =

Now, A B+ = {8,16,24} {4,16,20,28} { }16+ =

Thus, ( )A B+ l = \( ) {4,8,12,16,20,24,28} \ {16} {4,8,12,20,24,28}U A B+ = =

6. Given that U = { , , , , , , , }U a b c d e f g h= , { , , , } { , , },A a b f g B a b cand= =

verify De Morgan’s laws of complementation.Solution: De Morgan’s laws of complementation are ( )A B A B(i) , +=l l l ( )A B A B(ii) + ,=l l l

{ , , , , , , , }, { , , , ,}U a b c d e f g h A a b f g= = and { , , }B a b c= .


(i) A B, = { , , , } { , , } { , , , , }a b f g a b c a b c f g, =

( )A B, l = \ { , , , , , , , }\{ , , , , } { , , }U A B a b c d e f g h a b c f g d e h, = = g (1) Al = \ { , , , , , , , }\{ , , , } { , , , }U A a b c d e f g h a b f g c d e h= =

Bl = \ { , , , , , , , }\{ , , } { , , , , }U B a b c d e f g h a b c d e f g h= =

A B+l l = { , , , } { , , , , } { , , }c d e h d e f g h d e h+ = g (2)From (1) and (2), we get ( )A B A B, +=l l l.(ii) A B+ = { , , , } { , , } { , }a b f g a b c a b+ =

( )A B+ l = \ { , , , , , , , } \ { , } { , , , , , }U A B a b c d e f g h a b c d e f g h+ = = g (3) Al = \ { , , , }; \ { , , , , }U A c d e h B U B d e f g h= = =l

A B,l l = { , , , } { , , , , } { , , , , , }c d e h d e f g h c d e f g h, = g (4)From (3) and (4), we get ( )A B A B+ ,=l l l.

7. Verify De Morgan’s laws for set difference using the sets given below:

{1, 3, 5, 7, 9, 11, 13, 15}, {1, 2, 5, 7} {3,9,10, 12, 13}A B Cand= = = .Solution: De Morgan’s laws for set difference are \( ) ( \ ) ( \ )( ) A B C A B A Ci , += , \( ) ( \ ) ( \ )( ) A B C A B A Cii + ,=

Now, A = , , , , , , , }, { , , , }B1 3 5 7 9 11 13 15 1 2 5 7= and C = { , , , , }3 9 10 12 13

(i) B C, = { , , , } { , , , , } { , , , , , , , , }1 2 5 7 3 9 10 12 13 1 2 3 5 7 9 10 12 13, =

So, \( )A B C, = {1,3,5,7,9,11,13,15}\{1,2,3,5,7,9,10,12,13} { , }11 15= g (1) \A B = {1,3,5,6,7,9,11,13,15}\{1,2,5,7} {3, ,9,11,13,15}6=

\A C = { , , , , , , , }\{ , , , , } { , , , , }1 3 5 7 9 11 13 15 3 9 10 12 13 1 5 7 11 15=

( \ ) ( \ )A B A C+ = { , , , , } { , , , , } { , }3 9 11 13 15 1 5 7 11 15 11 15+ = g (2)From (1) and (2), we get \( ) ( \ ) ( \ )A B C A B A C, += .(ii) B C+ = {1,2,5,7} {3,9,10,12,13}+ Q=

\( )A B C+ = {1,3,5,7,9,11,13,15}\ {1,3,5,7,9,11,13,15}Q = g (3) ( \ ) ( \ )A B A C, = { , , , , } { , , , , } { , , , , , , , }3 9 11 13 15 1 5 7 11 15 1 3 5 7 9 11 13 15, = g (4)

From (3) and (4), we get \( ) ( \ ) ( \ )A B C A B A C+ ,= .

8. Let { , , , , , , , , }U 10 15 20 25 30 35 40 45 50= , {1, 5, 10, 15, 20, 30}B = and

{7, 8, 15,20,35,45, 48}C = . Verify \ \ \A B C A B A C+ ,=^ ^ ^h h h.Solution: Given, { , , , , , , , , }, { , , , , , }A B10 15 20 25 30 35 40 45 50 1 5 10 15 20 30= = and { , , , , , , }C 7 8 15 20 35 45 48= . B C+ = { , , , , , } { , , , , , , } { , }1 5 10 15 20 30 7 8 15 20 35 45 48 15 20+ =

\( )A B C+ = { , , , , , , , , }\{ , }10 15 20 25 30 35 40 45 50 15 20

= { , , , , , , }10 25 30 35 40 45 50 g (1) \A B = { , , , , , , , , }\{ , , , , , }10 15 20 25 30 35 40 45 50 1 5 10 15 20 30

= { , , , , }25 35 40 45 50

\A C = { , , , , , , , , }\{ , , , , , , }10 15 20 25 30 35 40 45 50 7 8 15 20 35 45 48

= { , , , , }10 25 30 40 50


( \ ) ( \ )A B A C, = {25,35,40,45,50} {10,25,30,40,50},

= { , , , , , , }10 25 30 35 40 45 50 g (2)From (1) and (2), we get, \ ( ) ( \ ) ( \ )A B C A B A C+ ,= .

9. Using Venn diagram, verify whether the following are true: (i) A B C A B A C, + , + ,=^ ^ ^h h h

(ii) A B C A B A C+ , + , +=^ ^ ^h h h

(iii) A B A B, += l ll^ h

(iv) \ \ \A B C A B A C, +=^ ^ ^h h h

Solution: (i)

From (2) and (5), we get ( ) ( ) ( )A B C A B A C, + , + ,= .

(ii)

From (2) and (5), we get ( ) ( ) ( )A B C A B A C+ , + , += .


(iii)

From (2) and (5), we get ( )A B A B, +=l l l.(iv)

From (2) and (5), we get \( ) ( \ ) ( \ )A B C A B A C, += .

Exercise 1.3 1. If Aand B are two sets and U is the universal set such that 700n U =^ h , ,n A n B n A B n A B200 300 100and , find+ += = = l l^ ^ ^ ^h h h h.

Solution: ( )n A B, = ( ) ( ) ( )n A n B n A B 200 300 100++ - = + -

= 500 100 400- =

( )n A B+l l = ( ) ( ) ( )n A B n U n A B 700 400 300, ,= - = - =l

Aliter: ( )n A B+l l = ( ) ( ) ( ) ( ) ( ) ( )n A n B n A B n A n B n A B, ++ - = + -l l l l l l l

= 500 400 600 003+ - =

2. Given 285, 195, 500, 410,n A n B n U n A B n A Bfind, ,= = = = l l^ ^ ^ ^ ^h h h h h.Solution: ( )n A B+ = ( ) ( ) ( ) 2n A n B n A B 85 195 410 70,+ - = + - =

( )n A B,l l = [( ) ] ( ) ( )n A B n U n A B 500 70 430+ += - = - =l


3. For any three sets A, B and C if 17n A =^ h , 17, 17, 7n B n C n A B+= = =^ ^ ^h h h , ( ) 6 , 5 2n B C n A C n A B Cand+ + + += = =^ ^h h , find n A B C, ,^ h.

Solution: ( )n A B C, , = ( ) ( ) ( ) ( )n A n B n C n A B++ + - -

( ) ( ) ( )n B C n A C n A B C+ + + +- + . = 17 17 17 7 6 5 2 53 18+ + - - - + = - = 35.

4. Verify n A B C n A n B n C n A B, , += + + - -^ ^ ^ ^ ^h h h h h n B C n A C n A B C+ + + +- +^ ^ ^h h h for the sets given below:

(i) {4,5,6}, {5,6,7,8} {6,7,8,9}A B Cand= = =

(ii) { , , , , }, { , , } { , , }A a b c d e B x y z C a e xand= = = .Solution: (i) {4,5,6}, ( ) 3, {5,6,7,8}, ( ) 4A n A B n B= = = = and { , , , }, ( )C n C6 7 8 9 4= =

Now, A B C, , = { , , , , , }, ( )n A B C4 5 6 7 8 9 6, , = g (1) A B+ = { , , } { , , , } { , }, ( )n A B4 5 6 5 6 7 8 5 6 2+ += =

B C+ = { , , , } { , , , } { , , }, ( )n B C5 6 7 8 6 7 8 9 6 7 8 3+ += =

A C+ = { , , } { , , , } { }, ( )n A C4 5 6 6 7 8 9 6 1+ += =

Thus, A B C+ + = {4,5,6} {5,6,7,8} {6,7,8,9} {6}, ( ) 1n A B C+ + + += =

Now, ( ) ( ) ( ) ( ) ( ) ( ) ( )n A n B n C n A B n B C n A C n A B C+ + + + ++ + - - - +

= 3 4 4 2 3 1 1 6+ + - - - + = g (2)From (1) and (2), we get,( )n A B C, , = ( ) ( ) ( ) ( ) ( ) ( ) ( )n A n B n C n A B n B C n A C n A B C+ + + + ++ + - - - +

(ii) { , , , , }, { , , }A a b c d e B x y z= = and { , , }C a e x= ; ( ) , ( ) , ( )n A n B n C5 3 3= = =

A B+ = { , , , , } { , , } { }, ( ) 0a b c d e x y z n A B+ += =

B C+ = { , , } { , , } { }, ( ) 1x y z a e x x n B C+ += =

A C+ = { , , , , } { , , } { , }, ( ) 2a b c d e a e x a e n A C+ += =

A B C, , = { , , , , } { , , } { , , } { , , , , , , , }a b c d e x y z a e x a b c d e x y z, , =

Thus, ( )n A B C, , = 8 g (1)( ) ( ) ( ) ( ) ( ) ( ) ( )n A n B n C n A B n B C n A C n A B C+ + + + ++ + - - - +

= 5 3 3 0 1 2 0 8+ + - - - + = g (2)From (1) and (2), we get ( )n A B C, ,

( ) ( ) ( ) ( ) ( ) ( ) ( )n A n B n C n A B n B C n A C n A B C+ + + + += + + - - - +

5. In a college, 60 students enrolled in chemistry, 40 in physics, 30 in biology, 15 in chemistry and physics, 10 in physics and biology, 5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects. Solution: Let C, P and B be the set of students enrolled in Chemistry, Physics and Biology respectively. Thus, ( ) 60, ( ) 40, ( ) 30n C n P n B= = = ,

( ) , ( ) , ( ) , ( )n C P n P B n C B n C P B15 10 5 0+ + + + += = = =( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )n C P B n C n P n B n C P n P B n B C n C P B, , + + + + += + + - - - +

= 60 40 30 15 10 5 0 100+ + - - - + =

Hence, Number of students enrolled in atleast one of the subjects = 100.


6. In a town 85% of the people speak Tamil, 40% speak English and 20% speak Hindi. Also, 32% speak English and Tamil, 13% speak Tamil and Hindi and 10% speak English and Hindi, find the percentage of people who can speak all the three languages.Solution: Let E denote the set of people speaking English, T denote the people speaking Tamil and H denote the people speaking Hindi. Thus, ( ) 85, ( ) 40, ( ) 20n T n E n H= = =

Also, ( ) , ( ) , ( ) , ( )n E T n T H n E H n E T H32 13 10 100+ + + , ,= = = = .( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )n E T H n E n T n H n E T n T H n H E n E T H, , + + + + += + + - - - +

100 = ( )n E T H40 85 20 32 13 10 90 + ++ + - - - = +

Thus, ( )n E T H+ + = 100 90 10- =

Therefore, the percentage of people who can speak all the three languages = 10%.

7. An advertising agency finds that, of its 170 clients, 115 use Television, 110 use Radio and 130 use Magazines. Also, 85 use Television and Magazines, 75 use Television and Radio, 95 use Radio and Magazines, 70 use all the three. Draw Venn diagram to represent these data. Find (i) how many use only Radio?

(ii) how many use only Television? (iii) how many use Television and magazine but not radio?

Solution: Let, T - Clients of Television

R - Clients of Radio

M - Clients of Magazines

Then, ( )n U = 170, ( ) 115n T =

( )n R = 110, ( ) 130n M =

( )n T M+ = 85, ( ) 75n T R+ =

( )n R M+ = 95, ( ) 70n T R M+ + =

From the Venn diagram (i) Number of clients using only radio = 10

(ii) Number of clients using only television = 25

(iii) Number of clients using only magazines but not radio = 15

8. In a school of 4000 students, 2000 know French, 3000 know Tamil and 500 know Hindi, 1500 know French and Tamil, 300 know French and Hindi, 200 know Tamil and Hindi and 50 know all the three languages.

(i) How many do not know any of the three languages? (ii) How many know at least one language? (iii) How many know only two languages?


Solution:

Let F be the set of students who know French.

T - Students who know Tamil

H - Students who know Hindi

Now, ( )n U = , ( )n F4000 2000=

( )n T = , ( )n H3000 500=

( )n F T+ = 5 , ( )n F H1 00 300+ =

( )n T H+ = , ( )n F T H200 50+ + =

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )n F T H n F n T n H n F T n T H n F H n F T H, , + + + + += + + - - - +

= 2000 3000 500 1500 300 200 50+ + - - - + =3550.Thus, ( )n F T H, , l = ( ) ( )n U n F T H 4000 3550 450, ,- = - = .

(i) Number of students do not know any of the three languages = 450

(ii) Number of students who know atleast one language = 3550

(iii) Number of students who know only two languages = 1850

9. In a village of 120 families, 93 families use firewood for cooking, 63 families use kerosene, 45 families use cooking gas, 45 families use firewood and kerosene, 24 families use kerosene and cooking gas, 27 families use cooking gas and firewood. Find how many use firewood, kerosene and cooking gas.Solution: Let, F denote the Families using Firewood.

K - Families using Kerosene.

G - Families using Gas.

( )n F` = 93, ( ) 63, ( )n K n G 45= =

( )n F K+ = 5, ( )n K G4 24+ =

( )n T H+ = 27, ( ) 120n F K G, , =

( )n F K G, ,

( ) ( ) ( ) ( ) ( ) ( ) ( )n F n K n G n F K n K G n F G n F K G+ + + + += + + - - - +

93 63 45 45 24 27 ( )n F K G& + ++ + - - - + = 120.

( )n F K G+ + = 120 105 15- = .

Thus, the number of families using all the three is 15.


Exercise 1.4

1. State whether each of the following arrow diagrams define a function or not. Justify your answer.Solution: (i) (ii)

The element c in P has no image inQ.Thus , it is not a function.

Every element in the domain L has a unique image in M. Therefore, it is a function.

2. For the given function F= { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }, write the domain and the range.Solution: Domain = { , , , , }1 2 3 4 5 , Range { , , , , }1 3 5 7 9= .

3. Let A = { 10, 11, 12, 13, 14 }; B = { 0, 1, 2, 3, 5 } and :f A Bi " , i = 1,2,3. State the type of function for the following (give reason): (i) f

1 = {(10, 1), (11, 2),

(12, 3), (13, 5), (14, 3)}, (ii) f2 = {(10, 1), (11, 1), (12, 1), (13, 1), (14, 1)},

(iii) f3 = {(10, 0), (11, 1), (12, 2), (13, 3), (14, 5)}.

Solution:{(10,1), (11,2),(12,3), (13,5), (14,3)}f(i)

1=

The elements 12 and 14 in A have same image 3 in B.Hence, it is not one-one. The element B0 ! has no pre-image in A. Thus, it is not onto. Hence, it is neither one-one nor onto.

{(10,1), (11,1), (12,1), (13,1), (14,1)}f(ii)2=

Since ( ) 1,f x2

= for every x A! , f2 is a constant function.

{(10,0),(11,1), (12,2),(13,3), (14,5)}f(iii)3=

Distinct elements in A have distinct images in B under f3.

Thus, f3 is one-one.

Also, ( )f A B3

= .Thus, f

3 is onto.

Hence, f3 is one-one and onto. ( f

3 is a bijective function)


4. If X = { 1, 2, 3, 4, 5 }, Y = { 1, 3, 5, 7, 9 } determine which of the following relations from X to Y are functions? Give reason for your answer. If it is a function, state its type. (i) R

1 = { ,x y^ h| 2y x= + , x X! , y Y! }, (ii) R

2 = { (1, 1), (2, 1), (3, 3),

(4, 3), (5, 5) }, (iii) R3 = { (1, 1), (1, 3), (3, 5), (3, 7), (5, 7) }, (iv) R

4 = { (1, 3), (2, 5),

(4, 7), (5, 9), (3, 1)}.Solution: {( , ) | 2, , }R x y y x x X y Y(i)

1! != = +

Given that y = x 2+

When 2x = , y = 4 Yg . Thus, it is not a function.Also, note that element 4 in X has no image in Y.

{(1,1), (2,1), (3,3), (4,3), (5,5)}R(ii)2=

Every element in X has unique image. Thus, R2

is a function.Since the elements 1 and 2 have same image 1, it is not one-one.Since the element 7 has no pre-image, it is not onto.Thus, it is neither one-one nor onto.

{(1,1), (1,3), (3,5), (3,7), (5,7)}R(iii)3=

The element X1 ! has two images 1, 3 in Y.Thus, it is not a function.

{(1,3), (2,5), (4,7), (5,9), (3,1)}R(iv)4=

Every element has unique image. Thus, R4

is a function.Distinct elements in X have distinct images in Y .Thus, it is one-one. Each element in Y has atleast one pre-image in X.Thus, it is onto. Hence, it is a bijective function.

5. If {( , 2), ( 5, ), (8, ), ( , 1)}R a b c d= - - - represents the identity function, find the values of , ,a b c and d .Solution: Given R is an identity function. Thus, ( ) ,R x x x6= .Hence, 2, 5, 8a b c=- =- = and d 1=- .

6. A = { –2, –1, 1, 2 } and , :f xx

x A1 != ` j$ .. Write down the range of f .

Is f a function from A to A ?

Solution: Given that f = , 1xx

` j. So, ( ) .f xx1=

Thus, ( 2)f - = ; (1) 1.f21

21

11

-=- = =

( )f 1- = 1; (2) .f11

21

-=- =

Thus, the range of f = , , ,21 1 1

21- -$ ..

However, , A21

21 g- . Hence, it is not a function from .A Ato .


7. Let f = { (2, 7), (3, 4), (7, 9), (–1, 6), (0, 2), (5, 3) } be a function from A = { –1, 0, 2, 3, 5, 7 } to B = { 2, 3, 4, 6, 7, 9 }. Is this (i) an one-one function

(ii) an onto function (iii) both one-one and onto function?Solution: (i) Distinct elements in A have distinct images in B. Thus, it is an one-one function. (ii) Every element in B has atleast one pre-image in A Thus, f is onto function. (iii) From (i) and (ii) we have, it is a bijective.

8. Write the pre-images of 2 and 3 in the function f = { (12, 2), (13, 3), (15, 3), (14, 2), (17, 17) }.

Solution: Pre-images of 2 are 12 and 14. Pre-images of 3 are 13 and 15.

9. The following table represents a function from {5,6,8,10}A = to { , , , }B 19 15 9 11= where f x^ h = 2 1x - . Find the values of a and b .

x 5 6 8 10f(x) a 11 b 19

Solution: Given, ( )f x = ,x x A2 1 6 !-

( )f 5 = 2 5 1 10 1 9# - = - =

( )f 8 = 2 8 1 16 1 15# - = - =

Thus, the values of a and b are 9 and 15 respectively.Note: A function of the form ( ) 1, ,f x px x R!= + is called linear function. Such function are always one to one.

10. Let A= { 5, 6, 7, 8 }; B = { –11, 4, 7, –10,–7, –9,–13 } and f = {( ,x y ) : y = 3 2x- , x A! , y B! } (i) Write down the elements of f (ii) What is the co-domain? (iii) What is the range? (iv) Identify the type of function.Solution: Given, {5,6,7,8}, { 11,4,7, 10, 7, 9, 13}A B= = - - - - -

Here, y ( )f x= = 3 2 ,x x A6 !-

Now, ( )f 5 = 7, (6) 9f- =-

Now, ( )f 7 = 11, (8) 13f- =-

(i) {( , ), ( , ), ( , ), ( , )}f 5 7 6 9 7 11 8 13= - - - -

(ii) Co-domain = { , , , , , , }11 4 7 10 7 9 13- - - - -

(iii) Range = { , , , }7 9 11 13- - - -

(iv) Distinct elements have distinct images under f . Thus, f is one-one.

(Here, the function is not onto.)


11. State whether the following graphs represent a function. Give reason for your answer.

(i) Given graph satisfies the vertical line test. Thus, it is a function.

(ii) Given graph satisfies the vertical line test. Thus, it is a function.

(iii) In this graph, the vertical line l cuts the graph at two points A and B .

Thus, it is not a function.

(iv) The vertical line l cuts the graph at three points ,A B and C.

Thus, it is not a function.

(v) Given graph satisfies the vertical line test. Thus, it is a function.

12. Represent the function f = { (–1, 2), (– 3, 1), (–5, 6), (– 4, 3) } as (i) a table (ii) an arrow diagram.Solution: {( , ), ( , ), ( , ), ( , )}f 1 2 3 1 5 6 4 3= - - - -

(i) Table

x 1- 3- 5- 4-

( )f x 2 1 6 3 (ii) Arrow diagram

13. Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and :f A B" be defined by

f x^ h = x33- . Represent f by (i) an arrow diagram, (ii) a set of ordered pairs,


(iii) a table (iv) a graph .Solution: Given { , , , , }, { , , , , }A B6 9 15 18 21 1 2 4 5 6= = .Let us find the images under f .

Now, ( )f x = ,x x A33 !-

Thus, ( )f 6 = 1, (9) 2f3

6 33

9 3- = = - =

( )f 15 = 4, (1 ) 5, (21) 6f f3

15 3 83

18 33

21 3- = = - = = - =

(i) An arrow diagram

(ii) Set of ordered pairs {(6,1), (9,2), (15,4), (18,5), (21,6)}f =

(iii) Table

x 6 9 15 18 21( )f x 1 2 4 5 6

(iv) Graph

Here, the graph is the collection of all points (6,1), (9,2), (15,4),(18,5), (21,6) in x -y plane.

14. Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If :f A B" is defined by 1f x x21= +^ h

then represent f by (i) an arrow diagram (ii) a set of ordered pairs and (iii) a table.

Solution: ( )f x = ,x x A2

1 !+

Thus, ( )f 4 = 1 3, (6) 1 4f24

26+ = = + =

( )f 8 = 1 5, (10) 1 6f28

210+ = = + = .

(i) An arrow diagram:


(ii) Set of ordered pairs: {( , ), ( , ), ( , ), ( , )}f 4 3 6 4 8 5 10 6= . (iii) Table:

x 4 6 8 10( )f x 3 4 5 6

15. A function f : ,3 7- h6 " R is defined as follows f x^ h = ;

;

;

x x

x x

x x

4 1 3 2

3 2 2 4

2 3 4 7

2 1

1 1

#

# #

- -

-

-

*

Find (i) f f5 6+^ ^h h (ii) f f1 3- -^ ^h h (iii) f f2 4- -^ ^h h (iv) ( ) ( )

( ) ( )

f ff f2 6 1

3 1

-

+ - .Solution: Let us find the values of the function at the required points.When 3, 2, 1 1x and=- - - , the function is ( ) 4 1f x x2= - . Thus, ( 3) 35, ( 2) 15, ( 1) 3, (1) 3.f f f f- = - = - = =

When, x = ,3 4 , the function is ( )f x = 3 2,x -

Thus, ( )f 3 = 7 (4) 10fand = .When 5 6x and= , the function ( )f x = 2 3x - . Thus, ( )f 5 = 7, (6) 9f =

(i) ( ) ( )f f5 6+ = 7 9 16+ =

(ii) ( ) ( )f f1 3- - = 3 35 32- =-

(iii) ( ) ( )f f2 4- - = 15 10 5- =

(iv) ( ) ( )

( ) ( )f f

f f2 6 13 1

-+ - =

( )2 9 37 3

1510

32

-+ = = .

16. A function f : ,7 6- h6 " R is defined as follows ( )f x = ;

;

; .

x x x

x x

x x

2 1 7 5

5 5 2

1 2 6

2 1

1 1

#

# #

+ + - -

+ -

-

*

Find (i) 2 ( 4) 3 (2)f f- + (ii) ( 7) ( 3)f f- - - (iii) ( ) ( )

( ) ( )

f ff f

6 3 1

4 3 2 4

- -

- + .Solution: When ,x 7 6=- - , ( ) 2 1f x x x2

= + +

Thus, ( )f 7- = ( ) ( )7 2 7 1 49 14 1 362- + - + = - + = and

( )f 6- = ( ) ( )6 2 6 1 36 12 1 252- + - + = - + = .

When 4, 3, 2 1 ( ) 5x f x xand the function is=- - = +

Thus, ( )f 4- = 4 5 1,- + = ( )f 3- 2 , (1) 6f= = and ( )f 2 = 7. When 4, ( ) 1.x f x xthe function is= = - So, (4) 3.f =

(i) Now, 2 ( 4) 3 (2)f f- + = 2 1 3 7 23# #+ = .(ii) Now, ( ) ( ) .f f7 3 36 2 34- - - = - =

(iii) Now, ( ) ( )( ) ( )

f ff f

6 3 14 3 2 4

- -- + =

25 3 64 2 2 3

25 188 6

714 2

## #

-+ =

-+ = =


Exercise 1.5

1. For two sets A and B , A B, = A if and only if

(A) B A3 (B) A B3 (C) A B! (D) A B+ z=

Solution:

( Ans. (A) ) 2. If A B1 , then A B+ is

(A) B (B) \A B (C) A (D) \B A

Solution:

( Ans. (C) ) 3. For any two sets Pand Q , P Q+ is

(A) :x x P x Qor! !" , (B) :x x P x Qand b!" ,

(C) :x x P x Qand! !" , (D) :x x P x Qandb !" ,

Solution: By definition, { : }andP Q x x P x Q+ ! != ( Ans. (C) )

4. If A= { p, q, r, s }, B = { r, s, t, u }, then \A B is

(A) { , }p q (B) { , }t u (C) { , }r s (D) { , , , }p q r s

Solution: \A B is the set of elements in A but not in B . ( Ans. (A) )

5. If ( )n P A6 @ = 64, then n A^ h is

(A) 6 (B) 8 (C) 4 (D) 5

Solution: [ ( )] ( )n P A n A2 64 2 6( )n A 6 `= = = = . ( Ans. (A) )

6. For any three sets A, B and C, A B C+ ,^ h is

(A) A B B C, , +^ ^h h (B) A B A C+ , +^ ^h h

(C) ( )A B C, + (D) A B B C, + ,^ ^h h

Solution: ( ) ( ) ( )A B C A B A C+ , + , += . ( Ans. (B) ) 7. For any two sets A Band , {( \ ) ( \ )} ( )A B B A A B, + + is

(A) z (B) A B, (C) A B+ (D) A B+l l

Solution:

( Ans. (A) )


8. Which one of the following is not true ?

(A) \A B = A B+ l (B) \A B A B+=

(C) \ ( )A B A B B, += l (D) \ ( ) \A B A B B,=

Solution: We know that \A B A B+= l ; \A B A B+! ( Ans. (B) )

9. For any three sets ,A B and C , \B A C,^ h is

(A) \ \A B A C+^ ^h h (B) \ \B A B C+^ ^h h

(C) \ \B A A C+^ ^h h (D) \ \A B B C+^ ^h h

Solution: De Morgan’s Law : \( ) ( \ ) ( \ )B A C B A B C, += ( Ans. (B) )

10. If n(A) = 20 , n(B) = 30 and ( )n A B, = 40, then ( )n A B+ is equal to

(A) 50 (B) 10 (C) 40 (D) 70.Solution: ( ) ( ) ( ) ( )n A B n A n B n A B 20 30 40 10+ ,= + - = + - = ( Ans. (B) )

11. If { ( x , 2), (4, y ) } represents an identity function, then ( , )x y is

(A) (2, 4) (B) (4, 2) (C) (2, 2) (D) (4, 4)Solution: In an identity function, each element is associated with itself. ( Ans. (A) )

12. If { (7, 11), (5, a ) } represents a constant function, then the value of ‘a ’ is

(A) 7 (B) 11 (C) 5 (D) 9Solution: In a constant function all the images are same ( Ans. (B) )

13. Given ( )f x = 1 x-^ h is a function from N to Z . Then the range of f is

(A) { 1} (B) N (C) { 1, – 1 } (D) Z

Solution: For , ( )x f xN! = ( 1) .1x !- = ( Ans. (C) )

14. If f = { (6, 3), (8, 9), (5, 3), (–1, 6) }, then the pre-images of 3 are

(A) 5 and –1 (B) 6 and 8 (C) 8 and –1 (D) 6 and 5.Solution: The pre-images of 3 is 5 and 6. ( Ans. (D) )

15. Let A = { 1, 3, 4, 7, 11 }, B = {–1, 1, 2, 5, 7, 9 } and :f A B" be given by

f = { (1, –1), (3, 2), (4, 1), (7, 5), (11, 9) }. Then f is

(A) one-one (B) onto

(C) bijective (D) not a function ( Ans. (A) )

Solution:


16. The given diagram represents

(A) an onto function (B) a constant function

(C) an one-one function (D) not a function

Solution: 2 has two images 4 and 2. It is not a function. ( Ans. (D) )

17. If A = { 5, 6, 7 }, B = { 1, 2, 3, 4, 5 }and :f A B" is defined by ( )f x x 2= - , then the range of f is

(A) { 1, 4, 5 } (B) { 1, 2, 3, 4, 5 }

(C) { 2, 3, 4 } (D) { 3, 4, 5 }

Solution: (5)f = , ( ) , ( )f f3 6 4 7 5= = . ( Ans. (D) )

18. If ( )f x x 52= + , then ( )f 4- =

(A) 26 (B) 21

(C) 20 (D) – 20

Solution: ( )f x = 5x2 +

( 4)f& - = ( 4) 5 21.2- + = ( Ans. (B) )

19. If the range of a function is a singleton set, then it is

(A) a constant function (B) an identity function (C) a bijective function (D) an one-one function

Solution: Constant function. ( Ans. (A) )

20. If :f A B" is a bijective function and if n(A) = 5 , then n(B) is equal to

(A) 10 (B) 4

(C) 5 (D) 25

Solution: If A and B are finite sets and f is bijective, then ( ) ( )n A n B= ( Ans. (C) )


Exercise 2.1 1. Writethefirstthreetermsofthefollowingsequenceswhosenth termsaregiven

by (i) a n n

3

2n=

-^ h (ii) 3c 1n

n n 2= -

+^ h (iii) z n n

4

1 2n

n

=- +^ ^h h .

Solution: (i) Given that an n

3

2n=

-^ h for n= 1,2,3,g .

When n = 1, 2 and 3, we get

( ) ( )a

31 1 2

31 1

31

1=

-=

-= -

( ) ( )0a

32 2 2

32 0

2=

-= =

( ) ( )1a

33 3 2

33 1

3=

-= =

Hence, the first three terms of the sequence are , 0 131 and- .

(ii) Given that ( 1) 3cn

n n 2= -

+ . When n = 1, 2 and 3, we get

( 1) (3) (3) 27c1

1 1 2 3= - =- =-

+

( 1) (3) (3) 81c2

2 2 2 4= - = =

+

( 1) (3) (3) 243c3

3 3 2 5= - =- =-

+

Hence, the first three terms of the sequence are –27, 81, –243.

(iii) Given that ( ) ( )z

n n4

1 2n

n

=- + . If we put n = 1, 2 and 3, we get

( ) ( )( ) ( )( )z

41 1 1 2

41 3

43

1

1

=- +

=-

= -

( ) ( )( ) ( )( )2z

41 2 2 2

42 4

2

2

=- +

= =

( ) ( )( ) ( )( )z

41 3 3 2

4

1 3 5

415

3

3

=- +

=-

= -^ h .

Hence, the first three terms of the sequence are , 243

415and- - .

2. Findtheindicatedtermsineachofthesequenceswhosenth termsaregivenby

(i) ; ,ann a a2 3

2n 7 9=

++ (ii) 2 ; ,a n a a1 1

n

n n 3

5 8= - +

+^ ^h h

(iii) 2 3 1; ,a n n a a.n

2

5 7= - + (iv) ( 1) (1 ); ,a n n a an

n 2

5 8= - - +

Solution: (i) Given that 1, 2, 3, .ann n2 3

2 forn

g=++ =

When 7,( )

n a2 7 37 2

179

7= =

++ =

When 9,( )

n a2 9 39 2

2111

9= =

++ = .

Sequences and Series of Real Numbers2

Solution - Sequences and Series of Real Numbers 23

(ii) Given that, ( 1) 2 ( 1) 1, 2, 3,a n nforn

n n 3g= - + =

+

When 5, ( 1) (2) (5 1) ( 1)(256)(6) 1536n a5

5 8= = - + = - =-

When 8, ( 1) (2) (8 1) (2) (9) (2048)(9) 18432n a8

8 8 3 11= = - + = = =

+ .

(iii) Given that, 2 3 1 1, 2, 3,a n n nforn

2g= - + =

When 5, 2(5) 3(5) 1 2(25) 15 1 50 15 1 36n a5

2= = - + = - + = - + =

When 7, 2(7) 3(7) 1 2(49) 21 1 98 20 78n a7

2= = - + = - + = - = .

(iv) Given that, ( 1) (1 ) 1, 2, 3, .a n n nforn

n 2g= - - + =

When n = 5, ( 1) (1 ) ( )( ) ( )( )a 5 5 1 4 25 1 21 215

5 2= - - + = - - + = - =-

When n = 8, ( 1) (1 8 8 ) ( 7 64) 57a8

8 2= - - + = - + = .

3. Find the 18th and 25th terms of the sequence defined by( ) ,

, .

and even

and odda

n n n n

n

n n n

3

1

2isis

N

Nn 2

!

!=

+

+*

Solution: Given that ( ),

, .

and even

and oddan n n n

n

n n n

3

1

2isis

N

Nn 2

!

!=+

+*

If n is even, ( 3)a n nn= + .

So, a18 = 18(18 3) 18(21) 378+ = = .

If n is odd, an

n

1

2n 2=

+.

So, a25

= ( )

25 1

2 25625 1

5062650

31325

2+

=+

= = .

4. Findthe13thand16thtermsofthesequencedefinedby

,

( ), .

if and even

if and oddb

n n n

n n n n2

isis

N

Nn

2!

!=

+)

Solution: Given that bn = ,

( ), .

if and even

if and odd

n n n

n n n n2

isis

N

N

2!

!+)

So, b13

= ( ) 13(13 2) 195n n 2+ = + = and 16 256.b n16

2 2= = =

5. Findthefirstfivetermsofthesequencegivenby ,a a a2 3

1 2 1= = + and 2 5 2a a nfor

n n 12= +

-.

Solution: Given that 2, 3a a a1 2 1= = + and 2 5 2a a nfor

n n 12= +

-.

So, a2

= 2 53 + =

a3 = 2 5 2 5 15a 5

2+ = + =^ h

a4

= 2 5 2 5 35a 153+ = + =^ h

a5

= 2 5 7535 + =^ h

Hence, the first five terms of the sequence are 2, 5, 15, 35 and 75.


6. Findthefirstsixtermsofthesequencegivenby 1a a a

1 2 3= = = and a a a

n n n1 2= +

- - for 3n 2 .

Solution: Given that 1a a a1 2 3= = = and a a a

n n n1 2= +

- - for n 32 .

So, a4

= 1 1 2a a3 2+ = + =

a5

= 2 1 3a a4 3+ = + =

a6 = 3 2 5a a

5 4+ = + =

Hence, the first six terms of the sequence are 1, 1, 1, 2, 3 and 5.

Exercise 2.2

1. ThefirsttermofanA.Pis6andthecommondifferenceis5.FindtheA.Panditsgeneralterm.

Solution: The general form of an A.P is , , 2 , 3 ,a a d a d a d g+ + + .

Given that a 6= and .d 5= Thus, the A.P is 6, , 6 2 , 6 3 ,6 5 5 5 g+ + +^ ^ ^h h h

That is, the required A.P. is 6, 11, 16, 21, g

The general term tn = ( )a n d1+ -

= n6 1 5+ -^ ^h h

= 6 5 5 5 1 , , , ,n n n 1 2 3 g+ - = + = .

2. Findthecommondifferenceand15thtermoftheA.P.125,120,115,110,g .

Solution: Given that the sequence 125, 120, 115, 110, g is an A.P.

Here, a = 125, d = t t2 1- = 120 125- = 5-

The general term is

tn = a n d1+ -^ h .

Thus, t15

= 125 125 14 125 70 5515 1 5 5+ - - = + - = - =^ ^ ^h h h .

3. Whichtermofthearithmeticsequence24, 23 , 22 , 21 , .41

21

43 g is 3?

Solution: Given arithmetic sequence is , , , ,24 2341 22

21 21

23 g

Here, 24, 23 24a d41

43= = - = - and the last term l 3= .

If n is the number of terms in an A.P, then

n = 1d

l a- +

& n = 1

43

3 24-- + =

321 4 1 28 1 29#

-- + = + =` j

Thus, 29th term of the sequence is 3.


4. Findthe12thtermoftheA.P. , 3 , 5 , .2 2 2 g

Solution: Given A.P is , 3 , 5 ,2 2 2 g . Here, , 3 2a d2 2 2 2= = - =

The nth term of the A.P is ( )t a n d1n= + -

Thus, t12

= 2 12 1 2 2+ -^ h

= 2 11 2 2+ ^ h = 2 22 2+

t12

= 23 2 .

5. Findthe17thtermoftheA.P.4,9,14,g .Solution: Given A.P is , , ,4 9 14 g . Here, 4, 9 4 5a d= = - =

The general term tn of an A.P is

tn = a n d1+ -^ h

Thus, t17

= 4 17 1 5 4 16 5 84+ - = + =^ ^ ^h h h .

6. HowmanytermsarethereinthefollowingArithmeticProgressions?

(i) 1, , , , .65

32

310g- - - (ii)7,13,19,g ,205.

Solution: (i) The given arithmetic sequence is 1, , , , .65

32

310g- - -

Here, 1, 1a d65

61=- = - + = and l

310= .

Now, n = 1d

l a- +

Thus, n = 310 1

61

1+

+ = ( ) 1 26 1 27313 6 + = + = .

Hence, the sequence has 27 terms.

(ii) Given A.P. is 7,13,19, ,205g .

Here, 7, 13 7 6a d= = - = and l = 205.

The number of terms in the given A.P. is n = d

l a 1- +

= 6

205 7 1- + = 6

198 1+ = 34

Hence, there are 34 terms in the given sequence.

7. If9thtermofanA.P.iszero,provethatits29th termisdouble(twice)the19thterm.

Solution: Given that t 09= & 8 0 8a d a dor+ = =- .

Now, t29

= a d d d d28 8 28 20+ =- + =

= d2 106 @ = d d2 8 18- +6 @ = a d2 18+6 @ = 2t

19

Hence, t29

= 2 t19

.


8. The10thand18thtermsofanA.P.are41and73respectively.Findthe27thterm.

Solution: Given that t 4110

= and t 7318

= .

& a d10 1 41+ - =^ h and & a d18 1 73+ - =^ h

& 9a d+ = 41 g (1) ; & 17a d+ = 73 g (2)

Now, (2) (1) 8d&- = 32 4d& = .

Also, (1) 9(4)a& + = 41

& 41 36a = - = 5

Thus, t27

= a d27 1 5 26 4 5 104 109+ - = + = + =^ ^h h .

Hence, the 27th term is 109.

9. Find nsothatthenthtermsofthefollowingtwoA.P.’sarethesame.1,7,13,19,g and100,95,90,g .

Solution: First consider the A.P. 1, 7, 13, 19, g .

Here, ,a d1 7 1 6= = - = . So, the nth term of the A.P. is 1t n 1 6n= + -^ ^h h.

Now, consider the A.P. 100, 95, 90, g . Here, ,a d100 95 100 5= = - =- .

Thus, the nth term of this A.P. is 100s n 1 5n= + - -^ ^h h

Given that t sn n

=

& n1 1 6+ -^ ^h h = n100 1 5+ - -^ ^h h

& 1 6 6n+ - = n100 5 5- +

& 11n = 110 10.n& =

10. Howmanytwodigitnumbersaredivisibleby13?

Solution: Two digit numbers divisible by 13 are 13,26,39, 91.g . This is an A.P.

Here, ,a d13 13= = and l 91= .

Now, n = d

l a 1- + .

= 13

91 13 11378 1 6 1 7- + = + = + = .

Hence, there are 7 two-digit numbers divisible by 13. 11. ATVmanufacturerhasproduced1000TVsintheseventhyearand1450TVs

inthetenthyear.Assumingthattheproductionincreasesuniformlybyafixed

numbereveryyear,findthenumberofTVsproducedinthefirstyearandinthe15thyear.

Solution: Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st year, 2nd year, 3rd year ... will form an A.P.

Let us denote the number of TV sets manufactured in the nth year by tn.


Then, t 10007= and t 1450

10=

Thus, we have t7 = 6 1000a d+ = g (1)

t10

= 9 1450a d+ = g (2)

(2) (1) &- 3d = 450

& d = 150.

Substituting d 150= in (1), we get

( )a 6 150+ = 1000

a = 1000 900-

= 100.

Therefore, the production of TV sets in the first year is 100.

Now, 14 100 14 100 2100 2200.t a d 15015

= + = + = + =^ h

So, the production of TV sets in the 15th year is 2200.

12. Amanhassaved` 640duringthefirstmonth,` 720inthesecondmonthand` 800inthethirdmonth.Ifhecontinueshissavingsinthissequence,whatwillbehissavingsinthe25thmonth?

Solution: The amount saved by the man during first month, second month, third month, ... are 640, 720, 800, ... . This is an A.P. with a = 640, d = 720 – 640 = 80.

Thus, t25

= 640 25 1 80+ -^ ^h h

= 640 24 80+ ^ h = 640 1920+

= 2560 Hence, his saving in the 25th month is ` 2560.

13. ThesumofthreeconsecutivetermsinanA.P.is6andtheirproductis–120.Findthethreenumbers.

Solution: Let , ,a d a a d- + be the three consecutive terms. Now, 6 3 6 2a d a a d a a& &- + + + = = =^ ^h h

Also a d a a d- +^ ^ ^h h h = 120-

a a d2 2& -^ h = 120-

2 d4 2& -^ h = 120 4 60d2&- - =-

& d2 = 64 & d 8!=

When ,d 8= the three numbers are , ,6 2 10- .

When d 8=- , the three numbers are , ,10 2 6- .


14. FindthethreeconsecutivetermsinA.P.whosesumis18andthesumoftheirsquaresis140.

Solution: Let the three consecutive terms be , , .a d a a d- + Then

a d a a d- + + +^ ^ ^h h h = 18

& 3a = 18 6a& =

Also, a d a a d 1402 2 2- + + + =^ ^h h

& a ad d a a ad d2 2 1402 2 2 2 2- + + + + + =

& a d3 2 1402 2+ =

& d3 36 2 1402+ =^ h

& 2 140 108 2 32 4d d d2 2& & != - = =

When d 4= , the terms are, 2, 6, 10.

When d 4=- , the terms are, 10, 6, 2.

15. Ifmtimesthemth termofanA.P.isequaltontimesitsnthterm,thenshowthatthe (m + n)thtermoftheA.P.iszero.

Solution: Given that m t n tm n

=

& m a m d n a n d1 1+ - = + -^ ^h h6 6@ @& am m m d na n n d1 1+ - = + -^ ^h h

& 0m n a m m n n d1 1- + - - - =^ ^ ^h h h6 @& m n a m m n n d 02 2

- + - - + =^ h 6 @

& m n a m n m n d 02 2- + - - - =^ ^ ^h h h6 @

& m n a m n m n d1 0- + - + - =^ ^h h6 @& a m n d1 0+ + - =^ h

& 0tm n

=+

.

16. Apersonhasdeposited 25,000inaninvestmentwhichyields14%simpleinterestannually.Dotheseamounts(principal+interest)formanA.P.?Ifso,determinetheamountofinvestmentafter20years.

Solution: I = P R T100# # = ,

10025 000 14 1# #

= 250 14# = 3500

Amount = P I+

= 25,000 5003+ = 28,500

So, the amount at the end of the first year = ` 28,500

The amount at the end of the second year = 28,500 + 3500 = ` 32,000

Similarly, we can obtain the amount at the end of 3rd year, 4th year, 5th year and so on.


Therefore, the amount at the end of 1st, 2nd, 3rd, 4th ... year respectively are 28,500, 32,000, 35,500, g .

It is an A.P as the difference between any two consecutive terms in the list is 3500.

Here, , ,d a3500 28 500= =

Now, to find the amount at the end of 20 years, we shall find t20

.

Thus, t20

= a d19+

= 28,500 + 19 (3500) = 28500 + 66500 = ` 95,000.

17. Ifa,b,careinA.P.thenprovethat( ) 4( )a c b ac2 2

- = - .

Solution: Given that , ,a b c are in A.P.

So, 2b = a c+ g (1)

& 4b2 = a c 2+^ h

& 4b ac4 2- = 4a c ac2+ -^ h

& 4 b ac2-^ h = a c 2-^ h

Aliter: L.H.S. = b ac4 2-^ h

= b ac4 42-

= 4a c ac2+ -^ h ( using equation (1) )

= a c 2-^ h = R.H.S.

18. Ifa,b,careinA.P.,thenprovethat , ,bc ca ab1 1 1 arealsoinA.P.

Solution: Given that , ,a b c are in A.P. So , ,abca

abcb

abcc are also in A.P.

& , ,bc ca ab1 1 1 are in A.P.

19. If , ,a b c2 2 2 areinA.P.thenshowthat , ,

b c c a a b1 1 1+ + +

arealsoinA.P.

Solution: Given that , ,a b c2 2 2 are in A.P.

So, b a2 2- = c b2 2

- (common difference)

& b a b a+ -^ ^h h = c b c b+ -^ ^h h

& b c c a

b a+ +

-^ ^h h

= a b c a

c b+ +

-^ ^h h

[dividing by a b b c c a+ + +^ ^ ^h h h]

& b c c ab c c a+ ++ - -

^ ^h h =

c a a bc a a b+ ++ - -

^ ^h h

& b c c a

b c c a

+ +

+ - +

^ ^^ ^

h hh h =

c a a b

c a a b

+ +

+ - +

^ ^^ ^

h hh h

& c a b c1 1+

-+

= a b c a1 1+

-+

& c a2+

= a b b c1 1+

++

. Thus, , ,b c c a a b1 1 1+ + +

are in A.P.


20. If , 0, 0, 0a b c x y zx y z

! ! != = andb ac2= ,thenshowthat , ,

x y z1 1 1 areinA.P.

Solution: Given , 0, , 0a b c x y z0x y z ! ! != = . Let a b c kx y z= = = (say)

& , ,a k b k c kx y z1 1 1

= = =

Also given that b ac2=

Thus, /y1 2

k^ h = k k1 1x z

& k y2 = k x z

1 1+

& y2 =

x z1 1+

Hence, , ,x y z1 1 1 are in A.P.

Exercise 2.3

1. Findoutwhichof the followingsequencesaregeometric sequences.For thosegeometricsequences,findthecommonratio.

(i) 0.12, 0.24, 0.48,g . (ii) 0.004, 0.02, 0.1,g . (iii) , , , ,21

31

92

274 g .

(iv)12, 1, ,121 g . (v) , , ,2

2

1

2 2

1 g .(vi)4, 2, 1, ,21 g- - - .

Solution: (i) Here, ..

.

. 20 120 24

0 240 48 g= = =

Thus, the common ratio is 2. Therefore the given sequence is a geometric sequence.

(ii) The given sequence is 0.004, 0.02, 0.1,g

Now, ..

..

0 0040 02

0 020 1 5g= = =

Thus, the common ratio is 5. Therefore, the given sequence is a geometric sequence.

(iii) Considering the ratio of the consecutive terms, we see that

2131

3192

92274

32g= = = =

So, the common ratio is .32 Therefore, the given sequence is a geometric sequence.

(iv) We have 121

1121

121g= = = . Thus, the common ratio is .

121

Hence, the given sequence is a geometric sequence.

(v) We have 2

2

1

2

12 2

1

21g= = =

Thus, the common ratio is 21 .

Hence, the given sequence is a geometric sequence.


(vi) Since 42

21!-

-- , the given sequence has no common ratio.

Hence, it is not a geometric sequence.

2. Findthe10thtermandcommonratioofthegeometricsequence , ,1, 2,41

21 g- - .

Solution: The common ratio of the sequence,

r4121

211

12 2g=

-

=-

= - = =-

The first term of the sequence is 41 .

The general term of the sequence is , , ,t a r n 1 2 3n

n 1 g= =-

Thus, t41 2

1010 1= - -` ^j h = 2

2

1 22

9 7- =-^ h .

3. Ifthe4thand7thtermsofaG.P.are54and1458respectively,findtheG.P.

Solution: Given that t 544= and t 1458

7= .

Using the formula , , ,t a r n 1 2 3n

n 1 g= =- for the general term we have,

a r 543= and a r 14586

= & a r

a r54

14583

6

= & r r27 33&= =

Now, a r3 = 54

& a 33^ h = 54

& a2754= = 2

Hence, the required geometric sequence is , , ,2 2 3 2 3 2 32 3g^ ^ ^ ^ ^ ^h h h h h h

i.e., 2, 6, 18, 54,g

4. Inageometricsequence,thefirsttermis31 andthesixthtermis

7291 ,findthe

G.P.

Solution: First term is 31 and sixth term is

7291 . So a

31= and a r

72915

=

Now, a r5 = 7291 & r

31

72915

=

& r5 = 2431 & r

315 5

= ` j & r31=

Thus, the required G.P. is , , ,31

31

31

31

31 2

g` ` ` ` `j j j j j . i.e., , , ,31

91

271 g

5. Whichtermofthegeometricsequence,

(i) 5, 2, , ,54

258 g ,is

15625128 ? (ii) 1, 2, 4, 8,g ,is1024?

Solution: (i) The given Geometric sequence is 5, 2, , ,54

258 g

15625128


Here, 5,a r52

254

52g= = = = =

The general term is , 1,2,3,t a r nn

n 1 g= =-

Thus, ( )( )

a r15625128

625 252n 17

= =-

& 552

5

2n 1

6

7

=-

` j

& 52

52n 1 7

=-

` `j j

& 1 7n - = & n 8=

(ii) The given geometric sequence is 1, 2, 4, 8,g1024

First term a 1= and the common ratio r12

24 2g= = = =

The general term is , , ,t a r n 1 2 3n

n 1 g= =- we get

Thus, a rn 1- = 1024 = 210

& 2 n 1-^ h = 210

& n 1- = 10 & n 11= . Then, the 11th term of the given geometric sequence is 1024.

6. Ifthegeometricsequences162,54,18,g and , , ,812

272

92 g havetheirnthterm

equal,findthevalueofn.

Solution: Consider the geometric sequence 162, 54, 18, g

Here, ,a r16216254

5418

31g= = = = = .

So 162t31

n

n 1=

-` j g (1)

Now, consider the geometric sequence , , ,812

272

92 g

Here, ,a r812

812272

27292

3g= = = = =

t812 3

nn 1

= -^ h g (2)

Given that nth term of the two sequence are equal.

Thus, 16231 n 1-

` j = 812 3 1n-^ h

& 3

162n 1-

= 3812 n 1-

& 3n 1 2-^ h = 2

162 81#


& 3n 1 2-^ h = 812

& 3 1n- = 81 34=

& n 1- = 4 & n 5=

Hence, 5th term of the above G.P’s are equal.

7. ThefifthtermofaG.P.is1875.Ifthefirsttermis3,findthecommonratio.

Solution: Given the first term and fifth term of the G.P are 3 and 1875 respectively.

Therefore, ,a t a r3 18755

4= = =

& r3 4 = 1875

& r4 = 625 = 54 & r 5=

Thus, the common ratio is 5.

8. Thesumofthreetermsofageometricsequenceis1039 andtheirproductis1.Find

thecommonratioandtheterms.

Solution: Let the three terms of the G.P be , ,ra a ar . Then

ra a ar+ + =

1039

ar

r1 1+ +` j = 1039

& r

r r 12+ +c m =

1039 g (1)

Also, ra a ar 1=` ^ ^j h h

& a 13= or a 1=

Substituting a 1= in equation (1) we get,

(1) & r

r r 12+ +^ h =

1039

& r r10 10 102+ + = 39r

& r r10 29 102- + = 0

& r r2 5 5 2- -^ ^h h = 0

& r25= or

52

When r25= , the terms are , ,

52 1

25 ; When ,r

52= the terms are , ,

25 1

52

9. IftheproductofthreeconsecutivetermsinG.P.is216andsumoftheirproductsinpairsis156,findthem.

Solution: Let the three terms of the G.P be , ,ra a ar .

Then 216ra a ar =` ^ ^j h h

& a 2163= & a 6=


Also, ra a a ar ar

ra 156+ + =` ^ ^ ^ ^ `j h h h h j

& ra a r a2

2 2+ + = 156

& ar

r1 12+ +` j = 156

& 36 1r

r r2+ +c m = 156

& r

r r 12+ + =

313

& r r3 3 32+ + = 13r

& r r3 10 32- + = 0

& r r3 1 3- -^ ^h h = 0

Thus, r 3= or 31

When ,r 3= the terms are 2, 6, 18; When r31= , the terms are 18, 6, 2

10. FindthefirstthreeconsecutivetermsinG.P.whosesumis7andthesumoftheirreciprocalsis

47 .

Solution: Assume that the three consecutive terms of the G.P are , ,ra a ar . Then

ra a ar+ + = 7

& ar

r1 1+ +` j = 7

& ar

r r 12+ +c m = 7 g (1)

Also, ar

a ar1 1+ + =

47

& a

rr

1 1 1+ +8 B = 47

& a r

r r1 12+ +; E =

47 g (2)

On dividing (1) by (2) we get a 7742

#=

& a 42= & a 2!=

Since the sum of the terms are positive, a 2=

Now, 2r

r r 12+ +c m = 7

& 2 2 2r r2+ + = r7

& r r2 5 22- + = 0

& r r2 1 2- -^ ^h h = 0

& r 2= or 21

By taking ,a r2 2= = the three terms of the G.P. are 1, 2, 4

By taking ,a r221= = , the three terms of the G.P. are 4, 2, 1.


11. ThesumofthefirstthreetermsofaG.P.is13andsumoftheirsquaresis91.DeterminetheG.P.

Solution: Let the first three terms of the G.P be , ,a ar ar2 .

Given that a r r1 2+ +^ h = 13 g (1)

a r r12 2 4+ +^ h = 91 g (2)

Using (1) and (2) we get,

a r r

a r r

1

12 2 2

2 2 4

+ +

+ +

^

^

h

h = 16991

137=

& r r

r r r r

1

1 12 2

2 2

+ +

+ + - +

^

^ ^

h

h h = 137

& r r

r r

1

12

2

+ +

- + = 137

& 3 10 3r r2- + = 0

& r r3 3 1- -^ ^h h = 0

& r 3= or 31

When r 3= , the three terms are 1, 3, 9; when r31= , the three terms are 9, 3, 1.

12. If `1000 isdeposited inabankwhichpaysannual interestat therateof5%compoundedannually,findthematurityamountattheendof12years.

Solution: The principal is ` 1000. Interest for the first year is 1000 1005` j

Amount at the end of the first year is

100 100001005+ ` j = 1000 1

1005+` j

Interest for second year = 1000 15005

1005+` `j j

So, the amount at the end of the second year

= 1000 11005 1000 1

1005

1005+ + +` ` `j j j

= 1000 11005 1

1005+ +` `j j

= 1000 11005 2

+` j

Continuing in this way, we see that the amount at the end of 12th year

= ` 1000 11005 12

+` j = ` 1051000100

12` j

Note: Without using above method, one can easily find that total amount using the formula A P i1 n= +^ h Where A is the amount, P is the principal ,i r r

100= is the

annual interest rate and n is the number of years.

A = 1000 11005 12

+` j = ` 1000100105 12` j


13.Acompanypurchasesanofficecopiermachinefor` 50,000.Itisestimatedthatthecopierdepreciates in itsvalueatarateof15%peryear.Whatwillbethevalueofthecopierafter15years?

Solution: ` Value of the copier at the end of first year 50,000t10085

1#= ` j

Value of the machine at the end of second year

t2

= 5000010085

10085

# #` j

= 5000010085 2

# ` j

Thus, the value of the machine at the end of 15th year, t15

= ` 5000010085 15

` j .

14. If , , ,a b c d areinageometricsequence,thenshowthat .a b c b c d ab bc cd- + + + = + +^ ^h h

Solution: Given , , ,a b c d are in geometric sequence. Let r be the common ratio of the given sequence. Then, , ,b ar c ar d ar2 3

= = =

Now, a b c b c d- + + +^ ^h h = a ar ar ar ar ar2 2 3- + + +^ ^h h

= a r r r r r1 12 2 2- + + +^ ^h h

= a r a r a r2 2 3 2 4+ +

= ab bc cd+ + .

Aliter: Given that , , ,a b c d are in G.P. So, b ac2= , ad bc= and c bd2

=

Now, a b c b c d- + + +^ ^h h

= ab ac ad b bc bd bc c cd2 2+ + - - - + + +

= ab ac b ad bc bd c bc cd2 2+ - + - + - + + +^ ^ ^h h h

= ab bc cd+ + .

15. If , , ,a b c d areinaG.P.,thenprovethat , , ,a b b c c d+ + + arealsoinG.P.

Solution: Given that , , ,a b c d are in G.P.

Let r be the common ratio of the given G.P.

Then, , ,b ar c ar d ar2 3= = =

We have a bb c

++ =

a arar ar2

++

= a r

ar r

1

1

+

+

^^

hh = r

Also, b cc d

++ =

ar ar

ar ar2

2 3

+

+ = 1

ar r

ar rr

1

2

+

+=

^^

hh

Hence, , ,a b b c c d+ + + are in G.P.


Exercise 2.4

1. Findthesumofthefirst(i)75positiveintegers(ii)125naturalnumbers.

Solution: (i) 1 2 3 75g+ + + + is an Arithmetic series

Here, ,a d1 2 1 1= = - = and n 75=

Now, Sn = n a n d

22 1+ -^ h6 @

& s75

= ( ) ( )275 2 1 75 1 1+ -^ h6 @

= 275 2 74+6 @ =

275 766 @ = 75 38#6 @

Thus, S75

= 2850.

Remarks: Also note that the above problem can be solved by using the formula S n a l

2n= +6 @. Here, 75, 1n a= = .

` S75

= 275 1 75+6 @ =

275 76 2850=^ h

(ii) 1 2 3 125g+ + + + is an Arithemetic series

Here, , ,a d n l1 1 125= = = =

Now, S125

= 2

125 1 125+6 @ S n a l2n

` = +` j6 @

= 2

125 126# = 125 63#

Thus, S125

= 7875.

2. Findthesumofthefirst30termsofanA.P.whosenth termis3 2n+ .

Solution: Given that the nth term of an A.P is n3 2+ .

Now, tn = 3 2 ( )( )n n5 1 2+ = + - of the form ( )a n d1+ -

S30

= 230 2 5 30 1 2+ -^ ^ ^h h h6 @

= 15 10 58+6 @ = 15 68#

S30

= 1020

3. Findthesumofeacharithmeticseries

(i) 38 35 32 2g+ + + + . (ii) 6 5 4 2541

21 g+ + + terms.

Solution: (i) 38 35 32 2g+ + + + is an A.P

Here, , ,a d l38 35 38 3 2= = - =- =

Now, ( 1)l a n d &= + - nd

l a 13

2 38 1= - + =-- + =13

Thus, S13

= 213 2 38+6 @ ( )S n a l

2n= +6 @

= 213 406 @ = 260=

Aliter: t a 51= =

d t t 42 1

= - =

S30

= 1020


(ii) Given arithmetic series is 6 5 4 2541

21 g+ + + terms

Here, 6, 5 6a d41

43= = - = - and n 25= .

Now, Sn = n a n d

22 1+ -^ h6 @

Thus, S25

= 225 2 6 25 1

43+ - -^ ^ `h h j8 B

= 3225 12 24

4+ -` j8 B =

225 12 18-6 @ 75=- .

4. FindtheSnforthefollowingarithmeticseriesdescribed.

(i) 5,a = 30,n = 121l = (ii) 50,a = 25,n = 4d =-

Solution: (i) Given that , ,a n l5 30 121= = =

Now, Sn = n a l

2+6 @

& S30

= 230 5 121+6 @ = 15 1266 @ = 1890.

(ii) Given that ,a n50 25= = and d 4=-

Now, Sn = n a n d

22 1+ -^ h6 @

& S25

= 225 2 50 25 1 4+ - -^ ^ ^h h h6 @

= 225 100 24 4+ -^ h6 @ =

225 100 96-6 @ =

225 4 50=6 @

5. Findthesumofthefirst40termsoftheseries1 2 3 42 2 2 2

g- + - + .Solution: The given series can be written in the form of

1 4 9 16 25 36- + - + - +^ ^ ^h h h . . . . . to 20 terms.

= 3 7 11- + - + - +^ ^ ^h h h . . . . . to 20 terms. This is an A.P with 3 ,a d 4=- =- and n 20= .

We have Sn = ( )n a n d

22 1+ -6 @

& S20

= ( )220 6 19 4- + -^ h6 @ = 10 82-^ h = – 820.

Aliter: 1 2 3 4 39 402 2 2 2 2 2g- + - + + -

= 1 2 3 4 39 402 2 2 2 2 2g+ + + + + +

2 2 4 6 402 2 2 2g- + + + +^ h

= 1 2 3 40 2 2 1 2 202 2 2 2 2 2 2 2g g+ + + + - + + +^ ^h h

= 40

6

41 818

6

20 21 41-

^ ^ ^ ^ ^ ^h h h h h h

= 20 41 27 28 820- =-^ ^h h6 @ .


6. Inanarithmeticseries,thesumoffirst11termsis44andthatofthenext11termsis55.Findthearithmeticseries.

Solution: We have Sn = ( )n a n d

22 1+ -6 @

Thus, S 4411

= 44a d211 2 11 1& + - =^ ^h h6 @ & a d5 4+ = g (1)

Also, S S 5522 11

= + = 44 55 99+ =

& a d222 2 22 1 99+ - =^ h6 @ & a d2 21 9+ = g (2)

Solving (1) and (2), we get a1139= and d

111= .

The required arithmetic series is ( ) ( )a a d a d2 g+ + + + +

= 1139

1139

111

1139

112 g+ + + + +` `j j =

1139

1140

1141

1142 g+ + + + .

7. Inthearithmeticsequence60,56,52,48,g ,startingfromthefirstterm,howmanytermsareneededsothattheirsumis368?

Solution: Given arithmetic sequence is 60, 56, 52, 48,g

Here, ,a d60 56 60 52 56 4g= = - = - = =-

Also, S 368n= .

Let us find the number of terms needed.

Now, Sn = n a n d

22 1+ -^ h6 @

& 368 = n n2

2 60 1 4+ - -^ ^ ^h h h6 @

& n n2120 4 4- +6 @ = 368

& n n2124 4-6 @ = 368 & n n62 2-^ h = 368

& n n2 62 3682- + = 0 & n n31 1842

- + = 0& n n8 23- -^ ^h h = 0 & n 8= or 23Hence, 8 terms or 23 terms are needed to get the sum 368.

8. Findthesumofall3digitnaturalnumbers,whicharedivisibleby9.

Solution: The sequence of 3 digit numbers which are divisible by 9 are , , , ,108 117 126 999g .

This is an A.P., where ,a d108 9= = and l 999= .

Also, ( )l a n d1 &= + - n = d

l a 1- +

& = 9

999 108 1- + = 9

891 1+ = 99 1 100+ =

Now, ( )S n a l2n

&= + S100

= 2

100 999 108+6 @

= 50 1107 55350=^ h .


9. Findthesumoffirst20termsofthearithmeticseriesinwhich3rdtermis7and7th termis2morethanthreetimesits3rdterm.

Solution: Given that t 73= and t t2 3 23

7 3= + =

Now, t a n d1n

= + -^ h

Thus, a d2+ = 7 g (1)

a d6+ = 23 g (2)

( ) ( )2 1- & d4 = 16 & d = 4

Substituting d 4= in (1), we get a 2 4+ ^ h = 7 & a = 1-

Now, { 2 ( 1) }S n a n d2n

&= + - S20

= 220 2 1 19 4- +^ ^h h6 @ = 10 740.2 76- + =6 @

10.Findthesumofallnaturalnumbersbetween300and500whicharedivisibleby11

Solution: The natural numbers between 300 and 500 which are divisible by 11 are 308, 319, 330, , .495g They form an A.P.

Here, the first term, a = 308, last term, l = 495 and the common difference, d =11.

Also, ( )l a n d1 &= + - n = d

l a 1- + = 111

495 308 18- + =

Now, S n a l2n

= +6 @ & S18

= 218 308 495+6 @ = 9 8036 @ = 7227.

11. Solve:1 6 11 16 148xg+ + + + + = .

Solution: The terms of the series form an A.P. with first term, 1a = and

the common difference, 5d =

Given that 1 6 11 16 148xg+ + + + + =

Now, Sn = 148 & n a n d

22 1+ -^ h6 @ = 148

& 2 5n n2

1 1+ -^ ^h h6 @ = 148

& n n2

5 3-^ h = 148 & n n5 3 2962- - = 0

& n n5 37 8+ -^ ^h h = 0 & n 8= or 537-

Thus, n 8= [Here, n537= - is not possible]

Hence, x = t8 = a +7d = 1+7(5) = 36.

12. Findthesumofallnumbersbetween100and200whicharenotdivisibleby5.

Solution: Let T be the sum of all natural numbers between 100 and 200.

Thus, 101 102 103 199T g= + + + + , where the terms form an A.P. with

101 , 199 .a l n 99and= = =

Then, 99 150 14850T299 101 199 #= + = =^ h .

Let S be the sum of natural numbers between 100 and 200 which are divisible by 5.Then, 105 110 115 195S g= + + + + .


Here, the terms form an A.P. with , ,a d l105 5 195= = = .

Also, ( )l a n d1 &= + - n = d

l a 1- + = 1955105 1- + = 18 1 19+ = .

Now, S n a l2n

= +6 @ & S = 219 195 105+6 @ = 19 150 2850# = .

Hence, the sum of the natural numbers between 100 and 200 which are not divisible by 5 is 14850 2850 12000.T S- = - =

13. Aconstructioncompanywillbepenalisedeachdayfordelayinconstructionofabridge.Thepenaltywillbe`4000forthefirstdayandwillincreaseby`1000foreach followingday.Basedon itsbudget, thecompanycanafford topayamaximumof`1,65,000towardspenalty.Findthemaximumnumberofdaysbywhichthecompletionofworkcanbedelayed.

Solution: Penalty amounts to be levied for consecutive days form an Arithmetic series with ,a d4000 1000= = .

Let n be the maximum number of days for which the work can be delayed.

Then, Sn = 1,65,000 ( given )

& n n2

2 4000 1 1000+ -^ ^ ^h h h6 @ = 1,65,000

& n n8000 1000 1000+ -6 @ = 3,30,000

& 7 330n n2+ - = 0

& 15n n22+ -^ ^h h = 0 & n 15= or 22-

Maximum number of days for which the work can be delayed is 15. 14. A sum of `1000 is deposited every year at 8% simple interest.Calculate the

interestattheendofeachyear.DotheseinterestamountsformanA.P.?

Ifso,findthetotalinterestattheendof30years.Solution: Every year `1000 is deposited at 8% simple interest.

Interest for the first year, t 10001008 80

1#= =

Interest for the second year, t 20001008 160

2#= =

Thus, the interest amounts 80 , 160 , 240, g at the end of each year form an A.P. with a 80= and d 80= .

The total interest is { ( ) }S n a n d2

2 1n

&= + - S230 160 29 80

30= + ^ h6 @ = `7200

15. Thesumoffirst ntermsofacertainseriesisgivenas3 2n n2- .

Showthattheseriesisanarithmeticseries.

Solution: Given that S n n3 2n

2= - .

So, Sn 1-

= n n3 1 2 12- - -^ ^h h


= n n n3 2 1 2 22- + - +6 @ = n n3 8 52

- +

Now, the nth term is tn = S S

n n 1-

-

= n n n n3 2 3 8 52 2- - - +6 @

= n n n6 5 6 6 1 1 1 6- = - + = + -^ h .

So, tn is of the term a n d1+ -^ h .

Hence, the given series is an Arithmetic series with ,a d1 6= = .

16. Ifaclockstrikesonceat1o’clock,twiceat2o’clockandsoon,howmanytimeswillitstrikeinaday?

Solution: Number of times the clock strikes each hour form an A.P.

Then, for the first 12 hours, the arithmetic series is 1 2 3 12g+ + + + .

Thus, S n a l2n

= +6 @ & S12

= 212 1 12 6 13 78+ = =^ h6 @ .

Hence, the clock strikes in a day ( in 24 hours) = 2 78# = 156 times. 17. Showthatthesumofanarithmeticserieswhosefirsttermisa ,secondtermb and

the lasttermisc ,isequaltob a

a c b c a

2

2

-

+ + -

^^ ^

hh h .

Solution: Given that ,t a t b1 2= = and the last term t l c

n= =

Now, the common difference, d = t t2 1- = b a-

Thus, tn = a n d c1+ - =^ h & a n b a c1+ - - =^ ^h h

& n 1- = b ac a

-- & n =

b ab c a2

-+ -

Hence, Sn = n a l

2+6 @

= b a

b c aa c

2

2

-

+ -+

^^

^hh

h

18. Ifthereare n2 1+^ htermsinanarithmeticseries,thenprovethattheratioofthesumofoddtermstothesumofeventermsis :n n1+^ h .Solution: Given that the arithmetic series has n2 1+^ h terms. Let ,T S denote the sum of odd terms and even terms respectively.Now, T = t t t t

n1 3 5 2 1g+ + + +

+

= 21n t t

n1 2 1+ +

+` j 6 @ (There are n 1+ terms)

= 21 { ( ) }n a a n d2+ + +` j 6 @

= na nd

2

12

++

^^

hh6 @ = n a nd1+ +^ ^h h.

Now, S = t t t tn2 4 6 2

g+ + + +

= n t t2 n2 2

+6 @ ( There are n terms)


= { } { }n a d a n d2

2 1+ + + -^ h6 @

= n a nd2

2 2+6 @ = n a nd+^ h

Hence, ST =

n a nd

n a nd

nn1 1

+

+ += +

^^ ^

hh h .

19. Theratioofthesumsoffirstmandfirst ntermsofanarithmeticseriesis

:m n2 2 .Showthattheratioofthemth andnth termsis :m n2 1 2 1- -^ ^h h.

Solution: Given the ratio of the sums of first m and first n terms of arithmetic series is :m n2 2 .

That is, S

S

n

m

n

m2

2

= & 2n a n d

m a m d

n

m

22 1

2 1

2

2

+ -

+ -=

^

^

h

h

6

6

@

@

& a n d

a m d

nm

2 1

2 1

+ -

+ -=

^^

hh

& an mnd nd am mnd md2 2+ - = + -

& an nd am md2 2- = -

& a n m n m d2 - = -^ ^h h & a d2 =

Thus, t

t

n

m = a n d

a m d

1

1

+ -

+ -

^^

hh

= a n a

a m a

1 2

1 2

+ -

+ -

^ ^^ ^

h hh h

= a n

a m

1 2 2

1 2 2

+ -

+ -

66

@@ =

nm2 12 1

-- .

20. Agardenerplanstoconstructatrapezoidalshapedstructureinhisgarden.Thelongersideoftrapezoidneedstostartwitharowof97bricks.Eachrowmustbedecreasedby2bricksoneachendandtheconstructionshouldstopat25throw.Howmanybricksdoesheneedtobuy?

Solution: The number of bricks to be used in each row form an A.P.

Thus, the arithmetic series is 97 93 89 g+ + + to 25 terms.

Here, ,a d97 4= =- and n 25= .

Now, Sn = n a n d

22 1+ -^ h6 @

& S25

= 225 2 97 25 1 4+ - -^ ^ ^h h h6 @

= 225 194 96-6 @

= 225 986 @

Thus, the number of bricks needed = 1225 bricks.


Exercise2.5

1. Findthesumofthefirst20termsofthegeometricseries25

65

185 g+ + + .

Solution: Given geometric series is 25

65

185 g+ + +

Here, ,a r25

2565

65185

31g= = = = = and n = 20.

We have, Sn =

ra r11 n

--^ h

Thus, S20

= 1

31

25 1

31 20

-

- ` j; E = 2

5

32

131 20

- ` j; E =

415 1

31 20

- ` j; E

2. Findthesumofthefirst27termsofthegeometricseries91

271

811 g+ + + .

Solution: For the given geometric series 91

271

811 g+ + + ,

we have, a = , r91

91271

31= = and n = 27.

Thus, Sn=

ra r11 n

--^ h & S

27=

131

91 1

31 27

-

- ` j; E =

32

91 1

31 27

- ` j; E =

61 1

31 27

- ` j; E.

3. FindSnforeachofthegeometricseriesdescribedbelow.

(i) 3,a = 384,t8= 8n = . (ii) 5,a = 3r = , 12n = .

Solution: (i) Given ,a 3= 384,t8= n 8= .

Now, t8 = a r8 1$ =- a r7

& a r7 = 384 & r3 7^ h = 384

& r7 = 128 & r 27 7= & r = 2

Thus, Sn =

ra r11 n

--^ h & S

8 =

1 23 1 28

--6 @ = 3 256 1 765- =6 @ .

(ii) Given that ,a 5= r 3= , n 12=

Thus, Sn=

ra r

11n

--^ h & S

12 =

3 15 3 112

--6 @ =

25 3 112

-6 @.

4. Findthesumofthefollowingfiniteseries (i) 1 0.1 0.01 0.001 .0 1 9g+ + + + +^ h (ii) 1 11 111 g+ + + to20 terms.

Solution: (i) Given series 1 0.1 0.01 0.001 .0 1 9g+ + + + + ^ h is a geometric series

Here, , .a r1 0 1= = and .t 0 1n

9= ^ h


Now, tn = a rn 1

#- & ( . )0 1 9 = .1 0 1 n 1-^ ^h h

& 1n - = 9 & n 10= .

Thus, Sn=

ra r11 n

--^ h & S

10 =

.

.

1 0 1

1 1 0 1 10

-

-^

^ ^h

h h6 @ =

.

.

0 9

1 0 1 10- ^ h .

(ii) Given series is 1 11 111 g+ + + to 20 termsLet S

20 = 1 11 111 g+ + + to 20 terms.

Multiplying and dividing by 9, we get

S20

= [91 9 99 999 g+ + + to 20 terms]

= [91 10 1 100 1 1000 1 g- + - + - +^ ^ ^h h h 20 terms]

= 91 10 10 102 3 g+ + +^6 20 terms) 20- @

= 91

10 110 10 1

2020

#-

--

^ h; E' 1.

( Here, sum of n terms of G.P. is r

a r11n

--^ h )

= 10 18110

92020

- -^ h6 @ .

5. Howmanyconsecutivetermsstartingfromthefirsttermoftheseries (i) 3 9 27 g+ + + wouldsumto1092? (ii)2 6 18 g+ + + wouldsumto728?

Solution: (i) Let us find n such that 3 9 27 t 1092n

g+ + + + =

Now, the given series is a geometric series with 3, 3.a r39

927 g= = = = =

Given that S 1092n=

& r

a r11n

--^ h =1092 &

3 13 3 1n

--6 @ = 1092

& 3 1 728n- = & 3 729n

= = 36

Thus, the number of terms needed to get the sum is, n 6=

(ii) Given series 2 6 18 g+ + + is a geometric series with

, ,a r226 3= = = and S 728

n= .

Thus, Sr

a r11

n

n

=--^ h &

3 12 3 1n

--6 @ = 728 & 3 1n

- = 728

& 3n = 729 3 6n6&= =

Thus, the number of terms needed to get the sum is, n 6= .

6. The second term of a geometric series is 3 and the common ratio is .54

Findthesumoffirst23consecutivetermsinthegivengeometricseries.

Solution: Given that ,t r354

2= = and n 23=

Now, 3t2

&= ar = 3 .a

543

415& = =


Thus, Sr

a r11

n

n

=--^ h & S

23 =

154

415

154

23

-

- ` j; E =

1415

5

154 23

- ` j; E

= 475 1

54 23

- ` j; E. 7. Ageometricseriesconsistsoffourtermsandhasapositivecommonratio.The

sumofthefirsttwotermsis9andsumofthelasttwotermsis36.Findtheseries.

Solution: Let the four terms of the G.P. be a ,ar , ar2 , ar3 .

Given that 9 ( ) ;a ar 1g+ = ( )ar ar 36 22 3 g+ =

Thus, (2) & r a ar2+^ h = 36

& r 92^ h = 36 & r 42= & r 2!=

Since 0, 2r r> = . Thus, a + ar = 9 & a = 3.

Hence, the required series is 3 + 3(2) + 3(22 ) + 3(23 ) = 3 + 6 + 12 + 24.

8. Findthesumoffirstntermsoftheseries

(i) 7 77 777 g+ + + . (ii) 0.4 0.94 0.994 g+ + + .Solution: (i) Given series is 7 77 777 g+ + + to n terms

Taking 7 as a common factor and multipling and dividing by 9 , we get

Sn =

97 9 99 999 g+ + +6 to n terms]

= 97 10 1 100 1 1000 1 g- + - + - +^ ^ ^h h h6 to n terms]

= (97 10 100 1000 g+ + +6 n terms) (1 1 1 ng- + + terms) ]

= n97 10 10 102 3 g+ + +6 terms n- @

= 10 110 1 n

97 10

n

-- -c m; E


a r11n

--^ h )

= n8170 10 1

97n

- -^ h

(ii) Given series is 0.4 0.94 0.994 g+ + + .

Let Sn = . . .0 4 0 94 0 994 g+ + + to n terms

Then Sn = . . .1 0 6 1 0 06 1 0 006 g- + - + - +^ ^ ^h h h to n terms

= (1 1 1 g+ + + to n terms) ( . . .0 6 0 06 0 006 g- + + + to n terms)

= n 6101

10

1

10

12 3

g- + + +; to n terms]


= 101

n 61

101

1101 n

--

- ` j8>

BH = n 6

91 1

101 n

- - `` j j8 B = n32 1

101 n

- - ` j8 B


a r11n

--^ h )

9. Supposethatfivepeopleareillduringthefirstweekofanepidemicandeachsickperson spreads the contagious disease to four other people by the end of thesecondweekandsoon.Bytheendof15thweek,howmanypeoplewillbeaffectedbytheepidemic?

Solution: The number of people affected by the epidemic during each week form a

geometric series.

Thus, the total number of people affected by the epidemic in 15 weeks is,

S15

= 5 4 5 4 20 4 80# # # g+ + + +^ ^ ^h h h to 15 terms

= 5 20 80 g+ + + to 15 terms.

It is geometric series with 5, 4a r= = , n 15=

Thus, Sr

a r11

n

n

=--^ h & S

15 =

4 15 4 115

--6 @ =

35 4 115

-6 @.

10. Agardenerwantedtorewardaboyforhisgooddeedsbygivingsomemangoes.Hegavetheboytwochoices.Hecouldeitherhave1000mangoesatonceorhecouldget1mangoon thefirstday,2on thesecondday,4on the thirdday,8mangoesonthefourthdayandsoonfortendays.Whichoptionshouldtheboychoosetogetthemaximumnumberofmangoes?

Solution: If the boy receives mangoes daily for 10 days, then

the total number of mangoes is S10

= 1 2 4 8 g+ + + + to 10 terms.

The above series is a geometric series with ,a r1 2= = and n 10= .

Thus, Sr

a r11

n

n

=--^ h & S

10 = ( )

2 11 2 110

--6 @ = 2 110

- = 1023.

Hence, the boy should opt for getting mangoes daily for 10 days.

11. Ageometricseriesconsistsofevennumberofterms.Thesumofalltermsis3timesthesumofoddterms.Findthecommonratio.

Solution: Assume that the number of terms in the geometric series is n2 .

Given that Sn2 = 3 (sum of odd terms)

& Sn2 = 3 t t t t

n1 3 5 2 1g+ + + +

-6 @

& 3arr a ar ar ar

11 n

n2

2 4 2 2g-- = + + + + -c m 6 @


& = a r r r3 1n2 2 2 2 1

g+ + +-^ ^h h6 @

& = 3ar

r

1

1 2 n

2-

- ^ h= G) 3 & 2r

r13 1 &+

= =

Thus, the common ratio is 2.

12. If ,S S Sand1 2 3

arethesumoffirstn,2nand3ntermsofageometricseries

respectively,thenprovethatS S S S S1 3 2 2 1

2- = -^ ^h h .

Solution: Given that ,Sr

a rS

ra r

11

11n n

1 2

2

=--

=--^ ^h h and S

ra r11 n

3

3

=--^ h

Now, S S S1 3 2

-^ h = 11

ra r

ra r

ra r

11

11n n n3 2

--

--

---^ ^ ^h h h

' '1 1

= 1

1

r

a rr r1 1

2

2 nn n3 2

-

-- - +

^

^

h

h 6 @) 3 = 1

1

r

a rr r

2

2 nn n2 3

-

--

^

^

h

h 6 @) 3

= r

a r r r

1

1 1n n n

2

2 2

-

- -

^

^ ^

h

h h = r

a r r

1

1n n

2

2 2 2

-

-

^

^

h

h g (1)

Also S S2 1- = a

rr a

rr

11

11n n2

-- -

--c cm m

= r

a r r1

1 1n n2

-- - +6 @ =

ra r r1

1n

n

--6 @

& S S2 1

2-^ h = r

a r r

1

1n n

2

2 2 2

-

-

^

^

h

h = S S S1 3 2

-^ h . (using (1))

Exercise2.6

1. Findthesumofthefollowingseries.

(i)1+2+3+g +45(ii)16 17 18 252 2 2 2

g+ + + +

(iii)2+4+6+g +100 (iv)7+14+21g +490

(v)5 7 9 392 2 2 2

g+ + + + (vi)16 17 353 3 3

g+ + +

Solution: (i) We have, 1 + 2 + 3 + g nn n

2

1+ =

+^ h .

Thus, 1 + 2 + 3 + g + 45 = 2

45 45 1+^ h = 45 23 1035# = .

(ii) 16 17 18 252 2 2 2

g+ + + +

= (1 2 3 25 ) 1 2 3 152 2 2 2 2 2 2 2g g+ + + + - + + + +^ h

= k kkk

2 2

1

15

1

25-

==

// . { ( )( ).k

n n n6

1 2 1n2

1=

+ +/ }

= 6

25 25 1 50 1

6

15 15 1 30 1+ +-

+ +^ ^ ^ ^h h h h

= 6

25 26 51

6

15 16 31-

^ ^ ^ ^ ^ ^h h h h h h = 5525 1240 4285- = .


(iii) 2 + 4 + 6 + g + 100

= 2 1 2 3 50g+ + + +^ h = 2 k1

50/ . ( k

n n

2

1

k

n

1=

+

=

^ h/ )

= 25502

2 50 50 1+=

^ ^h h .

(iv) 7 + 14 +21 g + 490

= 7 1 2 3 70g+ + + +^ h = 7 7k2

70 70 1

1

70=

+^ h; E/ ( kn n

2

1

k

n

1=

+

=

^ h/ )

= 7 35 71 17395# # = .

(v) 5 7 9 392 2 2 2

g+ + + +

= 1 2 3 39 2 4 6 38 1 32 2 2 2 2 2 2 2 2 2g g+ + + + - + + + + - +^ ^ ^h h h

= 2 10k 1 2 3 192

1

392 2 2 2 2g- + + + + -6 @/ = 4 10k k2

1

392

1

19- -/ /

{ ( )( )k

n n n6

1 2 1n2

1=

+ +/ }

= 106

39 39 1 78 14

6

19 19 1 38 1+ +-

+ +-

^ ^ ^ ^h h h h; E

= 20540 9880 10 10650- - = .(vi) 16 17 35

3 3 3g+ + +

= 1 2 3 35 1 2 153 3 3 3 3 3 3g g+ + + + - + + +^ ^h h

= k k3

1

353

1

15-/ / k

n n

2

1

k

n3

1

2

=+

=

^c

hm) 3/

= ( )2

35 35 12

15 15 12 2+

-+^ h; ;E E =

235 36

215 162 2# #-8 8B B

= 35 18 15 82 2# #-^ ^h h = 630 1202 2-^ ^h h

= ( )( ) 382500.630 120 630 120+ - =

2. Findthevalueofk if

(i) 1 2 3 6084k3 3 3 3

g+ + + + = (ii) 1 2 3 2025k3 3 3 3

g+ + + + =

Solution: (i) Given that 1 2 3 6084k3 3 3 3

g+ + + + =

& nk

3

1/ = 6084

& k k

2

1 2+^ h; E = 6084 = 782

& ( )k k 1+ = 156 = 1312 # & k = 12 .

(ii) Given that 1 2 3 2025k3 3 3 3

g+ + + + =

& nk

3

1/ = 2025

& k k

2

1 2+^ h; E = 2025 = 452 & k k 1+^ h= 109 #

Thus, k = 9


3. If1 2 3 171pg+ + + + = ,thenfind1 2 3 p3 3 3 3

g+ + + + .

Solution: Given that 1 2 3 171pg+ + + + =

& np

1/ = 171

& p p

2

1+^ h = 171

& p p

2

1 2+^ h; E = 1712

Thus, 1 2 3 p3 3 3 3g+ + + + = 1712 = 29241.

4. If1 2 3 8281k3 3 3 3

g+ + + + = ,thenfind1 2 3 kg+ + + + .

Solution: Given that 1 2 3 8281k3 3 3 3

g+ + + + =

& k k

2

1 2+^ h; E = 8281 = 912

& k k

2

1+^ h = 91

& 1 2 3 kg+ + + = 91 5. Findthetotalareaof12squareswhosesidesare12cm,13cm,g ,23cm.respectively.

Solution: Given that the sides of 12 squares are 12cm, 13cm, 14cm, g , 23cm.

Total area of 12 squares = 12 13 14 232 2 2 2g+ + + +

= 1 2 3 23 1 2 3 1122 2 2 2 2 2 2g g+ + + + - + + + +^ ^h h

= k k2

1

232

1

11/-/ / { ( )( )

kn n n

61 2 1n

2

1=

+ +/ }

= 6

23 23 1 46 1

6

11 11 1 22 1+ +-

+ +^ ^ ^ ^h h h h

= 6

23 24 476

11 12 23# # # #-

= 4324 506 3818 cm2- = . 6. Findthetotalvolumeof15cubeswhoseedgesare16cm,17cm,18cm,g ,30cm

respectively.

Solution: The edges of 15 cubes are 16 cm, 17 cm, 18 cm, g , 30 cm respectively.

Total volume of 15 cubes = 16 17 18 303 3 3 3g+ + + +

= 1 2 3 30 1 2 3 153 3 3 3 3 3 3 3g g+ + + + - + + + +^ ^h h

= k k3

1

303

1

15-/ / k

n n

2

1

k

n3

1

2

=+

=

^c

hm) 3/

= 2

30 30 1

2

15 15 12 2+-

+^ ^h h; ;E E = 15 31 15 82 2# #-^ ^h h = 465 1202 2

-^ ^h h

= 216225 14400 201825 cm3- = .


Exercise2.7

Choosethecorrectanswer.

1. Which one of the following is not true?

(A) A sequence is a real valued function defined on N . (B) Every function represents a sequence. (C) A sequence may have infinitely many terms. (D) A sequence may have a finite number of terms.

Solution: A real valued function defined on N is a sequence. ( Ans. (B) )

2. The 8th term of the sequence 1, 1, 2, 3, 5, 8, g is

(A) 25 (B) 24 (C) 23 (D) 21

Solution: In Fibonacci sequence, , .F F F n 2>n n n1 2= +

- - ( Ans. (D) )

3. The next term of 201 in the sequence , , , ,

21

61

121

201 g is

(A) 241 (B)

221 (C)

301 (D)

181

Solution: The general term is ( )

tn n 1

1n=

+ ( Ans. (C) )

4. If a, b, c, l, m are in A.P, then the value of 4 6 4a b c l m- + - + is

(A) 1 (B) 2 (C) 3 (D) 0

Solution: 4 6 4a b c l m- + - + ( ) ( ) ( ) 0a m b l c c c c4 6 2 4 2 6= + - + + = - + = ( Ans. (D) )

5. If a, b, c are in A.P. then b ca b

-- is equal to

(A) ba (B)

cb (C)

ca (D) 1 ( Ans. (D) )

6. If the nth term of a sequence is 100 n +10, then the sequence is

(A) an A.P . (B) a G.P.

(C) a constant sequence (D) neither A.P. nor G.P.

Solution: 100n + 10 = 110 + 100(n–1) is of the form a + (n–1)d . ( Ans (A) )

In fact, , 1,2,t an b nn

g= + = ,where a and b are constants, is nth term of an A.P

7. If , , ,a a a1 2 3

gare in A.P. such that ,a

a

23

7

4 = then the 13th term of the A.P. is

(A) 23 (B) 0 (C) a12 1 (D) a14 1

Solution: 2( a + 3d ) = 3( a + 6d ) & 3a + 18d – 2a – 6d = 0

& a + 12d = 0 ( Ans (B) )


8. If the sequence , , ,a a a1 2 3

g is in A.P. , then the sequence , , ,a a a5 10 15

g is

(A) a G.P. (B) an A.P.

(C) neither A.P nor G.P. (D) a constant sequence

Solution: , , ,a a a5 10 15

g = a + 4d, a+ 9d, a+ 14d,g .

This an A.P with common difference = 5d . ( Ans (B) )

(Terms of an A.P. selected at equal intervals consecutively, again form an A.P. )

9. If k+2, 4k–6, 3k–2 are the three consecutive terms of an A.P, then the value of k is

(A) 2 (B) 3 (C) 4 (D) 5

Solution: (k+2) + (3k–2) = 2(4k–6) & 4k = 8k – 12 & k = 3. ( Ans (B) )

10. If a, b, c, l, m. n are in A.P., then 3a+7, 3b+7, 3c+7, 3l+7, 3m+7, 3n+7 form

(A) a G.P. (B) an A.P.

(C) a constant sequence (D) neither A.P. nor G.P

Solution: If an A.P is Multiplied by a constant or Added by a constant, the resulting sequence is an A.P ( Ans (B) )

11. If the third term of a G.P is 2, then the product of first 5 terms is

(A) 52 (B) 25 (C) 10 (D) 15

Solution: G.P : , , , ,r

ara a ar ar

2

2 . The third term a = 2. So,the product a5= 25 .

(Here, the product of first (2n+1)) terms is 2( )n2 1+ ) ( Ans (B) )

12. If a, b, c are in G.P, then b ca b

-- is equal to

(A) ba (B)

ab (C)

ca (D)

bc

Solution: b ca b

-- =

1

1

bbc

aab

-

-

`

`

j

j =

( )( )

b ra r11-- =

ba . Note; ,b ar c ar2= = ( Ans (A) )

13. If ,x x2 2+ , 3 3x + are in G.P, then ,x5 x10 10+ , 15 15x + form

(A) an A.P. (B) a G.P.

(C) a constant sequence (D) neither A.P. nor a G.P.

Solution: The terms ,x5 x10 10+ , 15 15x + are obtained by multiplying the original sequence by 5. ( Ans (B) )


14. The sequence –3, –3, –3,g is

(A) an A.P. only (B) a G.P. only

(C) neither A.P. nor G.P (D) both A.P. and G.P.Solution: A constant sequence is both A.P and G.P. ( Ans (D) )

15. If the product of the first four consecutive terms of a G.P is 256 and if the common ratio is 4 and the first term is positive, then its 3rd term is

(A) 8 (B) 161

(C) 321 (D) 16

Solution: Let the G.P be , , ,r

ara ar ar

3

3 ; a4 = 256 & a = 4. ar = 16. ( Ans (D) )

16. In a G.P, t53

2= and t

51

3= . Then the common ratio is

(A) 51 (B)

31 (C) 1 (D) 5 ( Ans (B) )

17. If x 0! , then 1 sec sec sec sec secx x x x x2 3 4 5

+ + + + + is equal to

(A) (1 )( )sec sec sec secx x x x2 3 4

+ + + (B) (1 )( )sec sec secx x x12 4

+ + +

(C) (1 )( )sec sec sec secx x x x3 5

- + + (D) (1 )( )sec sec secx x x13 4

+ + +

Solution: Expression ( )( )secsec

secsec sec sec

xx

xx x x

11

11 16 2 2 4

=-- =

-- + + ( Ans (B) )

18. If the nth term of an A.P. is 3 5t nn= - , then the sum of the first n terms is

(A) n n21 5-6 @ (B) n n1 5-^ h (C) n n

21 5+^ h (D) n n

21 +^ h

Solution: ;a t 21

= =- Sn = ( ) ( )n a l n n2 2

2 3 5+ = - + - = ( )n n2

1 5- ( Ans (A) )

19. The common ratio of the G.P. am n- , am , am n+ is

(A) am (B) a m- (C) an (D) a n- ( Ans (C) )

20. If 1 + 2 + 3 +. . . + n = k then 13 n23 3

g+ + + is equal to

(A) k2 (B) k3 (C) k k

2

1+^ h (D) k 1 3+^ h

( Ans (A) )


3. ALGEBRA

Exercise 3.1

Solve each of the following system of equations by elimination method.

1. 2 7x y+ = , 2 1x y- = .

Solution: x y2+ = 7 g (1)

x y2- = 1 g (2)

Now, (1) + (2) & 2x = 8 & x = 4.

When x = 4, (1) & y4 2+ = 7 & y = 23 .

Thus, the solution is ,423` j.

2. 3 8x y+ = , 5 10x y+ = .

Solution: x y3 + = 8 g (1)

x y5 + = 10 g (2)

(1) – (2) & x2- = – 2 & x = 1

When x = 1, (1) & y3 + = 8 & y = 5

Thus, the solution is (1, 5).

3. 4xy2

+ = , 2 5x y3+ = .

Solution: The given equations are

xy2

+ = 4 g (1)

x y3

2+ = 5 g (2)

( )1 2 &# x y2 + = 8 g (3)

( )2 3 &# x y6+ = 15 g (4)

( ) ( )3 4 2 &#- y11- = 22- & y = 2

When y 2= , (4) & ( )x 6 2+ = 15 & x = 3


4. 11 7x y xy- = , 9 4 6x y xy- = .

Solution: Clearly 0, 0x y= = satisfy both the equations.

So, (0,0) is a solution for the system.

Let us find the solution when 0, 0.x y^ ^

Dividing the given equations by xy, we get

y x11 7- = 1 and

y x9 4- = 6

Let ax1= and b

y1= .

Algebra3

Solution - Algebra 55

The given equations become,

a b7 11- + = 1 g (1)

a b4 9- + = 6 g (2)

( ) ( )1 4 2 7 &# #- b19- = – 38 & b = 2

When 2, (1)b &= ( )a7 11 2- + = 1

& – 7a = 1 – 22 & a = 3

Now, 3 ; 2 .a x b y31

21& &= = = =

Thus, the two solutions of the system are (0,0) , ,31

21` j.

5. x y xy3 5 20+ = ,

x y xy2 5 15+ = , 0, 0x y! ! .

Solution: Now, 0, 0.x y^ ^ The given equations are

x y3 5+ =

xy20 g (1)

x y2 5+ =

xy15 g (2)

Since 0, 0,x y^ ^ multiply both sides of the equations by xy.We get, x y5 3+ = 20 g (3) ` x y5 2+ = 15 g (4) ( ) ( )3 4 &- y = 5When 5, (3)y &= ( )x5 3 5+ = 20 & x5 = 5 1.x& =


6. 8 3 5x y xy- = , 6 5 2x y xy- =- .

Solution: Clearly 0, 0x y= = is a solution of the system.

Let us assume that x 0^ and y 0^ . x y8 3- = xy5 g (1) x y6 5- = xy2- g (2)Divide both sides of the equations by xy.

We get, y x8 3- = 5 5

x y3 8& - =- g (3)

and y x6 5- = 2

x y2 5 6&- - = g (4)

Let ax1= and b

y1= , then

(3) & a b3 8- = – 5 g (5)

( )4 & a b5 6- = 2 g (6)

( ) ( )5 5 6 3 &# #- b22- = – 31 & b = 2231


When , (5)b2231 &= a3 8

2231- ` j = – 5 & a =

1123 .

When ,ax

x1123 1

1123

2311we have &= = = .

When , .by

y2231 1

2231

3122we have &= = =

Thus, the two solutions are (0,0) , ,2311

3122` j.

7. 13 11 70x y+ = , 11 13 74x y+ = .


x y13 11+ = 70 g (1)

x y11 13+ = 74 g (2)

Adding (1) and (2), we get x y 6+ = g (3)

Subtracting (2) from (1), we get x y 2- =- g (4)

(3) (4) 2x&+ =

When 2, (4)x &= y2 - = – 2 & y = 4.


8. 65 33 97x y- = , 33 65 1x y- = .


x y65 33- = 97 g (1)

x y33 65- = 1 g (2)

( ) ( )1 2 &+ x y- = 1 g (3)

( ) ( )1 2 &- x y+ = 3 g (4)

Adding (3) and (4), we get x 2= . When 2, (4) .x y 1&= = Thus, the solution is (2, 1).

9. 17x y15 2+ = , , 0, 0

x yx y1 1

536 ! !+ = .


x y15 2+ = 17 g (1)

x y1 1+ =

536 g (2)

Let ax1= and b

y1= .

( )1 & 15a b2+ = 17 g (3)

( )2 & a b+ = 536 g (4)

( ) ( )3 4 15 &#- b13- = – 91 b& = 7


When b = 7, ( )4 & ( )a5 5 7+ = 36 & a = 51

When , 5.ax

x51 1

51 &= = = When 7, 7 .b

yy1

71&= = =

Thus, the solution is ,571` j.

10. x y2

32

61+ = , 0, 0, 0

x yx y3 2 ! !+ = .

Solution: The given equations are,

x y2

32+ =

61 g (1)

x y3 2+ = 0 g (2)

Let ax1= and b

y1= . Now, (1) & a b2

32+ =

61 g (3)

(2) 3 2 0a b& + = g (4)

( ) 3 (4) 23 &# #- 2b- = 21 .b

41& =-

When , (4)b41 &=- a3 2

41+ -` j = 0 3 .a a

21

61& &= =

Thus, 6a x61 &= = and 4.b y

41 &=- =-

Hence, the solution is (6, – 4).

Exercise 3.2

1. Solve the following systems of equations using cross multiplication method.

(i) 3 4 24x y+ = , 20 11 47x y- = (ii) 0.5 0.8 0.44x y+ = , 0.8 0.6 0.5x y+ =

(iii) 2,x y x y23

3

5

3 2 613- =- + = (iv) 2, 13

x y x y5 4 2 3- =- + =

Solution: (i) The given system of equations is

x y3 4 24+ - = 0

x y20 11 47- - = 0. Using cross multiplication method, we have

x y 1

4 24 3 4

11 47 20 11

& -

- - -

& ( ) ( )( )

x4 47 11 24- - - -

= ( )( ) ( )( )

y24 20 47 3- - -

= ( ) ( )( )3 11 20 4

1- -

& x452-

= y339 113

1-

=-

& x = 4,y113452

113339 3

-- = =

-- =



(ii) The given system of equation is

. . .x y0 5 0 8 0 44+ - = 0

. . .x y0 8 0 6 0 5+ - = 0

Multiply both sides of the given equations by 100.

We get, x y50 80 44+ - = 0

x y80 60 50+ - = 0.

& x y 1

80 44 50 80

60 50 80 60

-

-

& 80( 50) 60( 44)

x- - -

= ( ) ( ) ( ) ( )

y80 44 50 50 50 60 80 80

1- - -

=-

& x1360-

= y1020 3400

1-

=-

& x = .34001360 0 4

-- = ; y = .

34001020 0 3

-- =

Thus, the solution is (0.4, 0.3).

(iii) 9 10x y6- = – 2, x y

62 3

613+

= .

The given system of equations can be written as

x y9 10 12- + = 0

x y2 3 13+ - = 0

& x y 1

10 12 9 10

3 13 2 3

- -

-

& x130 36-

= y24 117 27 20

1+

=+

& x = 2 ; 3.y4794

47141= = =


(iv) Let ax1= and b

y1= .

The given system of equations is

a b5 4 2- + = 0

a b2 3 13+ - = 0To solve for the unknowns a and b, let us write the coefficients as

& a b 1

4 2 5 4

3 13 2 3

- -

-


& a52 6-

= b4 65 15 8

1+

=+

& a46

= b69 23

1=

& 2 ; 3.a b2346

2369= = = =

Now, 2 ;a x21&= = 3 .b y

31&= =

Thus, the solution is ,21

31` j.

2. Formulate the followingproblemsasapairofequations,andhencefindtheirsolutions:

(i) One number is greater than thrice the other number by 2. If 4 times the smaller numberexceedsthegreaterby5,findthenumbers.

Solution: Let x be the greater number and y be the smaller number.

Given that x y3- = 2 3 2x y& - - = 0 g (1)

y x4 - = 5 4 5x y& - + - = 0 g (2)

Now, (1) + (2) y& = 7

When y 7= , (1) & x = 23.

Thus, the required numbers are 23 and 7.

(ii) The ratio of income of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3. If each of them manages to save ` 2000permonth,find theirmonthly

income.

Solution: Let x and y be the income of two persons.

Given that :x y = :9 7

x7& = y x y9 7 9 0& - = g (1)Also, it is given that

( ) : ( )x y2000 2000- - = 4 : 3

& x3 6000- = y4 8000-

& 3 4x y 2000- + = 0 g (2)

To solve (1) and (2), let us write the coefficients as

x y 1

9 0 7 9

4 2000 3 4

- -

- -

& x18000 0- +

= y0 14000 28 27

1-

=- +

& x = 18,000 ; 14,000.y1

180001

14000-

- = =-

- =

Hence, their monthly incomes are ` 18,000 and ` 14,000.


(iii) A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

Solution: Let x be the digit in the tenth place and y be the digit in the unit place.

Thus, the number is x y10 + .

Given that x y10 + = ( )x y7 +

& 2x y- = 0 g (1)

Again, given that y x10 + = x y10 18+ -

& 2x y- + + = 0 g (2)

Adding (1) and (2), we get y 2= and hence, .x 4=

Thus, the required number is 42.

(iv) Three chairs and two tables cost `700andfivechairsandthreetablescost

` 1100. What is the total cost of 2 chairs and 3 tables?

Solution: Let the cost of a chair be ` x and cost of a table be ` y.

Given that x y3 2+ = 700 3 2 700 0x y& + - = g (1)

and x y5 3+ = 1100 5 3 1100 0x y& + - = g (2)

To solve (1) and (2), let us write the coefficients as follows

x y 1

2 700 3 2

3 1100 5 3

-

-

& x2200 2100- +

= y3500 3300 9 10

1- +

=-

Thus, x = 100 ; 200.y1

1001

200-

- = =-

- = .

Cost of 2 chairs and 3 tables = ( ) ( )2 100 3 200 200 600 800+ = + = .

Hence, Cost of 2 chairs and 3 tables = ` 800.

(v) In a rectangle, if the length is increased and the breadth is reduced each by 2 cm then the area is reduced by 28 cm2. If the length is reduced by 1cm and the breadth increased by 2 cm , then the area increases by 33 cm2. Find the area of the rectangle.

Solution: Let the length of the rectangle be x cm and breadth of the rectangle be y cm. Then, area = xyGiven that, ( )( )x y2 2+ - = xy 28-

& xy x y2 2 4- + - = xy 28-

& x y 12- + + = 0 g (1)


Given that ( )( )x y1 2- + = xy 33+

& xy x y2 2+ - - = xy 33+

& x y2 35- - = 0 g (2)Adding (1) and (2), we get x = 23.

When x = 23, (1) & y 11= .

Thus, the length of the rectangle = 23 cm, breadth of the rectangle = 11 cm.

Hence, the area of the rectangle = 253 cm2 .

(vi) A train travelled a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken 4 hours less than the scheduled time. If the train were slower by 6 km/hr, then it would have taken 6 hours more than the scheduled time. Find the distance covered by the train.Solution: Let the speed of the train be x km/hr. and time taken be y hrs. Distance = Speed × Time = xyGiven that, ( )( )x y6 4+ - = xy

& xy x y4 6 24- + - = xy

& x y4 6 24- + - = 0

& x y2 3 12- + - = 0 g (1)

Given that ( )( )x y6 6- + = xy

& xy x y6 6 36+ - - = xy

& x y 6- - = 0 g (2)

(1) (2) 2 &#+ y = 24

When y 24= , (2) & x 30= .

Thus, the total distance covered by the train, 24 30 720xy km#= =

Exercise 3.3

1. Find the zeros of the following quadratic polynomials and verify the basic relationshipsbetweenthezerosandthecoefficients.

(i) 2 8x x2- - .

Solution: Let ( )p x = 2 8 ( 4)( 2)x x x x2- - = - +

Clearly, (4)p = 0 and ( )p 2 0- = .

Thus, the zeros of ( )p x = x x2 82- - are 4 and – 2.

Sum of zeros = 4 – 2 = 2 g (1)

Product of zeros = (4) (– 2) = – 8 g (2)


The basic relationships:

Sum of zeros = ( )

coefficient of

coefficient of

x

x12

22

- =- -

= g (3)

Product of zeros = 8coefficient of

constant term

x 18

2= - =- g (4)

From (1) and (3) and also from (2) and (4), the basic relationships are verified.

(ii) 4 4 1x x2- + .

Solution: Let ( )p x = ( )( )x x x x4 4 1 2 1 2 12- + = - -

Clearly, p x 0=^ h when ,x21

21= (twice) .

Thus, the zeros of ( )p x x x4 4 12= - + are

21 and

21 .

Sum of zeros = 21

21 1+ = g (1)

Product of zeros = 21

21

41

# = g (2)

Basic relationships: sum of zeros = ( )

coefficient of

coefficient of

x

x44

12

- =- -

= g (3)

Product of zeros = coefficient of

constant term

x 41

2= g (4)


(iii) 6 3 7x x2- - .

Solution: Let ( )p x = ( )( )x x x x6 7 3 2 3 3 12- - = - +

So, p23` j = 0 and p

31 0- =` j

Thus, the zeros of ( )p x x x6 7 32= - - are

23 and

31- .

Sum of zeros = 23

31

67- = g (1)


31

21

#- =- g (2)


coefficient of

coefficient of

x

x67

67

2- =

- -= g (3)

Product of zeros = .coefficient of

constant term

x 63

21

2= - = - g (4)


(iv) 4 8x x2+ .

Solution: Let ( )p x = 4 8 ( )x x x x4 22+ = +

Thus, (0)p = 0 and ( ) 0p 2- = .

Hence, the zeros of ( )p x x x4 82= + are 0 and – 2.


Sum of zeros = 0 – 2 = – 2 g (1)

Product of zeros = (0) (– 2) = 0 g (2)

Basic relationships: sum of zeros = 2coefficient of

coefficient of

x

x48

2- = - =- g (3)


constant term

x 40

2= = g (4)


(v) 15x2- .

Solution: Let ( )p x = ( )( )x x x15 15 152- = + - .

Thus, ( )p 15- = 0 and ( ) 0p 15 = .

Hence, the zeros of ( )p x x 152= - are 15- and 15 .

Sum of zeros = 15 15 0- + = g (1)

Product of zeros = ( )( )15 15 15- =- . g (2)

Basic relationships: sum of zeros = coefficient of

coefficient of

x

x10 0

2- = = g (3)


constant term

x 115

2= - =- g (4)


(vi) 3 5 2x x2- + .

Solution: Let ( )p x = ( )( )x x x x3 5 2 3 2 12- + = - -

Clearly, p32` j = 0 and ( ) 0p 1 = .

Thus, the zeros of ( )p x x x3 5 22= - + are

32 and 1.

Sum of zeros = 1 .32

35+ = g (1)

Product of zeros = (1)32

32=` j g (2)


coefficient of

coefficient of

x

x35

35

2- =

- -= g (3)


constant term

x 32

2= g (4)


(vii) 2 2 1x x22- + .

Solution: Let ( )p x = 2 2 1 ( )( )x x x x2 2 1 2 12- + = - -


Clearly, p2

1c m = 0 ( twice)

Thus, the zeros of ( )p x x x2 2 2 12= - + are

2

1 and 2

1 .

Sum of zeros = 2

1

2

1

2

2 2+ = = g (1)


1

2

121=c cm m g (2)


coefficient of

coefficient of

x

x22 2

22

- =- -

= g (3)


constant term

x 21

2= g (4)


(viii) 2 143x x2+ - .

Solution: Let ( )p x = ( )( )x x x x2 143 13 112+ - = + -

Clearly, ( 13)p - = 0 and ( ) 0p 11 = .

Thus, the zeros of ( )p x x x2 1432= + - are – 13 and 11.

Sum of zeros = – 13 + 11 = – 2 g (1)

Product of zeros = (– 13) (11) = – 143 g (2)

Basic relationships: sum of zeros = 2coefficient of

coefficient of

x

x12

2- = - =- g (3)


constant term

x 1143 143

2= - =- g (4)


2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

(i) 3, 1.

Solution: Let a and b be the zeros of a quadratic polynomial ( )p x .

Given that a b+ = 3 and 1ab = .

Thus, ( )p x = ( )x x2 a b ab- + + = x x3 12- + .

(ii) 2, 4.



Thus, ( )p x = ( )x x2 a b ab- + + = x x2 42- + .


(iii) 0, 4.



Thus, ( )p x = ( )x x2 a b ab- + + = (0) 4 4x x x2 2- + = + .

(iv) ,251 .


Given that a b+ = 2 and 51ab =

Thus, ( )p x = ( )x x2 a b ab- + + = x x2512

- + . (v) ,1

31 .



Thus, ( )p x = ( )x x2 a b ab- + + = x x31 12

- + .

(vi) , 421 - .


Given that a b+ = 21 and 4ab =-

Thus, ( )p x = ( )x x2 a b ab- + + = x x21 42

- - .

(vii) ,31

31- .


Given that a b+ = 31 and

31ab =-

Thus, ( )p x = ( )x x2 a b ab- + + = .x x31

312

- -

(viii) , 23 .



Thus, ( )p x = ( )x x2 a b ab- + + = x x3 22- + .

Exercise 3.4

1. Find the quotient and remainder using synthetic division.

(i) ( 3 5x x x3 2+ - + ) ' ( 1x - ).

Solution: Let ( )p x = x x x3 53 2+ - + .

The zero of the divisor ( )x 1- is 1.1 1 1 –3 5

0 1 2 –11 2 –1 4 " Remainder

Thus, Quotient x x2 12= + - , Remainder = 4.


(ii) (3 2 7 5x x x3 2- + - ) ' ( 3x + ).

Solution: Let ( )p x = x x x3 2 7 53 2- + - .

The zero of the divisor ( )x 3+ is – 3.– 3 3 – 2 7 – 5

0 – 9 33 – 1203 – 11 40 – 125 " Remainder

Thus, Quotient x x3 11 402= - + , Remainder = – 125.

(iii) (3 4 10 6x x x3 2+ - + )' ( 3 2x - ).

Solution: Let ( )p x = x x x3 4 10 63 2+ - + .

The zero of the divisor ( )x3 2- is 32 .

32 3 4 – 10 6

0 2 4 – 43 6 – 6 2 " Remainder

So, x x x3 4 10 63 2+ - + = 3 6 6 2.x x x

32 2

- + - +` ^j h

= ( ) ( )x x x3 231 3 6 6 22

- + - + .

Thus, the required quotient = ( )x x x x31 3 6 6 2 22 2

+ - = + - .

The remainder = 2.

(iv) (3 4 5x x3 2- - ) ' (3 1x + ).

Solution: Let ( )p x = x x3 4 53 2- - .

The zero of the divisor ( )x3 1+ is 31- .

31- 3 – 4 0 – 5

0 – 135

95-

3 – 535

950- " Remainder

So, ( )x x3 4 53 2- - = x x x

31 3 5

35

9502

+ - + -` `j j

= ( )x x x3 131 3 5

35

9502

+ - + -` j

Thus, the required quotient = x x x x31 3 5

35

35

952 2

- + = - +` j .

The remainder = 950- .

(v) (8 2 6 5x x x4 2- + - )' (4 1x + ).


The zero of the divisor ( 1)x4 + is 41- .


41- 8 0 – 2 6 – 5

0 – 221

83

3251-

8 – 223-

851

32211- " Remainder

So, x x8 2 64 54 2- + - = x x x x

41 8 2

23

851

322113 2

+ - - + -` `j j

= ( )x x x x4 141 8 2

23

851

322113 2

+ - - + -` j

Thus, the required quotient = x x x x x x41 8 2

23

851 2

21

83

32513 2 3 2

- - + = - - +` j


(vi) (2 7 13 63 48x x x x4 3 2- - + - )' (2 1x - ).

Solution: Let ( )p x = x x x x2 7 13 63 484 3 2- - + - .

The zero of the divisor ( 1)x2 - is 21 .

21 2 – 7 – 13 63 – 48

0 1 – 3 – 8255

2 – 6 – 16 55241- " Remainder

So, x x x x2 7 13 63 484 3 2- - + - = ( )x x x x

21 2 6 16 55

2413 2

- - - + -` j

= ( ) ( )x x x x2 121 2 6 16 55

2413 2

- - - + -

Thus, the required quotient = (2 6 16 55) 3 8x x x x x x21

2553 2 3 2

- - + = - - + .


2. If the quotient on dividing 10 35 50 29x x x x4 3 2+ + + + by 4x + is

6x ax bx3 2- + + ,thenfinda, b and also the remainder.

Solution: Let ( )p x = 10 35 50 29x x x x4 3 2+ + + + .

The zero of the divisor ( 4)x + is – 4.– 4 1 10 35 50 29

0 – 4 – 24 – 44 – 241 6 11 6 5 " Remainder

So, ( )( )x x x x x x x x10 35 50 29 4 6 11 6 54 3 2 3 2+ + + + = + + + + +

Thus, the quotient is x x x6 11 63 2+ + + .

But, x x x6 11 63 2+ + + = x ax bx 63 2

- + + . Comparing the respective coefficients, we get a 6=- and b 11= . The remainder is 5.


3. If the quotient on dividing, 8 2 6 7x x x4 2- + - by 2 1x + is 4 3x px qx

3 2+ - + ,

thenfindp , q and also the remainder.


The zero of the divisor (2 1)x + is 21- .

21- 8 0 – 2 6 – 7

0 – 4 2 0 – 38 – 4 0 6 – 10 " Remainder

So, x x x8 2 6 74 2- + - = ( )x x x

21 8 4 6 103 2

+ - + -` j

= (2 1) ( 4 6) 10x x x21 8 3 2

+ - + -

Thus, the quotient is ( )x x x x21 8 4 6 4 2 33 2 3 2

- + = - + . The remainder is –10.

But, 4 2 3x x3 2- + = 4 3x px qx3 2

+ - + . Comparing the respective coefficients, we get p 2=- and q 0= . The remainder is – 10.

Exercise 3.5 1. Factorize each of the following polynomials.

(i) 2 5 6x x x3 2- - + .

Solution: Let ( )p x = x x x2 5 63 2- - + .

Sum of the coefficients of ( )p x is 1 2 5 6 0- - + = .So, ( 1)x - is a factor of ( )p x .

1 1 – 2 – 5 60 1 – 1 – 61 – 1 – 6 0 " Remainder

The other factor is x x 62- - = 3 2 6 ( )( )x x x x x2 32

- + - = + - .Thus, 2 5 6x x x3 2

- - + = ( )( )( )x x x1 2 3- + - .

(ii) 4 7 3x x3- + .

Solution: Let ( )p x = x x4 7 33- + .

Sum of the coefficients of ( )p x = 4 7 3 0- + = .

So, ( 1)x - is a factor of ( )p x .1 4 0 – 7 3

0 4 4 – 34 4 – 3 0 " Remainder

The other factor is x x4 4 32+ - = ( )( )x x x x x4 6 2 3 2 3 2 12

+ - - = + - .Thus, 4 7 3x x3

- + = ( )( )( )x x x1 2 3 2 1- + - .


(iii) 23 142 120x x x3 2- + - .

Solution: Let ( )p x = x x x23 142 1203 2- + - .

Sum of the coefficients of ( )p x = 1 23 142 120 0- + - = .

So, ( 1)x - is a factor of ( )p x .1 1 – 23 142 – 120

0 1 – 22 1201 – 22 120 0 " Remainder

The other factor is x x22 1202- + = ( )( )x x x x x12 10 120 12 102

- - + = - - .

Thus, 23 142 120x x x3 2- + - = ( )( )( )x x x1 12 10- - - .

(iv) 4 5 7 6x x x3 2- + - .


Sum of the coefficients of ( )p x = 4 5 7 6 0- + - = .

Thus, ( 1)x - is a factor of ( )p x .1 4 – 5 7 – 6

0 4 – 1 64 – 1 6 0 " Remainder

Thus, the other factor is x x4 62- + .

Hence, x x x4 5 7 63 2- + - = ( )( )x x x1 4 62

- - + .

(v) 7 6x x3- + .

Solution: Let ( )p x = x x7 63- + .

Sum of the coefficients of ( )p x = 1 7 6 0- + = .

So, ( 1)x - is a factor of ( )p x .1 1 0 – 7 6

0 1 1 – 61 1 – 6 0 " Remainder

The other factor is x x 62+ - = ( )( )x x x x x3 2 6 3 22

+ - - = + - .Thus, 7 6x x3

- + = ( )( )( )x x x1 2 3- - + .

(vi) 13 32 20x x x3 2+ + + .

Solution: Let ( )p x = x x x13 32 203 2+ + + .

Sum of the coefficients of odd powers of 1 32 33.x is + =

Sum of the coefficients of even powers of 13 20 33x is + =

Since they are equal, ( )x 1+ is a factor of ( )p x .– 1 1 13 32 20

0 – 1 – 12 – 201 12 20 0 " Remainder


The other factor is x x12 202+ + = x x x10 2 202

+ + + .

= ( ) ( ) ( )( )x x x x x10 2 10 10 2+ + + = + + .

Thus, 13 32 20x x x3 2+ + + = ( )( )( )x x x1 10 2+ + + .

(vii) 2 9 7 6x x x3 2- + + .

Solution: Let ( )p x = x x x2 9 7 63 2- + + .

Since (1) 0, ( 1) 0,p p! !- neither ( 1) ( 1)norx x- + is a factor of ( )p x .

So, we have to search for other values of x by trial and error method.

When 2, (2) 0.x p= = Thus, ( )x 2- is a factor of ( )p x .2 2 – 9 7 6

0 4 – 10 – 62 – 5 – 3 0 " Remainder

The other factor is x x2 5 32- - = ( )( )x x3 2 1- + .

Hence, 2 9 7 6x x x3 2- + - = ( )( )( )x x x2 3 2 1- - + .

(viii) 5 4x x3- + .

Solution: Let ( )p x = x x5 43- + .

Sum of the coefficients of ( )p x is 1 5 4 0- + = .

So, ( )x 1- is a factor of ( )p x .1 1 0 – 5 4

0 1 1 – 41 1 – 4 0 " Remainder

The other factor is x x 42+ - .

Thus, 5 4x x3- + = ( )( )x x x1 42

- + - .

(ix) 10 10x x x3 2- - + .

Solution: Let ( )p x = x x x10 103 2- - + .

Sum of the coefficients of ( )p x is 1 10 1 10 0- - + = .

So, ( )x 1- is a factor of ( )p x .1 1 – 10 – 1 10

0 1 – 9 – 101 – 9 – 10 0 " Remainder

The other factor is 9 10 ( )( )x x x x10 12- - = - +

Thus, 10 10x x x3 2- - + = ( )( )( )x x x1 10 1- - + .

(x) 2 11 7 6x x x3 2+ - - .

Solution: Let ( )p x = x x x2 11 7 63 2+ - - .

Sum of the coefficients of ( )p x is 2 11 7 6 0+ - - = .


So, ( )x 1- is a factor of ( )p x .1 2 11 – 7 – 6

0 2 13 62 13 6 0 " Remainder

The other factor is ( )( )x x x x2 13 6 6 2 12+ + = + + .

Thus, 2 11 7 6x x x3 2+ - - = ( 1)( )( )x x x6 2 1- + + .

(xi) 14x x x3 2+ + - .

Solution: Let ( )p x = x x x 143 2+ + - .

Clearly, (1) 0. ( 1)p xSo,! - is not a factor.

Also, ( )p 1 0!- . Thus, ( )x 1+ is not a factor.

However, ( )p 2 0= . Thus, ( )x 2- is a factor of ( )p x .2 1 1 1 – 14

0 2 6 141 3 7 0 " Remainder

The other factor is x x3 72+ + .

Hence, 14x x x3 2+ + - = ( )( )x x x2 3 72

- + + .

(xii) 5 2 24x x x3 2- - + .


Sum of the coefficients of ( )p x is 1 5 2 24 0!- - + .Thus, ( )x 1- is not a factor of ( )p x .Also, ( )p 1- = 1 5 2 24 0!- - + + . Thus, 1x + is not a factor.By trial and error, we see that ( ) .p 2 0- = Thus, ( )x 2+ is a factor of ( )p x .

– 2 1 – 5 – 2 240 – 2 14 – 241 – 7 12 0 " Remainder

The other factor is ( )( )x x x x7 12 3 42- + = - - .

Thus, 5 2 24x x x3 2- - + = ( )( )( )x x x2 3 4+ - - .

Exercise 3.6

1. Find the greatest common divisor of

(i) 7x yz2 4 , 21x y z

2 5 3 .

Solution: x yz7 2 4 = x yz7 2 4#

x y z21 2 5 3 = 7 3 x y z2 5 3# #

GCD = x yz7 2 3 .


(ii) x y2 , x y

3 , x y2 2 .

Solution: GCD = x y2 .

(iii) 25bc d4 3 , 35b c

2 5 , 45c d3 .

Solution: bc d25 4 3 = bc d52 4 3#

b c35 2 5 = 5 7b c2 5#

c d45 3 = c d5 32 3#

GCD = c5 3 .

(iv) 35x y z5 3 4 , 49x yz

2 3 , 14xy z2 2 .

Solution: x y z35 5 3 4 = x y z7 5 5 3 4# #

x yz49 2 3 = 7 x yz7 2 3# #

xy z14 2 2 = xy z7 2 2 2# #

GCD = xyz7 2 .

2. Find the GCD of the following

(i) c d2 2- , c c d-^ h.

Solution: c d2 2- = ( )( )c d c d+ -

( )c c d- = ( )c c d-

GCD = ( )c d- .

(ii) 27x a x4 3- , x a3 2-^ h .

Solution: x a x274 3- = ( ) ( )( )x x a x x a x ax a3 3 3 93 3 3 2 2

- = - + +

( )x a3 2- = ( )( )x a x a3 3- -

GCD = ( )x a3- .

(iii) 3 18m m2- - , 5 6m m

2+ + .

Solution: m m3 182- - = ( )( )m m6 3- +

m m5 62+ + = ( )( )m m2 3+ +

GCD = ( )m 3+ .

(iv) 14 33x x2+ + , 10 11x x x

3 2+ - .

Solution: x x14 332+ + = ( )( )x x11 3+ +

10 11x x x3 2+ - = ( )( )x x x11 1+ -

GCD = ( )x 11+ .

(v) 3 2x xy y2 2+ + , 5 6x xy y

2 2+ + .

Solution: x xy y3 22 2+ + = ( )( )x y x y2+ +

x xy y5 62 2+ + = ( )( )x y x y3 2+ +

GCD = ( )x y2+ .


(vi) 2 1x x2- - , 4 8 3x x

2+ + .

Solution: x x2 12- - = ( )( )x x2 1 1+ -

x x4 8 32+ + = ( )( )x x2 1 2 3+ +

GCD = ( )x2 1+ .

(vii) 2x x2- - , 6x x

2+ - , 3 13 14x x

2- + .

Solution: x x 22- - = ( )( )x x2 1- +

x x 62+ - = ( )( )x x2 3- +

x x3 13 142- + = ( )( )x x2 3 7- -

GCD = ( )x 2- .

(viii) 1x x x3 2- + - , 1x

4- .

Solution: x x x 13 2- + - = ( )( )x x1 12

+ -

x 14- = ( )( )( )x x x1 1 12

+ - +

GCD = ( )( )x x1 12+ - .

(ix) 24 x x x6 24 3 2- -^ h, 20 x x x2 3

6 5 4+ +^ h.

Solution: ( )x x x24 6 24 3 2- - = 4 6 ( 1)(3 2) .x x x22

# # + -

( )x x x20 2 36 5 4+ + = ( )( )x x x4 5 2 1 14

# # + +

GCD = ( )x x4 2 12+ .

(x) a a1 35 2- +^ ^h h , a a a2 1 32 3 4- - +^ ^ ^h h h .

Solution: ( ) ( )a a1 35 2- + = ( 1) ( 3)a a a12 3 2

- - +^ h

( ) ( ) ( )a a a1 3 23 4 2- + - = ( 1) ( 3) ( 2)a a a a3 23 2 2

- + + -^ h

GCD = ( ) ( )a a1 33 2- +

3. Find the GCD of the following pairs of polynomials using division algorithm.

(i) 9 23 15x x x3 2- + - , 4 16 12x x

2- + .

Solution: Let ( )f x = x x x9 23 153 2- + - .

and ( )g x = ( )x x x x4 16 12 4 4 32 2- + = - + .

So, the divisor is 4 3.x x2- +

x 5-

x x4 32- + x x x9 23 153 2

- + -

x x x4 33 2- +

x x5 20 152- + -

x x5 20 152- + -

0Remainder is zero.Thus, GCD ( ( ), ( ))f x g x = x x4 32

- + .


(ii) 3 18 33 18x x x3 2+ + + , 3 13 10x x

2+ + .

Solution: Let ( )f x = x x x3 18 33 183 2+ + + .

and ( )g x = x x3 13 102+ + .

So, the divisor is ( )g x = x x3 13 102+ + .

x35+

x x3 13 102+ + x x x3 18 33 183 2

+ + +

x x x3 13 103 2+ +

(–) (–) (–)

5 23 18x x2+ +

5x x365

3502

+ +

x34

34+

( 1)x34& + ! 0. Note that

34 is not a divisor of ( ) .g x

x3 10+

x 1+ x x3 13 102+ +

x x3 32+

(–) (–)

x10 10+

x10 10+ 0

Remainder is zero.Thus, GCD ( ( ), ( ))f x g x = x 1+ .

(iii) 2 2 2 2x x x3 2+ + + , 6 12 6 12x x x

3 2+ + + .

Solution: Let ( )f x = ( )x x x2 13 2+ + + .

and ( )g x = ( )x x x6 2 23 2+ + + .

Note that 2 is a common factor of ( ) ( ) .f x g xand

Here, the divisor is x x x 13 2+ + + .

1

x x x 13 2+ + + x x x2 23 2

+ + +

x x x 13 2+ + +

1 0x2 !+


x 1+

1x2 + x x x 13 2+ + +

x3 x+

x 12+

x 12+

0

Remainder is zero.

Thus, GCD ( ( ), ( ))f x g x = ( 1)x2 2+ .

(iv) 3 4 12x x x3 2- + - , 4 4x x x x

4 3 2+ + + .

Solution: Let ( )f x = x x x3 4 123 2- + -

and ( )g x = ( )x x x x4 43 2+ + +

Thus, the divisor is x x x4 43 2+ + +

1

x x x4 43 2+ + + x x x3 4 123 2

- + -

x x x4 43 2+ + +

x4 2- 16-

( )x4 42& - + ! 0. Note that 4- is not a factor of ( ) .f x

x 1+

x 42+ x x x4 43 2

+ + +

x3 x4+(–) (–)

x2 + 4

x2 + 4 (–) (–) 0

Remainder is zero.

Thus, GCD ( ( ), ( ) )f x g x = x 42+ .

Exercise 3.7

Find the LCM of the following.

1. x y3 2 , xyz .

Solution: , .x y xyz3 2 Hence, LCM x y z3 2= .


2. 3x yz2 , 4x y

3 3 .

Solution: 3 ; 4 .x yz x y2 3 3

Thus, LCM = 3 4 12x y z x y z3 3 3 3# = .

3. a bc2 , b ca

2 , c ab2 .

Solution: a bc2 = a bc2 ; b ca ab c2 2= ; c ab2 = abc2

Thus, LCM = a b c2 2 2 .

4. 66a b c4 2 3 , 44a b c

3 4 2 , 24a b c2 3 4 .

Solution: a b c66 4 2 3 = a b c11 2 3 4 2 3# # #

a b c44 3 4 2 = a b c11 2 2 3 4 2# # #

a b c24 2 3 4 = a b c2 3 2 2 2 3 4# # # #

Thus, LCM = a b c11 2 2 3 2 4 4 4# # # # # = a b c264 4 4 4 .

5. am 1+ , am 2+ , am 3+ .

Solution: am 1+ = a am# ; a a am m2 2

#=+

am 3+ = a am 3#

Thus, LCM = a a am m3 3# = + .

6. x y xy2 2

+ , x xy2+ .

Solution: x y xy2 2+ = ( )xy x y+

x xy2+ = ( )x x y+

Thus, LCM = ( )xy x y+ .

7. 3 a 1-^ h, 2 a 1 2-^ h , a 12-^ h.

Solution: a3 1-^ h, a2 1 2-^ h , a 12-^ h.

Now, a 12- = ( )( )a a1 1- + .

Thus, LCM = ( 1) ( 1) .a a6 2- +

8. 2 18x y2 2- , 5 15x y xy

2 2+ , 27x y

3 3+ .

Solution: x y2 182 2- = ( )( )x y x y2 3 3+ -

x y xy5 152 2+ = ( )xy x y5 3+

x y273 3+ = ( )( )x y x xy y3 3 92 2

+ - +

LCM = 2 5 ( 3 )( 3 )( 3 9 )xy x y x y x xy y2 2# # + - - +

Thus, LCM = ( )( )( )xy x y x y x xy y10 3 3 3 92 2+ - - + .

9. x x4 32 3+ -^ ^h h , x x x1 4 3 2- + -^ ^ ^h h h .Solution: ( ) ( )x x4 32 3

+ - ; ( )( )( )x x x1 4 3 2- + -

Thus, LCM = ( ) ( ) ( )x x x4 3 12 3+ - - .


10. 10 x xy y9 62 2+ +^ h, 12 x xy y3 5 2

2 2- -^ h, 14 x x6 2

4 3+^ h.

Solution: ( )x xy y10 9 62 2+ + = ( )x y2 5 3 2

# +

( )x xy y12 3 5 22 2- - = ( )( )x y x y2 3 3 22

# + -

( )x x14 6 24 3+ = 2 7 (3 1)x x2 3

# # +

LCM = 2 7 5 3 (3 ) (3 1)( 2 )x x y x x y2 3 2# # # + + -

Thus, LCM = ( ) ( )( )x x y x y x420 3 2 3 13 2+ - + .

Exercise 3.8

1. Find the LCM of each pair of the following polynomials.

(i) 5 6x x2- + , 4 12x x

2+ - whose GCD is 2x - .

Solution: Let ( )f x = , ( )x x g x x x5 6 4 122 2- + = + -

and GCD = x 2-

We know that LCM GCD# = ( ) ( )f x g x#

Thus, LCM = ( ) ( ) ( )( )G.C.D.

f x g xx

x x x x2

5 6 4 122 2#=

-- + + -

= ( )( )( )( )x

x x x x2

3 2 6 2-

- - + -

Hence, LCM = ( )( )( )x x x3 2 6- - + .

(ii) 3 6 5 3x x x x4 3 2+ + + + , 2 2x x x

4 2+ + + whose GCD is 1x x

2+ + .

Solution: Let ( )f x = x x x x3 6 5 34 3 2+ + + +

( )g x = x x x2 24 2+ + + and GCD is x x 12

+ +

Thus, LCM = ( ) ( )G.C.D.

f x g x#

Now, GCD divides both ( )f x and ( )g x . Let us divide ( )f x by GCD.

x x2 32+ +

x x 12+ + x x x x3 6 5 34 3 2

+ + + +

x x x4 3 2+ +

x x x2 5 53 2+ +

x x x2 2 23 2+ +

x x3 3 32+ +

x x3 3 32+ +

0

LCM = ( )

( )( )( )

x x

x x x x x x x

1

1 2 3 2 22

2 2 4 2

+ +

+ + + + + + +

Thus, LCM = ( )( )x x x x x2 3 2 22 4 2+ + + + + .


(iii) 2 15 2 35x x x3 2+ + - , 8 4 21x x x

3 2+ + - whose GCD is 7x + .

Solution: Let ( )f x = x x x2 15 2 353 2+ + -

( )g x = x x x8 4 213 2+ + - and GCD. is x 7+

Thus, LCM = ( ) ( )GCD

f x g x#

Now, GCD divides both ( )f x and ( )g x . Let us divide ( )f x by GCD.

x x2 52+ -

x 7+ x x x2 15 2 353 2+ + -

x x2 143 2+

x x22+

x x72+

x5 35- -

x5 35- -

0

LCM = ( )( )( )x

x x x x x x7

7 2 5 8 4 212 3 2

++ + - + + -

Hence, LCM = ( )( )x x x x x2 5 8 4 212 3 2+ - + + - .

(iv) 2 3 9 5x x x3 2- - + , 2 10 11 8x x x x

4 3 2- - - + whose GCD is 2 1x - .

Solution: Let ( )f x = x x x2 3 9 53 2- - + .

( )g x = x x x x2 10 11 84 3 2- - - + and GCD = x2 1-

Thus, LCM = ( ) ( )GCD

f x g x#

Now, GCD divides both ( )f x and ( )g x . Let us divide ( )g x by GCD.

x x5 83- -

x2 1- x x x x2 10 11 84 3 2- - - +

x x2 4 3-

x x10 112- -

x x10 52- +

x16 8- +

x16 8- +

0

LCM = ( )

( )( )( )x

x x x x x x2 1

2 1 5 8 2 3 9 53 3 2

-- - - - - +

Thus, LCM = ( )( )x x x x x5 8 2 3 9 53 3 2- - - - + .


2. Find the other polynomial q x^ h of each of the following, given that LCM and GCD and one polynomial p x^ h respectively.

(i) x x1 22 2+ +^ ^h h , x x1 2+ +^ ^h h, x x1 22+ +^ ^h h.

Solution: LCM = ( 1) ( 2)x x2 2+ + ; GCD = ( )( )x x1 2+ +

and ( )p x = ( ) ( )x x1 22+ +

We know that LCM GCD# = ( ) ( )p x q x# .

& ( )q x = ( ) ( ) ( )

( ) ( ) ( )( )LCM GCDp x x x

x x x x

1 2

1 2 1 22

2 2# =

+ +

+ + + +

Thus, ( )q x = ( )( )x x1 2 2+ + .

(ii) x x4 5 3 73 3+ -^ ^h h , x x4 5 3 7 2+ -^ ^h h , x x4 5 3 73 2+ -^ ^h h .

Solution: LCM = ( ) ( )x x4 5 3 73 3+ - ; GCD = ( )( )x x4 5 3 7 2

+ -

and ( )p x = ( ) ( )x x4 5 3 73 2+ -


& ( )q x = ( ) ( ) ( )

( ) ( ) ( )( )LCM GCDp x x x

x x x x

4 5 3 7

4 5 3 7 4 5 3 73 2

3 3 2# =

+ -

+ - + -

Thus, ( )q x = ( ) ( )x x3 7 4 53- + .

(iii) x y x x y y4 4 4 2 2 4- + +^ ^h h, x y

2 2- , x y

4 4- .

Solution: LCM = ( )( )x y x x y y4 4 4 2 2 4- + + ; GCD = ( )x y2 2

-

and ( )p x = x y4 4-


& ( )q x = ( )

( )( )( )LCM GCDp x x y

x y x x y y x y4 4

4 4 4 2 2 4 2 2# =

-

- + + -

Thus, ( )q x = ( )( )x x y y x y4 2 2 4 2 2+ + - .

(iv) x x x4 5 13- +^ ^h h, x x5

2+^ h, x x x5 9 2

3 2- -^ h.

Solution: LCM = ( 4 )(5 1) ( )( )( )x x x x x x x2 2 5 13- + = + - +

GCD = ( ) ( )x x x x5 5 12+ = +

and ( )p x = ( )( )x x x x x x5 9 2 5 1 23 2- - = + -


& ( )q x = = ( ) ( )( )

( )( )( )( )( )LCM GCDp x x x x

x x x x x x5 1 2

2 2 5 1 5 1# =+ -

+ - + +

Thus, ( )q x = ( )( )x x x2 5 1+ + .

(v) x x x x1 2 3 32

- - - +^ ^ ^h h h, x 1-^ h, x x x4 6 33 2- + -^ h.

Solution: LCM = ( )( )( )x x x x1 2 3 32- - - + ; GCD = x 1-

and ( )p x = 4 6 3 ( )( )x x x x x x1 3 33 2 2- + - = - - +



& ( )q x = ( ) ( )( )

( )( )( )( )LCM GCDp x x x x

x x x x x

1 3 3

1 2 3 3 12

2# =

- - +

- - - + -

Thus, ( )q x = ( )( )x x1 2- - .

(vi) 2 x x1 42

+ -^ ^h h, x 1+^ h, x x1 2+ -^ ^h h.

Solution: LCM = ( )( ) ( )( )( )x x x x x2 1 4 2 1 2 22+ - = + + -

GCD = x 1+ and ( )p x = ( )( )x x1 2+ -


& ( )q x = ( ) ( )( )

( )( )( )( )LCM GCDp x x x

x x x x1 2

2 1 2 2 1# =+ -

+ + - +

Thus, ( )q x = ( )( )x x2 1 2+ + .

Exercise 3.9

Simplify the following into their lowest forms.

(i) x x

x x

3 12

6 92

2

-

+ .

Solution: x x

x x

3 12

6 92

2

-

+ = ( )( )

x xx x

xx

3 43 2 3

42 3

-+

=-+ .

(ii) x

x

1

14

2

-

+ .

Solution: x

x

1

14

2

-

+ = ( )( )x x

x

x1 1

1

1

12 2

2

2+ -

+ =-

.

(iii) x x

x

1

12

3

+ +

- .

Solution: x x

x

1

12

3

+ +

- = ( )( )( )

x x

x x xx

1

1 11

2

2

+ +

- + += - .

(iv) x

x

9

272

3

-

- .

Solution: x

x

9

272

3

-

- = ( )( )

( )( )x x

x x xx

x x3 3

3 3 93

3 92 2

+ -- + +

=+

+ + .

(v) x x

x x

1

12

4 2

+ +

+ + . (Hint: 1x x4 2+ + = x x1

2 2 2+ -^ h )

Solution: x x

x x

1

12

4 2

+ +

+ + = ( )( )

x x

x x x xx x

1

1 11

2

2 22

+ +

+ + - += - + .

(vi) x x

x

4 16

84 2

3

+ +

+ .

Solution: x x

x

4 16

84 2

3

+ +

+ = ( ) ( ) ( )( )

( )( )

x x

x

x x x x

x x x

4 2

2

2 4 2 4

2 2 42 2 2

3 3

2 2

2

+ -

+ =+ + - +

+ - +

= x x

x

2 4

22+ +

+ .


(vii) x x

x x

2 5 3

2 32

2

+ +

+ - .

Solution: x x

x x

2 5 3

2 32

2

+ +

+ - = ( )( )( )( )

x xx x

xx

2 3 12 3 1

11

+ ++ -

=+- .

(viii) x x

x

9 2 6

2 1622

4

+ -

-^ ^h h

.

Solution: ( )( )x x

x

9 2 6

2 1622

4

+ -

- = ( ) ( )

(( ) )

x x

x

9 2 3

2 92

2 2 2

+ -

-

= ( )( )

( )( )( )( )

x x

x x xx

2 9 3

2 9 3 33

2

2

+ -

+ + -= + .

(ix) x x x

x x x

4 2 3

3 5 42

2

- - -

- - +

^ ^

^ ^

h h

h h.

Solution: ( )( )

( )( )

x x x

x x x

4 2 3

3 5 42

2

- - -

- - + = ( )( )( )( )( )( )x x xx x x

xx

4 3 13 4 1

11

- - +- - -

=+- .

(x) x x x

x x x

10 13 40

8 5 502

2

+ - +

- + -

^ ^

^ ^

h h

h h.

Solution: ( )( )

( )( )

x x x

x x x

10 13 40

8 5 502

2

+ - +

- + - = ( )( )( )( )( )( )x x xx x x

10 8 58 10 5

1+ - -- + -

= .

(xi) x x

x x

8 6 5

4 9 52

2

+ -

+ + .

Solution: x x

x x

8 6 5

4 9 52

2

+ -

+ + = ( )( )( )( )

x xx x

xx

4 5 2 14 5 1

2 11

+ -+ +

=-+ .

(xii) x x x

x x x x

7 3 2

1 2 9 142

2

- - +

- - - +

^ ^

^ ^ ^

h h

h h h.

Solution: ( )( )

( )( )( )

x x x

x x x x

7 3 2

1 2 9 142

2

- - +

- - - + = ( )( )( )

( )( )( )( )( )

x x xx x x x

x7 2 1

1 2 7 22

- - -- - - -

= - .

Exercise 3.10 1. Multiply the following and write your answer in lowest terms.

(i) x

x xxx

22

23 6

2

#+-

-+ .

Solution: x

x xxx

22

23 6

2

#+-

-+ = ( ) ( )

3x

x xxx

x22

23 2

#+-

-+

= .

(ii) x

x

x x

x x

4

81

5 36

6 82

2

2

2

#-

-

- -

+ +

Solution: x

x

x x

x x

4

81

5 36

6 82

2

2

2

#-

-

- -

+ + = ( )( )( )( )

( )( )( )( )

x xx x

x xx x

2 29 9

9 44 2

#+ -+ -

- ++ +

= xx

29

-+ .


(iii) x x

x x

x

x x

20

3 10

8

2 42

2

3

2

#- -

- -

+

- +

Solution: x x

x x

x

x x

20

3 10

8

2 42

2

3

2

#- -

- -

+

- + = ( )( )( )( )

( )( )x xx x

x x x

x x5 45 2

2 2 4

2 42

2

#- +- +

+ - +

- +

= x 41+

.

(iv) x x

xxx

x xx x

3 216

644

2 84 16

2

2

3

2

2

2

# #- +-

+-

- -- +

Solution: 4 16x x

xxx

x xx x

3 216

644

2 8

2

2

2

3

2

2# #- +-

+-

- -- +

= ( )( )( )( )

( )( )

( )( )( )( )x x

x x

x x x

x xx xx x

2 14 4

4 4 16

2 24 24 16

2

2

# #- -+ -

+ - +

+ -- +- +

= x 11-

.

(v) x x

x x

x x

x x

2

3 2 1

3 5 2

2 3 22

2

2

2

#- -

+ -

+ -

- -

Solution: x x

x x

x x

x x

2

3 2 1

3 5 2

2 3 22

2

2

2

#- -

+ -

+ -

- - = ( )( )( )( )

( )( )( )( )

x xx x

x xx x

2 13 1 1

3 1 22 1 2

#- +- +

- ++ -

= xx

22 1++ .

(vi) x x

x

x x

x x

x x

x

2 4

2 1

2 5 3

8

2

32 2

4

2# #+ +

-

+ -

-

-

+

Solution: x x

x

x x

x x

x x

x

2 4

2 1

2 5 3

8

2

32 2

4

2# #+ +

-

+ -

-

-

+

= ( )( )

( )( )( )x x

xx x

x x x xx x

x

2 4

2 12 1 32 2 4

23

2

2

# #+ +

-- +

- + +-+ = 1.

2. Divide the following and write your answer in lowest terms.

(i) x

x

x

x1 1

2

2

'+ -

.

Solution: x

x

x

x1 1

2

2

'+ -

= ( ).

xx

x

x x

xx

1

1 1 12

#+

+ -= -^ h

(ii) x

xxx

49

3676

2

2

'-

-++

Solution: ( )( )( )( )( )( )

.x

xxx

x x xx x x

xx

49

3667

7 7 66 6 7

76

2

2

#-

-++ =

+ - ++ - +

=--

(iii) x

x x

x x

x x

25

4 5

7 10

3 102

2

2

2

'-

- -

+ +

- -

Solution:

( )

( )

( )

( )( 5)( )( )( )

( 5)( )( )( )

( )( )

.x

x x

x x

x xx xx x

x xx x

xx

25

4 5

3 10

7 105

5 12

5 251

2

2

2

2

# #-

- -

- -

+ +=

+ -- +

- ++ +

=-+


(iv) x x

x x

x x

x x

4 77

11 28

2 15

7 122

2

2

2

'- -

+ +

- -

+ +

Solution: x x

x x

x x

x x

4 77

11 28

2 15

7 122

2

2

2

'- -

+ +

- -

+ +

= ( )( )( )( )

( )( )( )( )

x xx x

x xx x

xx

11 77 4

3 45 3

115

#- ++ +

+ +- +

=-- .

(v) x x

x x

x x

x x

3 10

2 13 15

4 4

2 62

2

2

2

'+ -

+ +

- +

- -

Solution: x x

x x

x x

x x

3 10

2 13 15

4 4

2 62

2

2

2

'+ -

+ +

- +

- -

= ( )( )( )( )

( )( )( )( )

x xx x

x xx x

5 22 3 5

2 3 22 2

1#+ -+ +

+ -- -

= .

(vi) x

x x

x x

x

9 16

3 4

3 2 1

4 42

2

2

2

'-

- -

- -

-

Solution: x

x x

x x

x

9 16

3 4

3 2 1

4 42

2

2

2

'-

- -

- -

-

= ( )( )( )( )

( )( )( )( )

( )x xx x

x xx x

xx

3 4 3 43 4 1

4 1 13 1 1

4 3 43 1

#+ -- +

+ -+ -

=++ .

(vii) x x

x x

x x

x x

2 9 9

2 5 3

2 3

2 12

2

2

2

'+ +

+ -

+ -

+ -

Solution: x x

x x

x x

x x

2 9 9

2 5 3

2 3

2 12

2

2

2

'+ +

+ -

+ -

+ - = ( )( )( )( )

( )( )( )( )

x xx x

x xx x

2 3 32 1 3

2 1 12 3 1

#+ +- +

- ++ -

= xx

11

+- .

Exercise 3.11

1. Simplify the following as a quotient of two polynomials in the simplest form.

(i) x

xx2 2

83

-+

-.

Solution: x

xx2 2

83

-+

- =

xx

x xx

2 28

223 3 3

--

-=

--

= ( )( )x

x x xx x

22 2 4

2 42

2

-- + +

= + + .

(ii) x x

x

x x

x

3 2

2

2 3

32 2+ +

+ +- -

- .

Solution: x x

x

x x

x

3 2

2

2 3

32 2+ +

+ +- -

- = ( )( ) ( )( )x x

xx x

x2 1

23 1

3+ +

+ +- +

-

= x x x11

11

12

++

+=

+.


(iii) x

x x

x x

x x

9

6

12

2 242

2

2

2

-

- - +- -

+ - .

Solution: x

x x

x x

x x

9

6

12

2 242

2

2

2

-

- - +- -

+ -

= ( )( )( )( )

( )( )( )( )

x xx x

x xx x

xx

xx

3 33 2

4 36 4

32

36

+ -- +

+- ++ -

=++ +

++

= ( )x

x xxx

xx

32 6

32 8

32 4

++ + + =

++ =

++ .

(iv) x x

x

x x

x

7 10

2

2 15

32 2- +

- +- -

+ .

Solution: x x

x

x x

x

7 10

2

2 15

32 2- +

- +- -

+ = ( )( ) ( )( )x x

xx x

x5 2

25 3

3- -

- +- +

+

= x x x51

51

52

-+

-=

-.

(v) x x

x x

x x

x x

3 2

2 5 3

2 3 2

2 7 42

2

2

2

- +

- + -- -

- - .

Solution: x x

x x

x x

x x

3 2

2 5 3

2 3 2

2 7 42

2

2

2

- +

- + -- -

- -

= ( )( )( )( )

( )( )( )( )

x xx x

x xx x

xx

xx

2 12 3 1

2 1 22 1 4

22 3

24

- -- -

-+ -+ -

=-- -

--

= x

x xxx

22 3 4

21

-- - + =

-+ .

(vi) x x

x

x x

x x

6 8

4

20

11 302

2

2

2

+ +

- -- -

- + .

Solution: x x

x

x x

x x

6 8

4

20

11 302

2

2

2

+ +

- -- -

- +

= ( )( )( )( )

( )( )( )( )

x xx x

x xx x

xx

xx

2 42 2

5 46 5

42

46

+ ++ -

-- +- -

=+- -

+-

= x

x xx4

2 64

4+

- - + =+

.

(vii) xx

x

xxx

12 5

1

11

3 22

2

++ +

-

+ ---` j> H .

Solution: xx

x

xxx

12 5

1

11

3 22

2

++ +

-

+ ---` j= G

= ( )( )

( )xx

x xx

xx

12 5

1 11

13 22

++ +

+ -+ -

--

= ( )( )

( )( ) ( )( )x x

x x x x x1 1

2 5 1 1 3 2 12

+ -+ - + + - - +


= ( )( )x x

x x x x x x x1 1

2 2 5 5 1 3 3 2 22 2 2

+ -- + - + + - - + +

= ( )( ) ( )( )

( )x x

xx x

xx1 1

2 21 1

2 11

2+ -

- =+ -

-=

+.

(viii) x x x x x x3 2

15 61

4 32

2 2 2+ +

++ +

-+ +

.

Solution: x x x x x x3 2

15 61

4 32

2 2 2+ +

++ +

-+ +

= ( )( ) ( )( ) ( )( )x x x x x x1 2

12 31

3 12

+ ++

+ +-

+ +

= ( )( )( )

( )x x x

x x x1 2 3

3 1 2 2+ + +

+ + + - +

= ( )( )( )x x x

x x1 2 3

2 4 2 4 0+ + +

+ - - = .

2. Which rational expression should be added to x

x

2

12

3

+

- to get x

x x

2

3 2 42

3 2

+

+ + ?

Solution: Let ( )p x be the required rational expression.

Then, 1 ( )x

x p x2

3

2+

- + = x

x x

2

3 2 42

3 2

+

+ +

& ( )p x = x

x x

x

x

2

3 2 4

2

12

3 2

2

3

+

+ + -+

-

= x

x x x

2

3 2 4 12

3 2 3

+

+ + - +

Hence, ( )p x = x

x x

2

2 2 52

3 2

+

+ + .

3. Which rational expression should be subtracted from x

x x2 1

4 7 53 2

-- + to get

2 5 1x x2- + ?

Solution: Let ( )p x be the required rational expression.

Then, ( )x

x x p x2 1

4 7 53 2

-- + - = x x2 5 12

- +

& ( )p x = ( )

( )x

x x x x2 1

4 7 5 2 5 13 2

2

-- + - - +

= ( )x

x x x x x2 1

4 7 5 4 12 7 13 2 3 2

-- + - + - +

Hence, ( )p x = x

x x2 1

5 7 62

-- + .

4. If P = x yx+

, Q = x yy

+,thenfind

P Q P Q

Q1 22 2-

--

.

Solution: P Q P Q

Q1 22 2-

--

= ( )( )P Q P Q P Q

Q1 2

--

+ -


= ( )( ) ( )( )P Q P QP Q Q

P Q P QP Q2

+ -+ -

=+ -

-

= P Q

x yx

x yy

x yx y

1 1 1+

=

++

+

=

++

Thus, P Q P Q

Q1 22 2-

--

= 1.

Exercise 3.12

1. Find the square root of the following:

(i) 196a b c6 8 10 .

Solution: a b c196 6 8 10 = a b c a b c14 142 6 8 10 3 4 5=

(ii) 289 a b b c4 6- -^ ^h h .

Solution: 289 a b b c4 6- -^ ^h h = ( ) ( )a b b c172 4 6- -

= 17 ( ) ( )a b b c2 3- - .

(iii) 44x x11 2+ -^ h .

Solution: 11 44x x2+ -^ h = x x x x x22 121 44 22 1212 2+ + - = - +

= ( ) ( )x x11 112- = - .

(iv) 4x y xy2- +^ h .

Solution: 4x y xy2- +^ h = x xy y xy x xy y2 4 22 2 2 2- + + = + +

= ( ) ( )x y x y2+ = + .

(v) 121x y8 6 ' 81x y

4 8 .

Solution: 121 81x y x y8 6 4 8

' = x y

x y

y

xyx

81

121

9

11911

4 8

8 6

2 2

2 4 2

= = .

(vi) x y a b b c

a b x y b c

25

644 6 10

4 8 6

+ - +

+ - -

^ ^ ^

^ ^ ^

h h h

h h h .

Solution: 25

64

x y a b b c

a b x y b c4 6 10

4 8 6

+ - +

+ - -

^ ^ ^

^ ^ ^

h h h

h h h

= ( ) ( ) ( )

( ) ( ) ( )

x y a b b c

a b x y b c

5

82 4 6 10

2 4 8 6

+ - +

+ - - = ( ) ( ) ( )

( ) ( ) ( )

x y a b b c

a b x y b c58

2 3 5

2 4 3

+ - +

+ - - .

2. Find the square root of the following:

(i) 16 24 9x x2- + .

Solution: 16 24 9x x2- + = ( ) ( )x x4 3 4 32

- = - .


(ii) x x x x x25 8 15 2 152 2 2- + + - -^ ^ ^h h h.

Solution: 25 8 15 2 15x x x x x2 2 2- + + - -^ ^ ^h h h

= ( )( )( )( )( )( )x x x x x x5 5 3 5 5 3+ - + + - +

= ( ) ( ) ( ) ( )( )( )x x x x x x5 5 3 5 5 32 2 2+ - + = + - + .

(iii) 4 9 25 12 30 20x y z xy yz zx2 2 2+ + - + - .

Solution: 4 9 25 12 30 20x y z xy yz zx2 2 2+ + - + -

= ( ) ( ) ( ) ( )( ) ( )( ) ( )( )x y z x y y z z x2 3 5 2 2 3 2 3 5 2 5 22 2 2+ - + - + - + - - + -

= ( )x y z x y z2 3 5 2 3 52- - = - - .

(iv) 2xx

14

4+ + .

Solution: 1 2xx

4

4+ + = ( ) ( )xx

xx

1 2 12 2

2

22

2+ +d dn n

= xx

xx

1 122

2 22

+ = +c cm m .

(v) x x x x x x6 5 6 6 2 4 8 32 2 2+ - - - + +^ ^ ^h h h.

Solution: 6 5 6 6 2 4 8 3x x x x x x2 2 2+ - - - + +^ ^ ^h h h

= ( )( )( )( )( )( )x x x x x x2 3 3 2 3 2 2 1 2 1 2 3+ - - + + +

= ( ) ( ) ( ) ( )( )( )x x x x x x2 3 3 2 2 1 2 3 3 2 2 12 2 2+ - + = + - + .

(vi) x x x x x x2 5 2 3 5 2 6 12 2 2- + - - - -^ ^ ^h h h.

Solution: x x x x x x2 5 2 3 5 2 6 12 2 2- + - - - -^ ^ ^h h h

= ( )( )( )( )( )( )x x x x x x2 1 2 3 1 2 2 1 3 1- - + - - +

= ( ) ( ) ( ) ( )( )( )x x x x x x2 1 2 3 1 2 1 2 3 12 2 2- - + = - - + .

Exercise 3.13 1. Find the square root of the following polynomials by division method. (i) 4 10 12 9x x x x

4 3 2- + - + .

Solution: x x2 32- +

x2 x x x x4 10 12 94 3 2- + - +

x4

x x2 22- x x4 103 2

- +

x x4 43 2- +

x x2 4 32- + x x6 12 92

- +

6 12 9x x2- +

0

Thus, x x x x x x4 10 12 9 2 34 3 2 2- + - + = - + .


(ii) 4 8 8 4 1x x x x4 3 2+ + + + .

Solution: x x2 2 12+ +

x2 2 x x x x4 8 8 4 14 3 2+ + + +

x4 4

4 2x x2+ x x8 83 2

+

x x8 43 2+

4 1x x42+ + x x4 4 12

+ +

x x4 4 12+ +

0

Thus, x x x x x x4 8 8 4 1 2 2 14 3 2 2+ + + + = + + .

(iii) 9 6 7 2 1x x x x4 3 2- + - + .

Solution: x x3 12- +

x3 2 x x x x9 6 7 2 14 3 2- + - +

x9 4

x x6 2- x x6 73 2

- +

x x6 3 2- +

6 2 1x x2- + x x6 2 12

- +

x x6 2 12- +

0

Thus, x x x x x x9 6 7 2 1 3 14 3 2 2- + - + = - + .

(iv) 4 25 12 24 16x x x x2 3 4

+ - - + .

Solution: x x4 3 22- +

x4 2 x x x x16 24 25 12 44 3 2- + - +

x16 4

x x8 32- x x24 253 2

- +

x x24 93 2- +

x x8 6 22- + x x16 12 42

- +

x x16 12 42- +

0

Thus, x x x x x x16 24 25 12 4 4 3 24 3 2 2- + - + = - + .


2. Find the values of a and b if the following polynomials are perfect squares. (i) 4 12 37x x x ax b

4 3 2- + + + . (ii) 4 10x x x ax b

4 3 2- + - + .

(iii) 109 60 36ax bx x x4 3 2+ + - + . (iv) 40 24 36ax bx x x

4 3 2- + + + .

Solution: (i) x x2 3 72- +

x2 2 x x x ax b4 12 374 3 2- + + +

x4 4

x x4 32- 12 37x x3 2

- +

x x12 93 2- +

x x4 6 72- + 2 x ax b8 2

+ +

x x28 42 492- +

0Since the given polynomial is a perfect square, we must have a 42=- and b 49= . Thus, 42a =- and b 49= .

(ii)

x x2 32- +

x2 x x x ax b4 104 3 2- + - +

x4

x x2 22- x x4 103 2

- +

x x4 43 2- +

x x2 4 32- + x ax b6 2

- +

x x6 12 92- +

0 Since the given polynomial is a perfect square, we must have 2a 1= and 9b = .Thus, 12a = and 9b = .

x x6 5 7 2- +

(iii) 6 36 60 109x x bx ax2 3 4- + + +

36

x12 5- x x60 109 2- +

x x60 25 2- +

x x12 10 7 2- + x bx ax84 2 3 4

+ +

x x x84 70 492 3 4- +

0 Since the given polynomial is a perfect square, we must have 9a 4= and b 70=- .Thus, 9a 4= and b 70=- .


(iv)

x x6 2 3 2+ +

6 x x bx ax36 24 40 2 3 4+ + - +

36

12 2x+ x x24 40 2+

x x24 4 2+

12 4 3x x2+ + x bx ax36 2 3 4- +

x x x36 12 92 3 4+ +

0

Since the given polynomial is a perfect square, we must have a 9= and b 12=- .

Thus, a 9= and b 12=- .

Exercise 3.14

Solve the following quadratic equations by factorization method.

(i) 81x2 3 2+ -^ h = 0.

Solution: ( )x2 3 812+ - = 0 Aliter:

& x x4 12 9 812+ + - = 0 ( )x2 3 92 2

+ - = 0

& x x4 24 12 722+ - - = 0 (2 3 9)(2 3 9)x x& + + + - = 0

& ( )( )x x6 4 12+ - = 0 (2 12)(2 6)x x& + - = 0

& x 6 0+ = or x4 12 0- = 6x& =- or x = 3

6x& =- or x = 3

Hence, solution set is {– 6, 3}.

(ii) 3 5 12x x2- - = 0.

Solution: x x3 5 122- - = 0

& ( )( )x x3 3 4- + = 0 3x& = or x34=-

Hence, solution set is ,34 3-$ ..

(iii) 2 3x x5 52+ - = 0.

Solution: x x5 2 3 52+ - = 0

& ( )( )x x5 5 3+ - = 0 x 5& =- or x5

3=

Hence, solution set is ,55

3-' 1.


(iv) 3 x 62-^ h = 3x x 7+ -^ h .

Solution: ( )x3 62- = ( )x x 7 3+ -

& x x x3 7 18 32 2- - - + = 0

& x x2 7 152- - = 0

& ( )( )x x5 2 3- + = 0

5x& = or x23=-


(v) 3xx8- = 2.

Solution: xx

3 8- = 2

& 3 2 8x x2- - = 0

( )( )x x2 3 4- + = 0

2x& = or x34=-


(vi) xx1+ =

526 .

Solution: xx1+ =

526

& x x5 26 52- + = 0

& ( )( )x x5 5 1- - = 0

5x& = or x51=

Hence, solution set is ,51 5$ ..

(vii) x

xx

x1

1+

+ + = 1534 .

Solution: x

xx

x1

1+

+ + = 1534

& x x

x x x2 12

2 2

+

+ + + = 1534

& x x30 30 152+ + = x x34 342

+

& ( )( )x x2 5 2 3+ - = 0

x25& =- or x

23=

Hence, solution set is ,25

23-$ ..

(viii) 1a b x a b x2 2 2 2 2

- + +^ h = 0.

Ô®Î: ( )a b x a b x 12 2 2 2 2- + + = 0 & ( 1)( 1)a x b x2 2

- - = 0.

xa

12

& = or xb

12

= . Hence, the solution set is ,a b

1 12 2' 1.

(ix) 2 5x x1 12+ - +^ ^h h = 12.

Solution: ( ) ( )x x2 1 5 12+ - + = 12 Aliter: Let y x 1= + . Then,

& ( )x x x2 2 1 5 5 122+ + - - - = 0 y y2 5 122

- - = 0

& x x2 152- - = 0 ( )( )y y2 3 4& + - = 0

& ( )( )x x3 2 5- + = 0 ( )( )x x2 5 3& + - = 0

Thus, 3x = or x25=- Thus, 3x = or x

25=- .

Hence, the solution set is ,325-$ ..

(x) 3 5x x4 42- - -^ ^h h = 12.

Solution: ( ) ( )x x3 4 5 42- - - = 12 Aliter: Let y x 4= - . Then,

& x x x3 24 48 5 20 122- + - + - = 0 3 5 12y y2

- - = 0

& x x3 29 562- + = 0 ( )( )y y3 4 3& + - = 0

& ( )( )x x3 8 7- - = 0 ( )( )x x3 8 7& - - = 0

Thus, x38= or x 7= Thus, x

38= or x 7=

Hence, the solution set is ,38 7$ ..


Exercise 3.15

1. Solve the following quadratic equations by completing the square.

(i) 6 7x x2+ - = 0.

Solution: x x6 72+ - = 0

& ( )x x2 32+ = 7

& ( )x x2 3 92+ + = 7 + 9

& ( )x 3 2+ = 16

& x 3+ = 4! 1x& = or x 7=- .

Hence, the solution set is {– 7, 1}.

(ii) 3 1x x2+ + = 0.

Solution: x x3 12+ + = 0

& x x223

492

+ +` j = 149- +

& x23 2

+` j = 45

& x23+ =

25!

& x = 23

25!- & x =

23 5- - or x

23 5= - + .

Hence, the solution set is ,2

3 52

3 5- - - +' 1.

(iii) 2 5 3x x2+ - = 0.

Solution: x x2 5 32+ - = 0

& x x25

232

+ - = 0 (Divide both sides by 2)

& x x25

16252

+ + = 1625

23+ Note that

21

25

16252

=` j8 B .

& x45 2

+` j = 1649

& x45+ =

47! & x =

45

47!-

x21& = or x 3=- . Hence, the solution set is ,3

21-$ ..

(iv) 4 4x bx a b2 2 2+ - -^ h = 0.

Solution: ( )x bx a b4 42 2 2+ - - = 0 (Divide both sides by 4)

& x bx2+ = a b

4

2 2-

& x bx b4

22

+ + = a b b4 4

2 2 2- +


& x b2

2+` j = a

4

2

& x b2

+ = a2

! & x = b a2 2!-

Thus, x a b2

= - or ( )x

a b2

=-+

Hence, the solution set is ( ),

a b a b2 2

- + -' 1.

(v) x x3 1 32- + +^ h = 0.

Solution: ( )x x3 1 32- + + = 0

& ( )x x3 12

3 122

- + + +c m = 2

3 1 32

+ -c m

& x2

3 12

- +c m; E = 4

3 2 3 1 4 3+ + -

& x2

3 12

- +c m; E = 2

3 12

-c m

& x2

3 1- +c m = 2

3 1! -c m

& x = 2

3 12

3 1!+ -c cm m

Thus, x 3= or x 1= . Hence, the solution set is { , }1 3 .

(vi) xx

15 7

-+ = 3 2x + .

Solution: xx

15 7-+ = x3 2+

& x5 7+ = ( )( )x x3 2 1+ -

& x x3 6 92- - = 0

& x x2 32- - = 0 ( Note that 1

22 2- =` j )

& x x2 12- + = 1 + 3

& ( )x 1 2- = 4

& x 1- = 2! & x = 1 2!

Thus, x 3= or x 1=- .

Hence, the solution set is { , }1 3- .

2. Solve the following quadratic equations using quadratic formula.

(i) 7 12x x2- + = 0.

Solution: x x7 122- + = 0. The equation is of the form ax bx c 02

+ + = .

Here, , ,a b c1 7 12= =- =

x = a

b b ac2

42!- -


= 2

7 49 482

7 1! !- =

& x = or x28

26= & x = 4 or x = 3.

Thus, the solution set is , .4 3" ,

(ii) 15 11 2x x2- + = 0.

Solution: x x15 11 22- + = 0. It is of the form ax bx c 02

+ + = .

Here, 1 , ,a b c5 11 2= =- =

Thus, x = a

b b ac2

42!- -

= ( )2 15

11 121 12030

11 121 12030

11 1! ! !- = - =

& x = or x3012

3010= & x =

52 or x

31= .

Thus, the Solution set is ,52

31$ ..

(iii) xx1+ = 2

21 .

Solution: xx1+ = 2

21

& 1x

x2 + = 25

& x x2 5 22- + = 0 , which is of the form ax bx c 02

+ + = .

Here, , , 2a b c2 5= =- =

Thus, x = a

b b ac2

42!- -

= 4

5 25 164

5 94

5 3! ! !- = =

& x = or x221=

Thus, the Solution set is ,21 2$ ..

(iv) 3 2a x abx b2 2 2

- - = 0.

Solution: a x abx b3 22 2 2- - = 0. It is of the form 0Ax Bx C2

+ + = .

Here, , ,A a B ab C b3 22 2= =- =- .

Now, x = A

B B AC2

42!- -

= ( )

( )( )

a

ab a b a b

2 3

4 3 22

2 2 2 2! - -

= a

ab a b a b

a

ab ab

6

24

6

52

2 2 2 2

2! !+ =


& x = a

ab abab

6

52

+ = or x = .a

ab abab

6

532

2- = -

Thus, the Solution set is ,ab

ab

32-$ ..

(v) a x 12+^ h = x a 1

2+^ h.

Solution: ( )a x 12+ = ( )x a 12

+

& ax a2+ = ( )x a 12

+

& ( )ax x a a12 2- + + = 0. It is of the form 0Ax Bx C2

+ + =

Here, , ( ),A a B a C a12= =- + =

Now, x = A

B B AC2

42!- -

= ( ) ( )a

a a a2

1 1 42 2 2 2!+ + -

= ( )a

a a a a2

1 2 1 42 4 2 2!+ + + -

= ( ) ( ) ( )a

a a aa

a a2

1 2 12

1 12 4 2 2 2 2! !+ - +=

+ -

= ( ) ( )a

a a2

1 12 2!+ -

Thus, x = aa a22 2

= or x = a a22 1=

Hence, the Solution set is ,a

a1$ ..

(vi) 36 12x ax a b2 2 2- + -^ h = 0.

Solution: 3 12 ( )x ax a b6 2 2 2- + - = 0. It is of the form 0Ax Bx C2

+ + = .

Here, , , ( )A B a C a b36 12 2 2= =- = -

Now, x = A

B B AC2

42!- -

= ( )

( )( )a a a b2 36

12 144 4 362 2 2! - -

= a a a b72

12 144 144 1442 2 2! - +

= a b a b72

12 14472

12 122! !=

Thus, x = ( ) ( )a b a b72

126

+=

+ or x = ( ) ( )a b a b72

126

-=

-

Hence, the Solution set is ,a b a b6 6- +$ ..


(vii) xx

xx

11

43

+- +

-- =

310 .

Solution: xx

xx

11

43

+- +

-- =

310

& ( )( )

( )( ) ( )( )x x

x x x x1 4

1 4 3 1+ -

- - + - + = 310

& x x

x x x x

3 4

5 4 2 32

2 2

- -

- + + - - = 310

& x x

x x

3 4

2 7 12

2

- -

- + = 310

& x x6 21 32- + = x x10 30 402

- -

& x x4 9 432- - = 0 ( 0ax bx c2

+ + = form )

Here, , ,a b c4 9 43= =- =-

Now, x = a

b b ac2

42!- -

= ( )( )( )

2 49 81 4 4 43

89 769! !- -

=

Thus, x = 8

9 769+ or 8

9 769-

Hence, the Solution set is ,8

9 7698

9 769- +' 1.

(viii) a x a b x b2 2 2 2 2

+ - -^ h = 0.

Solution: ( )a x a b x b2 2 2 2 2+ - - = 0. This is of the form 0Ax Bx C2

+ + = .

Here, , ,A a B a b C b2 2 2 2= = - =-

Now, x = A

B B AC2

42!- -

= ( )

( ) ( ) ( )( )

a

a b a b a b

2

42

2 2 2 2 2 2 2!- - - - -

= ( ) ( ) ( )

a

b a a a b b

a

b a a b

2

2

22

2 2 4 2 2 4

2

2 2 2 2! !- + +=

- +

Thus, x = a

b a a b

a

b

2 2

2 2 2 2

2

2- + + = or x =

a

b a a b

21

2

2 2 2 2- - - =-

Hence, the Solution set is ,a

b12

2

-) 3.

Exercise 3.16

1. The sum of a number and its reciprocal is 865 . Find the number.

Solution: Let the number be x and its reciprocal be x1 .


Given that xx1+ =

865

& x

x 12+ =

865

& x x8 65 82- + = 0

& ( )( )x x8 1 8- - = 0

& x8 1- = 0 or x 8- = 0 & x = 81 or x = 8

Thus, the required number is 8.

2. The difference of the squares of two positive numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.

Solution: Let x and y be the two positive numbers. Given that x y<

Given that x2 = y4 g (1)

and y x2 2- = 45 g (2)

& y y4 452- - = 0 [ using the equation (1) ]

& ( )( )y y9 5- + = 0

& y 9- = 0 or y 5+ = 0 & y = 9 or y = – 5

Since the numbers are positive, we take y = 9.

Substituting y = 9 in (1), we get, x2 = 4 9, 6x&# =

Thus, the required numbers are 6 , 9.

3. A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences the sides of the rectangular garden letting his house compound wall act as the fourth side fence. Find the dimension of the garden.

Solution: Let AB = x metre and BC = y metre be the length and breadth of the rectangular garden. Let CD be the compound wall.

Given that the length of the fencing barbed wire = 30 m

& y x y+ + = 30

& x y2+ = 30

& y = x2

30 -` j g (1)

Area of the vegetable garden = 100m2

& xy = 100

& x x2

30 -` j = 100 ( using (1) )

& x x30 2- = 200

& 30 200x x2- + = 0


& ( )( )x x20 10- - = 0

& x 20- = 0 or x 10- = 0 & x = 20 or x = 10

Substituting x 10= in (1), we get, y = 2

30 10 10- =

Also, when x 20= , (1) & y = 2

30 20 5- = .

Thus, length and breadth of the rectangle are either 10m, 10m (or) 20m, 5m.

4. Arectangularfieldis20mlongand14mwide.Thereisapathofequalwidthall around it having an area of 111 sq. metres. Find the width of the path on the outside.Solution: Let AB = 20m and BC = 14m be the length and breadth of the rectangle ABCD. Let x metre be the width of uniform path around ABCD.Let PQ = x20 2+ and QR = x14 2+ be the length and breadth of the rectangle PQRS.Given that the area of the path = 111 sq.m.& Area of the rectangle PQRS – Area of the rectangle ABCD = 111m2

& (20 2 )(14 2 ) (20 14)x x #+ + - = 111

& x x x20 14 40 28 4 20 142# #+ + + - = 111

& 4 68 111x x2+ - = 0

& ( )( )x x2 37 2 3+ - = 0

& x = 237- or x

23=

Since length can’t be negative, we take x23= .

Thus, the width of the path on the outside is 1.5m 5. A train covers a distance of 90 km at a uniform speed. Had the speed been

15 km/hr more, it would have taken 30 minutes less for the journey.

Find the original speed of the train.

Solution: Let the original speed of the train be x km/hr.

Let T1 be the time taken to cover the distance of 90 km when the speed is x km/hr.

Let T2

be the time taken to cover the same distance when the speed is x 15+ km/hr.

Since time speeddistance= , we have T

1 =

x90 and T

2 =

x 1590+

.

Given that T T1 2- =

6030

& x x90

1590-+

= 21

& ( )

( )x xx x

1590 15 90

++ - =

21

& x x

x x

15

90 1350 902+

+ - = 21

& x x15 27002+ - = 0


& ( )( )x x60 45+ - = 0

& x = – 60 or x = 45

Since speed cannot be negative, we take x = 45.

Thus, the original speed of the train is 45 km/hr.

6. The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and return downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream.

Solution: Let the speed of the stream be x km/hr.

Given that the speed of the boat in still water is 15 km/hr.

Thus, the speed of the boat in the downstream and in the upstream are

( )x15 + km/hr and ( )x15 - km/hr respectively.

Let T1 be the time taken to cover the distance of 30 km in the downstream.

Let T2

be the time taken to cover the same distance in the upstream.


1 =

x1530+

and T2

= x15

30-

.

Now, given that T1 + T

2 = 4 hrs. 30 hours = 4

21 hrs.

& x x15

301530

-+

+ =

29

& ( )( )

( ) ( )x x

x x15 15

30 15 30 15- +

+ + - = 29

& ( )x9 225 2- = 1800

& x225 2- = 200

& x = 5!

Since the speed of the stream cannot be negative, we take .x 5=

Thus, the speed of the stream is 5 km/hr.

7. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

Solution:Son Father

Present age x years y yearsOne year ago their ages ( )x 1- years ( )y 1- years

Given that y = x2 g (1)

and y 1- = ( )x8 1-

y& = x8 7- g (2)

Using (2) in (1), we get x8 7- = x2


& x x8 72- + = 0 & ( )( )x x7 1- - = 0

& x 7= or x 1=

Since x 1= is inadmissible, we take x 7= .

Thus, Son’s age is 7 years ; Father’s age is = 7 492= years.

8. A chess board contains 64 equal squares and the area of each square is 6.25 cm2 . A border around the board is 2 cm wide. Find the length of the side of the chess board.Solution: Let the length of side of the chess board = x cm. Area of the each square = 6.25cm2

Thus, the area of 64 squares = .64 6 25#

& ( )x 4 2- = 400

& x 4- = 20! & x = 24 or – 16

Since, length of the sides of a board can’t be negative, so x = 24 cm.

9. A takes 6 days less than the time taken by Btofinishapieceofwork.IfbothA and B togethercanfinishitin4days,findthetimethatBwouldtaketofinishthiswork by himself.Solution: Let x be the number of days taken by B to finish the work .Then, A takes ( 6)x - days to finish the same work.

Amount of work done by A in 1 day = x 61-

Amount of work done by B in 1 day = x1

Amount of work done by A and B in 1 day = 41

Thus, x x61 1-

+ = 41

& ( )x x

x x66

-+ - =

41

& x x14 242- + = 0

& ( )( )x x12 2- - = 0

& x 12 0- = or x 2 0- = & x 12= or x 2=

Since x 2= is inadmissible, we take x 12= . Thus, B needs 12 days to finish the work by himself.

10. Twotrainsleavearailwaystationatthesametime.Thefirsttraintravelsduewestandthesecondtrainduenorth.Thefirsttraintravels5km/hrfasterthanthesecondtrain.Ifaftertwohours,theyare50kmapart,findtheaveragespeedof each train.


Solution: Let the speed of the second train be x km/hr.

Then, the speed of the first train = ( )x 5+ km/hr.

Let O be the position of railway station.

Distance travelled by first train in 2 hrs = 2 (x + 5) = OA (say)

Distance travelled by second train in 2 hrs = 2x = OB (say)

By Pythagoras theorem, OA OB2 2+ = AB2

& [ ( )] ( )x x2 5 22 2+ + = 502

& x x8 40 24002+ - = 0

& x x5 3002+ - = 0

& ( )( )x x20 15+ - = 0

& x 20 0+ = or x 15 0- = & x 20=- or x 15=

Since speed cannot be negative, we take 15x =

Thus, Speed of second train is 15 km/hr

and the Speed of the first train is 20 km/hr.

Exercise 3.17

1. Determine the nature of the roots of the equation.

(i) 8 12 0x x2- + = .

Solution: x x8 122- + = 0

Comparing the eqation with ax bx c 02+ + = , we get , ,a b c1 8 12= =- =

Now, D = b ac42- = ( ) ( )( )8 4 1 12 64 48 0>2

- - = -

Thus, the roots are real and distinct.

(ii) 2 3 4 0x x2- + = .

Solution: x x2 3 42- + = 0

Comparing the equation with ax bx c 02+ + = , we get , ,a b c2 3 4= =- =

Now, D = b ac42- = ( ) ( )( )3 4 2 4 0<2

- -

Thus, the equation has no real roots.

(iii) 9 12 4 0x x2+ + = .

Solution: x x9 12 42+ + = 0

Comparing the equation with ax bx c 02+ + = , we get , ,a b c9 12 4= = =

Now, D = b ac42- = ( ) ( )( )12 4 9 4 02

- =

Thus, the roots are real and equal.


(iv) 3 2 2 0x x62- + = .

Solution: x x3 2 6 22- + = 0

Comparing the equation with ax bx c 02+ + = , we get , ,a b c3 2 6 2= =- =

Now, D = b ac42- = ( ) ( )( ) ( )2 6 4 3 2 4 6 24 02

- - = - =

Thus, the roots are real and equal.

(v) 1 0x x53

322

- + = .

Solution: x x53

32 12

- + = 0

Comparing the equation with ax bx c 02+ + = , we get , , 1a b c

53

32= =- =

Now, D = b ac42- = 4 (1)

32

53

94

512

4588 0<

2- - = - =-` `j j

Thus, the equation has no real roots.

(vi) 4x a x b ab2 2- - =^ ^h h .

Solution: ( )( )x a x b2 2- - = ab4

( )x x a b ab2 2 42& - + + = ab4 ( )x x a b22

& - + = 0

Comparing the equation with ax bx c 02+ + = , we get 1, ( ),a b a b c2 0= =- + =

Now, D = b ac42- = [ ( )] ( )( ) ( )a b a b2 4 1 0 4 0>2 2

- + - = +

Thus, the roots are real and unequal. 2. Find the values of k for which the roots are real and equal in each of the following

equations.

(i) 2 10 0x x k2- + = .

Solution: x x k2 102- + = 0

Comparing with ax bx c 02+ + = , we get , ,a b c k2 10= =- =

Given that the equation has equal roots.

Thus, D = b ac42- = 0 & ( ) ( )( )k10 4 22

- - = 0

& k8 = 100 k225& =

(ii) 12 4 3 0x kx2+ + = .

Solution: x kx12 4 32+ + = 0

Comparing with ax bx c 02+ + = , we get 2, ,a b k c1 4 3= = =


Thus, D = b ac42- = 0

& ( ) ( )( )k4 4 12 32- = 0

& k2 = 16144 9=

Hence, k = 3! .


(iii) 2 5 0x k x 22+ - + =^ h .

Solution: ( )x k x2 2 52+ - + = 0

& 2 4 5x kx k2+ - + = 0

& ( )x kx k2 5 42+ + - = 0

Comparing with ax bx c 02+ + = , we get 1, ,a b k c k2 5 4= = = -

Given that, the equation has equal roots.

Thus, D = b ac42- = 0 & ( ) ( )( )k k2 4 1 5 42

- - = 0

& k k4 52+ - = 0

& ( )( )k k5 1+ - = 0 & 5k =- or k 1= .

(iv) 2 1 0k x k x1 12

+ - - + =^ ^h h .

Solution: ( ) ( )k x k x1 2 1 12+ - - + = 0

This is of the form ax bx c 02+ + = . Here, , 2( ),a k b k c1 1 1= + =- - =


Thus, D = b ac42- = 0

& [ ( )] ( )( )k k2 1 4 1 12- - - + = 0

& 4( 1) 4( 1)k k2- - + = 0

& ( ) ( )k k1 12- - + = 0 & k k32

- = 0

Hence, k = 0 , 3.

3. Show that the roots of the equation 2 2 0x a b x a b2 2 2+ + + + =^ ^h h are not real.

Solution: ( ) ( )x a b x a b2 22 2 2+ + + + = 0

This is of the form 0Ax Bx C2+ + = , we get 1, 2( ), 2( )A B a b C a b2 2

= = + = +

Now, D = 4B AC2-

= [2( )] 4(1)( )( )a b a b22 2 2+ - +

= ( )a ab b a b4 2 8 82 2 2 2+ + - -

= a ab b a b4 8 4 8 82 2 2 2+ + - -

= 4 8 4 4 ( 2 )a ab b a ab b2 2 2 2- + - =- - +

= 4( )a b 0<2- - , for all a, b R! .

Thus, the roots of the equation are not real numbers.

4. Show that the roots of the equation 3 2 0p x pqx q2 2 2

- + = are not real .

Solution: Given equation is 3 2p x pqx q2 2 2- + = 0

This is of the form ax bx c 02+ + = , we get , ,a p b pq c q3 22 2

= =- =

Now, D = b ac42-


= ( ) ( )( )pq p q2 4 32 2 2- -

= 4 12 8 0p q p q p q <2 2 2 2 2 2- =-

Thus, the roots of the equation are not real numbers.

5. If the roots of the equation 2 0a b x ac bd x c d2 2 2 2 2+ - + + + =^ ^h h ,

where , , ,b ca d are non-zero real numbers, are equal, then prove that ba

dc= .

Solution: ( ) ( )a b x ac bd x c d22 2 2 2 2+ - + + + = 0

This is of the form 0Ax Bx C2+ + = , we get ,

, 2( ),A a b B ac bd C c d2 2 2 2= + =- + = +

Now, D = 4B AC 02- = ( the roots are equal )

& [ ( )] ( )( )ac bd a b c d2 42 2 2 2 2- + - + + = 0

& ( ) ( )ac bd a c a d b c b d4 42 2 2 2 2 2 2 2 2+ - + + + = 0

& a c abcd b d a c a d b c b d22 2 2 2 2 2 2 2 2 2 2 2+ + - - - - = 0

& a d b c abcd22 2 2 2- - + = 0

& 2a d abcd b c2 2 2 2- + = 0

& ( )ad bc 2- = 0 & ad bc- = 0

& ad = bc

& ba =

dc . ( , , ,a b c d are non zero)

6. Show that the roots of the equation

0x a x b x b x c x c x a- - + - - + - - =^ ^ ^ ^ ^ ^h h h h h h are always real and they cannot be equal unless a b c= = .

Solution: ( )( ) ( )( ) ( )( )x a x b x b x c x c x a- - + - - + - - = 0

& x ax bx ab x bx cx bc x cx ax ca2 2 2- - + + - - + + - - + = 0

& ( )x a b c x ab bc ca3 22- + + + + + = 0

This is of the form 0Ax Bx C2+ + = .

Thus, we get , 2( ),A B a b c C ab bc ca3= =- + + = + + .

Now, D = 4B AC2-

= [ ( )] ( )( )a b c ab bc ca2 4 32- + + - + +

= ( ) ( )a b c ab bc ca4 122+ + - + +

= 4[( ) 3( )]a b c ab bc ca2+ + - + +

= [ ]a b c ab bc ca4 2 2 2+ + - - -

= [ ]a b c ab bc ca2 2 2 2 2 2 22 2 2+ + - - -


= [( ) ( ) ( ) ]a b b c c a2 2 2 2- + - + - > 0

Thus, the roots are always real.

If a b c= = , then 0D = . Thus, in this case the roots are equal.

7. If the equation 2 0m x mcx c a12 2 2 2

+ + + - =^ h has equal roots, then prove that c a m12 2 2= +^ h.

Solution: ( )m x mcx c a1 22 2 2 2+ + + - = 0

This is of the form 0Ax Bx C2+ + = , where

( ), 2 ,A m B mc C c a1 2 2 2= + = = -

Given that the equations has equal roots

Thus, D = 4B AC2- = 0

& ( ) ( )mc m c a2 4 12 2 2 2- + -^ h = 0

& 4 ( )m c c a m c a m42 2 2 2 2 2 2 2- - + - = 0

& m c c a m c a m2 2 2 2 2 2 2 2- + - + = 0

& c a a m2 2 2 2- + + = 0

Thus, c2 = ( )a m12 2+ .

Exercise 3.18

1. Find the sum and the product of the roots of the following equations.

(i) 6 5 0x x2- + = .

Solution: x x6 52- + = 0. This is of the form ax bx c 02

+ + = .

Here, , ,a b c1 6 5= =- = . Let a and b be the roots.

Thus, sum of the roots, a b+ = ( )6

ab

16

- =--

=

Product of the roots, ab = ac

15 5= =

Thus, sum and product of the roots are 6 and 5 respectively.

(ii) 0kx rx pk2+ + = .

Solution: kx rx pk2+ + = 0. This is of the form ax bx c 02

+ + = .

Here, , ,a k b r c pk= = = . Let a and b be the roots.

Now, sum of the roots, a b+ = ab

kr- =-


kpk

p= =

Thus, sum and product of the roots are kr- and p.

(iii) 3 5 0x x2- = .

Solution: x x3 52- = 0. This is of the form ax bx c 02

+ + = .


Here, , ,a b c3 5 0= =- = . Let a and b be the roots.

Thus, sum of the roots, a b+ = ( )ab

35

35- =-

-=


30 0= =

Thus, sum and product of the roots are 35 and 0.

(iv) 8 25 0x2- = .

Solution: x8 252- = 0. This is of the form ax bx c 02

+ + = .

Here, , ,a b c8 0 25= = = . Let a and b be the roots.

Thus, sum of the roots, a b+ = ab

80 0- = = Note: x

8252

= ,


825=- Thus. ,x

825

825= -

Thus, Sum and product of the roots are 0 and 825- .

2. Form a quadratic equation whose roots are

(i) 3 , 4 (ii) 3 7+ , 3 7- (iii) ,2

4 72

4 7+ -

Solution: (i) Given roots are 3, 4. Let ,a b be the roots.

Sum of the roots, a b+ = 3 + 4 = 7

Product of the roots, ab = 3 (4) = 12

The required equation is x2 - (sum of the roots) x + Product of the roots = 0.

Thus, the equation is x x7 12 02- + = .

(ii) Given roots are ,3 7 3 7+ - . Let 3 7a = + and 3 7b = - .

Sum of the roots, a b+ = 3 7 3 7 6+ + - =

Product of the roots, ab = ( )( )3 7 3 7 9 7 2+ - = - =



(iii) Let ,2

4 72

4 7a b= + = - be the given roots.

Sum of the roots, a b+ = 2

4 72

4 728 4+ + - = =

Product of the roots, ab = 2

4 72

4 74

16 749+ - = - =c cm m


(4) 0x x492

& - + = . Thus, the equation is 4 16 0x x 92- + = .

3. If a and b are the roots of the equation 3 5 2x x2- + =0,thenfindthevaluesof

(i) ba

ab

+ (ii) a b- (iii) 2 2

ba

ab

+

Solution: (i) 3 5 2x x 02- + = . It is in the form of ax bx c 02

+ + =

Here, , ,a b c3 5 2= =- = .


Thus, a b+ = ( )ab

35

35- =-

-= and ab =

ac

32= .

Now, ba

ab

+ = ( ) 22 2 2

aba b

aba b ab+

=+ -

= 32

32

35 2

925 12

23

613

2-

= - =` `

`j j

j .

(ii) Now, ( )2a b- = ( ) 42a b ab+ -

= 35 4

32

925 24

912

- = - =` `j j

Thus, a b- = 31!

(iii) 2

2 2

b

aab

+ = ( ) ( )33 3 3

aba b

aba b ab a b+

=+ - +

=

32

27125

930

27125 90

23

1835

#-

= - = .

4. If a and b are the roots of the equation 3 6 4x x2- + =0,findthevalueof .

2 2a b+

Solution: x x3 6 4 02- + = . It is in the form of ax bx c 02

+ + =

Here, 3, ,a b c6 4= =- = .

Thus, a b+ = ( )ab

36

2- =--

=

ab = ac

34=

Now, 2 2a b+ = ( ) 2 2 234

342 2a b ab+ - = - =` j

2 2a b+ = 34 .

5. If a , b are the roots of 2 3 5x x2- - = 0, form an equation whose roots are

2a and 2

b .Solution: x x2 3 5 02

- - = . It is in the form of ax bx c 02+ + =

Here, , ,a b c2 3 5= =- =- .

Now, a b+ = ( )ab

23

23- =-

-=

ab = ac

25= -

To form the required equation, let us find the sum and product of the roots.

Sum of the roots, 2 2a b+ = ( ) 22a b ab+ -

= 223

25

49

210

4292

- - = + =` `j j

Product of the roots, ( )( ) ( )25

4252 2 2 2

a b ab= = - =` j

Thus, required equation is, x2 - (Sum of the roots) x + Product of the roots = 0


& x x429

4252

- +` j = 0

Thus, the equation is x x4 29 25 02- + = .

6. If a , b are the roots of 3 2x x2- + = 0, form a quadratic equation whose roots

are a- and b- .Solution: 3 0x x 22

- + = . It is in the form of ax bx c 02+ + =

Here, , 3,a b c1 2= =- = .

a b+ = ( )ab

13

3- =--

=

ab = ac

12 2= =

Let us find the equation whose roots are a- and b- .

Now, sum of the roots = ( ) 3a b a b- - =- + =-

Product of the roots = ( )( ) 2a b ab- - = =

Now, the required equation is, x2 - (Sum of the roots) x + Product of the roots = 0

& 3 2x x2+ + = 0.

Note : In the above problem, 3 2x x2+ + = 0 can be obtained by replacing x by –x

in the given equation x x3 22- + = 0.

7. If a and b are the roots of 3 1x x2- - = 0, then form a quadratic equation

whose roots are 1 1and2 2a b

.

Solution: x x3 1 02- - = . It is in the form of ax bx c 02

+ + =

Here, 1, 3,a b c 1= =- =-

Now, a b+ = ( )3

ab

13- =-

-= and ab =

ac

11 1= - =-

Let us form an equation whose roots are 12a

and 12b

.

Sum of the roots, ( )

1 12 2 2

2 2

a b ab

a b+ =

+

= ( )

( )

( )

( )2

1

3 2 11

9 2 112

2

2

2

ab

a b ab+ -=

-

- -= + =

Product of the roots, 11 1 122

2

a b ab= =c c cm m m

Thus, the required equation is x2 - x (Sum of the roots) + (Product of the roots) = 0

& (11) 1x x2- + = 0



8. If a and b are the roots of the equation 3 6 1x x2- + = 0, form an equation whose

roots are (i) ,1 1a b

(ii) ,2 2a b b a (iii) 2 , 2a b b a+ +

Solution: 3 0x x6 12- + = . It is in the form of ax bx c 02

+ + = .

Here, 3, ,a b c6 1= =- =

Thus, a b+ = ( )ab

36

2- =--

= and ab = ac

31=

(i) Let us form an equation whose roots are 1a

and 1b

.

Sum of the roots, 1 1

312 6

a b aba b

+ =+

= =


311 3

a b ab= = =` cj m

Thus, the required equation is x2 - (Sum of the roots) x + Product of the roots = 0

x x6 32& - + = 0


(ii) Let us form an equation whose roots are 2a b and 2b a .

Sum of the roots, ( ) ( )31 2

322 2a b b a ab a b+ = + = =

Product of the roots, ( )( ) ( )31

2712 2 3 3 3 3

a b b a a b ab= = = =` j

Thus, the required equation is, x2 - (Sum of the roots) x + Product of the roots = 0

& x x32

2712

- +` j = 0


(iii) Let us form an equation whose roots are 2a b+ and 2b a+ .

Sum of the roots, ( ) ( ) ( ) ( )2 2 3 3 2 6a b b a a b+ + + = + = =

Product of the roots, (2 )(2 ) 4 2 22 2a b b a ab a b ab+ + = + + +

= [( ) ] (2) 2322 2 5 2 5

3122a b ab ab+ - + = - +` `j j8 B

= 3

12 2235

325- + =8 B .

Thus, the required equation is, x2 - (Sum of the roots) x + Product of the roots = 0

& (6)x x3252

- + = 0



9. Find a quadratic equation whose roots are the reciprocal of the roots of the equation 4 3 1x x

2- - = 0.

Solution: x x4 3 1 02- - = . It is in the form of ax bx c 02

+ + = .

Here, , ,a b c4 3 1= =- =- . Let a and b be the roots of x x4 3 1 02- - = .

Thus, a b+ = ( )ab

43

43- =-

-= and ab =

ac

41= -

Let us form an equation whose roots are 1a

and 1b

.

Sum of the roots, 31 1

4143

a b aba b

+ =+

=-

=-


411 4

a b ab= =

-=-` cj m

Now, the required equation is, x2 - (Sum of the roots) x + Product of the roots = 0

& ( 3) ( 4)x x2- - + - = 0

Thus, the equation is 3 4x x 02+ - = .

Note: Replace x by x1 in the equation x x4 3 1 02

- - = to get 3 4x x 02+ - = .

10. If one root of the equation 3 81x kx2+ - =0isthesquareoftheother,findk.

Solution: Let a and 2b a= be the roots of the equation x kx3 81 02+ - = .

It is in the form of ax bx c 02+ + = . Here, , ,a b k c3 81= = =- .

Now, sum of the roots a b+ = ab k

32

& a a- + = - g (1)

Product of the roots, ab = ( ) 27 ( 3)ac

3812 3 3

& &a a a= - =- = -

Thus, a = – 3

When 3a =- , (1) ( ) ( ) 18.k k3

3 3 2& &- = - + - =-

11. If one root of the equation x ax2 64 02- + = istwicetheother,thenfindthevalueofa.

Solution: Let a and 2b a= be the roots of the equation x ax2 64 02- + = .

It is in the form of 0Ax Bx C2+ + = . Here, , ,A B a C2 64= =- = .

Sum of the roots, a b+ = 2( )

AB a a

2 2& a a- + =-

-= .a

6& a = g (1)

Product of the roots, ab = (2 ) 2 32.AC

264 2

& &a a a= =

& a6

2` j = 16 (using (1))

Thus, 24.a !=

12. If a and b are the roots of 5 1x px2- + = 0 and a b- =1,thenfindp.

Solution: Given equation x px5 1 02- + = is of the form ax bx c 02

+ + = .Here, , ,a b p c5 1= =- = .


Now, a b+ = ab p

5- =` j g (1)

ab = 51 g (2)

Given that, a b- = 1 g (3)

Now, ( ) 42a b ab+ - = ( )2a b-

& p25 5

42

- = 1 ( using (1), (2) & (3) )

& p 202- = 25 & p2 = 45

Thus, p = 3 5! .

Exercise 3.19

Chose the correct answer.

1. If the system 6x – 2y = 3, kx – y = 2 has a unique solution, then

(A) k = 3 (B) k 3! (C) k = 4 (D) k 4!

Solution: ,x y kx y6 2 3 2- = - = . Here 6, 2, , 1a b a k b1 1 2 2= =- = =- .

For unique solution a

a

2

1 ! b

b

k6

12

2

1 & !-- k& ! 3 (Ans. (B) )

2. A system of two linear equations in two variables is inconsistent if their graphs

(A) coincide (B) intersect only at a point

(C) do not intersect at any point (D) cut the x-axis

Solution: If the system is inconsistent, it does not have solution. (Ans. (C) )

3. The system of equations x –4y = 8 , 3x –12y =24

(A) has infinitely many solutions (B) has no solution (C) has a unique solution (D) may or may not have a solution

Solution: , , , , ,a b c a b c1 4 8 3 12 241 1 1 2 2 2= =- =- = =- =- .

a

a

b

b

2

1

2

1= =c

c

31

2

1 = . (Here, the two equations are identical) (Ans. (A) )

4. If one zero of the polynomial p x^ h = (k +4) x2+13x+3k is reciprocal of the other, then

k is equal to (A) 2 (B) 3 (C) 4 (D) 5

Solution: ( )k x x k4 13 32+ + + . Here, 4, 3, 3a k b c k1= + = = .

Let a and 1a

be the zeros of the polynomial. ( )( ) .k

k k14

3 2&aa

=+

= (Ans. (A) )


5. The sum of two zeros of the polynomial 2 ( 3) 5f x x p x2

= + + +^ h is zero, then the value of p is

(A) 3 (B) 4 (C) –3 (D) –4Solution: ( 3) 5x p x2 2

+ + + . Here , ,a b p c2 3 5= = + = .

Sum of zeros = ( )0 3 0.

ab p

p2

3& &-

- += + = (Ans. (C) )

6. The remainder when x x2 72- + is divided by x+4 is

(A) 28 (B) 29 (C) 30 (D) 31Solution: By remainder theorem,

( )f 4- = ( ) ( )4 2 4 7 16 8 7 312- - - + = + + = (Ans. (D) )

7. The quotient when 5 7 4x x x3 2- + - is divided by x–1 is

(A) 4 3x x2+ + (B) 4 3x x

2- + (C) 4 3x x

2- - (D) 4 3x x

2+ -

Solution: By synthetic division,1 1 – 5 7 – 4

0 1 – 4 3Quotient " 1 – 4 3 – 1 " Remainder

(Ans. (B) ) 8. The GCD of x 1

3+^ h and 1x

4- is

(A) x 13- (B) 1x

3+ (C) x +1 (D) x 1-

Solution: Let ( )f x = ( ) ( )( )x x x x1 1 13 2+ = + - +

( )g x = ( )( )( )x x x x1 1 1 14 2- = + - + (Ans. (C) )

9. The GCD of 2x xy y2 2- + and x y

4 4- is

(A) 1 (B) x+y (C) x–y (D) x y2 2-

Solution: Let ( )f x = ( )x xy y x y22 2 2- + = -

( )g x = ( )( )( )x y x y x y x y4 4 2 2- = - + + (Ans. (C) )

10. The LCM of x a3 3- and (x – a)2 is

(A) ( )x a x a3 3- +^ h (B) ( )x a x a

3 3 2- -^ h

(C) x a x ax a2 2 2- + +^ ^h h (D) x a x ax a2 2 2

+ + +^ ^h h

Solution: Let ( )f x = ( )( )x a x a x ax a3 3 2 2- = - + +

( )g x = ( ) ( )x a x a2 2- = - (Ans. (C) )

Thus, LCM = x a x ax a( ) ( )2 22- + +

11. The LCM of , ,a a ak k k3 5+ + , where k Ne is

(A) ak 9+

(B) ak

(C) ak 6+

(D) ak 5+

Solution: Let ( )f x = ; ( ) .a g x a a ak k k3 3#= =+

( )h x = a a ak k5 5#=+ (Ans. (D) )

Thus, LCM is a 5k+ .


12. The lowest form of the rational expression 6x x

x x5 62

2

- -

+ + is

(A) xx

33

+- (B)

xx

33

-+ (C)

xx

32

-+ (D)

xx

23

+-

Solution: x x

x x

6

5 62

2

- -

+ + = ( )( )( )( )x xx x

3 23 2

- ++ + (Ans. (B) )

13. If a ba b-+ and

a b

a b3 3

3 3

+

- are the two rational expressions, then their product is

(A) a ab b

a ab b2 2

2 2

- +

+ + (B) a ab b

a ab b2 2

2 2

+ +

- + (C) a ab b

a ab b2 2

2 2

+ +

- - (D) a ab b

a ab b2 2

2 2

- -

+ +

Solution: a ba b

a b

a b3 3

3 3

#-+

+

- = ( )( )

( )( )

( )( )a ba b

a b a ab b

a b a ab b2 2

2 2

#-+

+ - +

- + + (Ans. (A) )

14. On dividing x

x325

2

+- by

9x

x 52-

+ is equal to

(A) (x –5)(x–3) (B) (x –5)(x+3) (C) (x +5)(x–3) (D) (x +5)(x+3)

Solution: x

x

x

x325

9

52

2'

+-

-

+ = ( )( ) ( )( )x

x xx

x x3

5 55

3 3#

++ -

++ - (Ans. (A) )

15. If a ba3

- is added with

b ab3

-, then the new expression is

(A) a ab b2 2+ + (B) a ab b

2 2- + (C) a b

3 3+ (D) a b

3 3-

Solution: a ba

b ab3 3

-+

-= ( )( )

a ba

a bb

a ba b a ab b3 3 2 2

--

-=

-- + + (Ans. (A) )

16. The square root of 49 ( 2 )x xy y2 2 2- + is

(A) 7 x y- (B) 7 x y x y+ -^ ^h h (C) 7( )x y2

+ (D) 7( )x y2

-

Solution: ( )x xy y49 22 2 2- + = ( ) ( )x y x y7 72 4 2

- = - ( Ans: (D))

17. The square root of 2 2 2x y z xy yz zx2 2 2+ + - + -

(A) x y z+ - (B) x y z- + (C) x y z+ + (D) x y z- -

Solution: x y z xy yz zx2 2 22 2 2+ + - + -

= ( ) ( ) ( )( ) ( )( ) ( )( )x y z x y x2 5 2 2 2 22 2 2+ - + - + - + - - + -

= ( ) | |x y z x y z2- - = - - ( Ans: (D))

18. The square root of 121 ( )x y z l m4 8 6 2

- is

(A) 11x y z l m2 4 4

- (B) 11 ( )x y z l m34 4

-

(C) 11x y z l m2 4 6

- (D) 11 ( )x y z l m32 4

-

Solution: ( )x y z l m121 4 8 6 2- = ( ) | |x y z l m x y z l m11 112 4 8 6 2 2 4 3

- = -

( Ans: (D))


19. If ax bx c 02+ + = has equal roots, then c is equal

(A) ab2

2

(B) ab4

2

(C) ab2

2

- (D) ab4

2

-

Solution: Equation has equal roots b ac42& - = 0

b2 = 4ac cab4

2& = ( Ans: (B))

20. If 5 16 0x kx2+ + = has no real roots, then

(A) k582 (B) k

582-

(C) k58

581 1- (D) k0

581 1

Solution: x kx5 16 02+ + = . Here , ,a b k c1 5 16= = = .

Equation has no real roots b ac4 0<2& -

( ) ( )( )k5 4 1 162- < 0 25 64

58k k< <2 2 2

& & ` j k58

58< <& - . ( Ans: (C))

21. A quadratic equation whose one root is 3 , is

(A) 6 0x x 52- - = (B) 0x x6 5

2+ - =

(C) 0x x5 62- - = (D) 0x x5 6

2- + =

Solution: (A) and (B) cannot be factorised.

(C) 5 6x x2- - = ( )( )x x6 1- + (D) 5 6x x2

- + = ( )( )x x3 2- - . ( Ans: (D)) 22. The common root of the equations 0x bx c

2- + = and x bx a 0

2+ - = is

(A) b

c a2+ (B)

bc a2- (C)

ac b2+ (D)

ca b2+

Solution: b c2a a- + = b a2a a+ -

Thus, a = b

c a2+ ( Ans: (A))

23. If ,a b are the roots of ax bx c 02

+ + = ,a 0=Y then the wrong statement is

(A) a

b ac22

2

22

a b+ = - (B) acab =

(C) aba b+ = (D)

cb1 1

a b+ =-

Solution: a b+ = ;ab

acab- = ( Ans: (C))

24. If anda b are the roots of ax bx c 02+ + = , then one of the quadratic

equations whose roots are 1 1anda b

, is (A) ax bx c 02

+ + = (B) 0bx ax c2+ + =

(C) 0cx bx a2+ + = (D) 0cx ax b2

+ + =

Solution: Replace x by x1 in ax bx c 02

+ + = , to get cx bx a 02+ + = ( Ans: (C))

25. Let b = a + c . Then the equation 0ax bx c2+ + = has equal roots, if

(A) a c= (B) a c=- (C) a c2= (D) a c2=-

Solution: 4 4 0 .b ac a c ac a c a c2 2 2& & &= + = - = =^ ^h h ( Ans: (A))

Solution - Matrices 115

Exercise 4.1 1. The rates for the entrance tickets at a water theme park are listed below:

Week Days rates (`)

Week End rates (`)

Adult 400 500Children 200 250

Senior Citizen 300 400 Write down the matrices for the rates of entrance tickets for adults, children and

senior citizens. Also find the dimensions of the matrices.Solution: In Matrix form, the rates of entrance tickets for adults, children and senior citizens is given in the following two ways.

(i) A = 400

200

300

500

250

400

f p. The dimension (order) of A is 3 # 2

(ii) B = 400

500

200

250

300

400c m. The dimension (order) of B is 2 # 3

2. There are 6 Higher Secondary Schools, 8 High Schools and 13 Primary Schools in a town. Represent these data in the form of 3 1# and 1 3# matrices.

Solution: We can represent the given data in the form of 3 # 1 matrix as A = 6

8

13

f p

We can also represent the given data in the form of 1 # 3 matrix as B = 6 8 13^ h

3. Find the order of the following matrices.

(i) 1

2

1

3

5

4-

-e o (ii)

7

8

9

f p (iii) 3

6

2

2

1

4

6

1

5

-

-f p (iv) 3 4 5^ h (v)

1

2

9

6

2

3

7

4

-

J

L

KKKKK

N

P

OOOOO

Solution: (i) The matrix 1

2

1

3

5

4-

-e o has 2 rows and 3 columns.

Hence, the order of the matrix is 2 # 3.

(ii) The given matrix 7

8

9

f p has 3 rows and 1 column.

Hence, the order of the matrix is 3 1.#

(iii) The given matrix 3

6

2

2

1

4

6

1

5

-

-f p has 3 rows and 3 columns.

Hence, the order of the matrix is 3 # 3.

Matrices 4


(iv) The given matrix 3 4 5^ h has 1 row and 3 columns. Hence, the order of the matrix is 1 # 3.

(v) Since the given matrix

1

2

9

6

2

3

7

4

-

J

L

KKKKK

N

P

OOOOO has 4 rows and 2 columns, the order is 4 # 2.

4. A matrix has 8 elements. What are the possible orders it can have?

Solution: The possible orders of the matrices having 8 elements are

1 # 8, 2 # 4, 4 # 2 and 8 # 1.

5. A matrix consists of 30 elements. What are the possible orders it can have?

Solution: The possible orders of the matrices having 30 elements are

1 # 30, 2 # 15, 3 # 10, 5 # 6, 6 # 5, 10 # 3, 15 # 2 and 30 # 1.

6. Construct a 2 2# matrix A aij

= 6 @ whose elements are given by

(i) a ijij= (ii) 2a i j

ij= - (iii) a

i ji j

ij=

+-

Solution: In general a 2 # 2 matrix is given by A = a

a

a

a11

21

12

22

e o

(i) Now, a iji j

= , where i = 1, 2 and j = 1, 2

(1)(1) 1, (1)(2) 2, (2)(1) 2, (2)(2) 4a a a a11 12 21 22= = = = = = = =

Hence, the required matrix A = 1

2

2

4c m

(ii) Now, 2 –a i ji j= , where i = 1, 2 and j = 1, 2

2(1) 1 1, 2(1) 2 0, 2(2) 1 3, 2(2) 2 2a a a a11 12 21 22= - = = - = = - = = - =


0

2c m

(iii) Now, ai ji j

i j=

+- , where i = 1, 2 and j = 1, 2

0, , , 0a a a a1 11 1

20

1 21 2

31

2 12 1

31

2 22 2

11 12 21 22=

+- = = =

+- =- =

+- = =

+- =


31

31

0

-J

L

KKK

N

P

OOO


= 6 @ whose elements are given by

(i) aji

ij= (ii) ( )

ai j

22

ij

2

=- (iii) a i j

2

2 3ij=

-

Solution: In general, a 3 × 2 matrix is given by A = a

a

a

a

a

a

11

21

31

12

22

32

f p


(i) Now, aji

i j= , where i = 1, 2, 3 and j = 1, 2

,a a11 1

21

11 12= = = , ,a a12 2

22 121 22= = = = , ,a a

13 3

23

31 32= = =

Hence, the required matrix A =

1

2

3

21

1

23

J

L

KKKKKK

N

P

OOOOOO

(ii) Now, a i j

2

2ij

2

=-^ h , where i = 1, 2, 3 and j = 1, 2

,a2

1 2

21

11

2

=-

=^ h a

2

1 4

29

12

2

=-

=^ h , 0,a

2

2 221

2

=-

=^ h

2a2

2 4

24

22

2

=-

= =^ h , ,a

2

3 2

21

31

2

=-

=^ h a

2

3 4

21

32

2

=-

=^ h

Hence, the required matrix A = 0 2

21

21

29

21

J

L

KKKKK

N

P

OOOOO

(iii) Now, 2 3a

i j

2ij=

- , where i = 1, 2, 3 and j = 1, 2

| | | | ,a2

2 321

21

11= - = - = | | | | 2a

22 6

24

12= - = - = , | | ,a

24 3

21

21= - =

| | 1a2

4 622

22= - = = , | | ,a

26 3

23

31= - = | | 0a

26 6

32= - =

Hence, the required matrix A =

2

1

0

21

21

23

J

L

KKKKKK

N

P

OOOOOO

8. If A1

5

6

1

4

0

3

7

9

2

4

8

=

-

-f p, (i) find the order of the matrix (ii) write down the elements

a24

and a32

(iii) in which row and column does the element 7 occur?

Solution: A = 1

5

6

1

4

0

3

7

9

2

4

8

-

-f p

(i) Since A has 3 rows and 4 columns, the order of A is 3 × 4 (ii) a24 = 4, which occurs in the second row and fourth column a32 = 0, which occurs in the third row and second column. (iii) The element 7 occurs in the second row and third column. i.e. 7a

23= .

9. If A

2

4

5

3

1

0

= f p, then find the transpose of A.


Solution:

The transpose AT of a matrix A, is obtained by interchanging rows and columns of A.

Thus, AT = 2

3

4

1

5

0c m

10. If A

1

2

3

2

4

5

3

5

6

=

-

-f p, then verify that ( )A AT T

= .

Solution: A = 1

2

3

2

4

5

3

5

6-

-f p. ... (1)

By interchanging rows and columns of the matrix A, we get

AT = 1

2

3

2

4

5

3

5

6-

-f p

Again by interchanging rows and columns of the matrixAT , we get

(AT)T = 1

2

3

2

4

5

3

5

6-

-f p ... (2)

From (1) and (2), we have (AT)T = A

Exercise 4.2 1. Find the values of x, y and z from the matrix equation

x y

z

5 2

0

4

4 6

12

0

8

2

+ -

+=

-e co m

Solution: Since the matrices are equal, the corresponding elements are equal.Equating the corresponding elements, we get 5 2 12x + = & 5 10 2x x&= =

4 8y - =- & 4y =-

4 6 2z + = & 4 4 1z z&=- =-

Thus, 2, 4, 1x y z= =- =-

2. Solve for x and y if x y

x y

2

3

5

13

+

-=e co m.

Solution: Since the matrices are equal, the corresponding elements are equal. Comparing the corresponding elements, we get

2 5 2 5 0x y x y&+ = + - = and 3 13 3 13 0x y x y&- = - - =

Solving the equations by the method of cross multiplication, we get

x y13 15 5 26 6 1

1- -

=- +

=- -

x y28 21 7

1&-

= =-

Thus, 4, 3x y728

721=

-- = =

-=- .


3. If A2

9

3

5

1

7

5

1=

--

-e eo o, then find the additive inverse of A.

Solution: A = 2

9

3

5

1

7

5

1--

-e eo o 2

9

3

5

1

7

5

1=

-+

-

-

-e eo o

= 2 1

9 7

3 5

5 1

1

16

2

6

-

- -

-

+=

-

-e eo o

The additive inverse of A is –A. Hence, the additive inverse is

A- = 1

16

2

6

1

16

2

6-

-

-=

-

-e eo o

4. Let A3

5

2

1= c m and B 8

4

1

3=

-c m. Find the matrix C if 2C A B= + .

Solution: Given that A3

5

2

1= c m and B 8

4

1

3=

-c m.

Now, C = 2A + B

= 23

5

2

1

8

4

1

3

6

10

4

2

8

4

1

3+

-= +

-c c c cm m m m

= 6 8

10 4

4 1

2 3

14

14

3

5

+

+

-

+=e co m

5. If A B4

5

2

9

8

1

2

3and=

-

-=

- -e eo o, find 6 3A B- .

Solution: Given that A4

5

2

9=

-

-e o and B 8

1

2

3=

- -e o

Now, A B6 3- = 6 34

5

2

9

8

1

2

3

-

--

- -e eo o

= 24

30

12

54

24

3

6

9

24 24

30 3

12 6

54 9

0

33

18

45

-

-+

- -=

-

+

- -

- +=

-

-e c e eo m o o

6. Find a and b if a b2

3

1

1

10

5+

-=c c cm m m.

Solution: Given that a b2

3

1

1

10

5+

-=c c cm m m

a

a

b

b

a b

a b

2

3

10

5

2

3

10

5& &+

-=

-

+=c c c e cm m m o m

Comparing the corresponding elements, we have

2a – b = 10 g (1)

3a + b = 5 g (2)

Adding (1) and (2), we get 5a = 15 & a = 3.

Substituting a = 3 in (2), we get 9 + b = 5 & b = – 4

Thus, 3, 4a b= =- .


7. Find X and Y if 2 3X Y2

4

3

0+ = c m and 3 2X Y

2

1

2

5+ =

-

-e o.

Solution: X Y2 3+ = 2

4

3

0c m ... (1)

X Y3 2+ = 2

1

2

5-

-e o ... (2)

We first eliminate Y. Now,

X Y1 2 4 6&# +^ h = 22

4

3

0

4

8

6

0=c cm m ... (3)

3 9 6X Y2 &# +^ h = 32

1

2

5

6

3

6

15-

-=

-

-e eo o ... (4)

Subtracting (3) from (4), we get

5X = 6

3

6

15

4

8

6

0-

--e co m

= 6

3

6

15

4

8

6

0

6

3

6

15

4

8

6

0-

-- =

-

-+

-

-

-e c e eo m o o

X5 = X2

11

12

1552

511

512

3&

-

-=

-

-J

L

KKK

e

N

P

OOO

o

Substituting X in (1), we get

Y3 = 2 2X2

4

3

0

2

4

3

0 3

2

4

3

0 652

511

512

54

522

524

- = --

-= +

-

-c c f c fm m p m p

= 2

4

3

0 654

522

524-

+

+

-f p =

656

542

539

-f p

Thus, Y = –3

1 56

542

539

6

J

L

KKK

N

P

OOO =

–52

514

513

2

J

L

KKK

N

P

OOO

8. Solve for x and y if 3x

y

x

y

2 9

4

2

2 +-

=-

e e co o m.

Solution: Given that 3–

x

y

x

y

2 9

4

2

2 + =-

e e co o m

–x

y

x

y

6

3

9

4

2

2& +

-=e e co o m x x

y y

6

3

9

4

2

2&

+

-=

-e co m

Equating the corresponding elements, we get

–x x6 92+ = y y3 42

- =

x x6 9 02& + + = 3 4 0y y2

& - - =


x 3 02& + =^ h –y y1 4 0& + =^ ^h h

,x 3 3& =- - ,–y 1 4& =

9. If ,A B O3

5

2

1

1

2

2

3

0

0

0

0and= =

-=c c cm m m, then verify: (i) A B B A+ = +

(ii) ( ) ( )A A O A A+ - = = - + .

Solution: Given that ,A B O3

5

2

1

1

2

2

3

0

0

0

0and= =

-=c c cm m m

(i) A + B = 3

5

2

1

1

2

2

3+

-c cm m = 3 1

5 2

2 2

1 3

4

7

0

4

+

+

-

+=e co m g (1)

B + A = 1

2

2

3

3

5

2

1

-+c cm m = 1 3

2 5

2 2

3 1

4

7

0

4

+

+

- +

+=e co m g (2)

From (1) and (2), we have A B B A+ = +

(ii) A + (–A) = 3

5

2

1

3

5

2

1

3 3

5 5

2 2

1 1+

-

-

-

-=

-

-

-

-c e em o o = O

0

0

0

0=c m g (3)

(–A) + A = 3

5

2

1

3

5

2

1

3 3

5 5

2 2

1 1

-

-

-

-+ =

- +

- +

- +

- +e c eo m o = O

0

0

0

0=c m g (4)

From (3) and (4), we have A+(–A) = (–A) + A = O

10. If ,A B

4

1

0

1

2

3

2

3

2

2

6

2

0

2

4

4

8

6

= - =f fp p and C1

5

1

2

0

1

3

2

1

=

-

-

f p,

then verify that ( ) ( )A B C A B C+ + = + + .Solution: We have

B + C = 2

6

2

0

2

4

4

8

6

1

5

1

2

0

1

3

2

1

+

-

-

f fp p =2 1

6 5

2 1

0 2

2 0

4 1

4 3

8 2

6 1

3

11

3

2

2

3

1

10

7

+

+

+

+

+

-

-

+

+

=f fp p

Thus, ( )A B C+ + = 4

1

0

1

2

3

2

3

2

3

11

3

2

2

3

1

10

7

- +f fp p

= 4 3

1 11

0 3

1 2

2 2

3 3

2 1

3 10

2 7

+

+

+

+

- +

+

+

+

+

f p = 7

12

3

3

0

6

3

13

9

f p g (1)

Now, A + B = 4

1

0

1

2

3

2

3

2

2

6

2

0

2

4

4

8

6

- +f fp p = 4 2

1 6

0 2

1 0

2 2

3 4

2 4

3 8

2 6

6

7

2

1

0

7

6

11

8

+

+

+

+

- +

+

+

+

+

=f fp p

Thus, (A + B) + C = 6

7

2

1

0

7

6

11

8

1

5

1

2

0

1

3

2

1

+

-

-

f fp p


= 6 1

7 5

2 1

1 2

0 0

7 1

6 3

11 2

8 1

7

12

3

3

0

6

3

13

9

+

+

+

+

+

-

-

+

+

=f fp p g (2)

From (1) and (2), we have ( ) ( )A B C A B C+ + = + + .

11. An electronic company records each type of entertainment device sold at three of their branch stores so that they can monitor their purchases of supplies. The sales in two weeks are shown in the following spreadsheets.

T.V. DVD Video games CD Players

Week IStore I 30 15 12 10Store II 40 20 15 15Store III 25 18 10 12

Week IIStore I 25 12 8 6Store II 32 10 10 12Store III 22 15 8 10

Find the sum of the items sold out in two weeks using matrix addition.Solution: The number of items sold by three stores during week I in matrix form is

TV DVD Video CD

A = 30

40

25

15

20

18

12

15

10

10

15

12

f p Store I

Store II

Store III

Similarly, the number of items sold by three stores during week II in matrix form is

TV DVD Video CD

B = 25

32

22

12

10

15

8

10

8

6

12

10

f p Store I

Store II

Store III

Thus, the matrix representation of the sum of the items sold by three stores during week I and II is

A B+ = 30

40

25

15

20

18

12

15

10

10

15

12

25

32

22

12

10

15

8

10

8

6

12

10

+f fp p

= 30 25

40 32

25 22

15 12

20 10

18 15

12 8

15 10

10 8

10 6

15 12

12 10

+

+

+

+

+

+

+

+

+

+

+

+

f p

TV DVD Video CD

= 55

72

47

27

30

33

20

25

18

16

27

22

f p Store I

Store II

Store III


12. The fees structure for one-day admission to a swimming pool is as follows:Daily Admission Fees in `

Member Children Adult Before 2.00 p.m. 20 30 After 2.00 p.m. 30 40

Non-Member Before 2.00 p.m. 25 35 After 2.00 p.m. 40 50

Write the matrix that represents the additional cost for non-membership.

Solution: Let the matrix A represents the fee structure for membership for one-day admission to a swimming pool. Then, Children Adult

A = 20

30

30

40c m Before 2pm

After 2pm Let the matrix B represents the fee structure for non-membership for one-day admission to a swimming pool. Then, Children Adult

B = 25

40

35

50c m Before 2pm

After 2pmHence, the matrix that represents the additional cost for non-membership is given by

B – A = –25

40

35

50

20

30

30

40c cm m = 25 20

40 30

35 30

50 40

-

-

-

-e o

Children Adult

= 5

10

5

10c m Before pm

After pm

2

2

Exercise 4.3 1. Determine whether the product of the matrices is defined in each case. If so, state

the order of the product. (i) AB, where ,A a B b

x xij ij4 3 3 2= =6 8@ B (ii) PQ, where ,P p Q q

x xij ij4 3 4 3= =6 6@ @

(iii) MN, where ,M m N nx xij ij3 1 1 5

= =6 6@ @ (iv) RS, where ,R r S sx xij ij2 2 2 2

= =6 6@ @

Solution: (i) Now, the number of columns in A and the number of rows in B are equal So, the product AB is defined and the order of AB is 4 × 2. (ii) Now, the number of columns in P and the number of rows in Q are not equal. So, the product PQ is not defined. (iii) Now, the number of columns in M and the number of rows in N are equal. So, the product MN is defined and the order of MN is 3×5. (iv) Now, the number of columns in R and the number of rows in S are equal. So, the product RS is defined and the order of RS is 2 × 2.


2. Find the product of the matrices, if exists,

(i) 2 15

4-^ ch m (ii) 3

5

2

1

4

2

1

7

-c cm m

(iii) 2

4

9

1

3

0

4

6

2

2

7

1-

--

-

e fo p (iv) 6

32 7

--e ^o h

Solution: (i) Let A = (2 –1) and B = 5

4c m.

The order of A is 1 × 2 and the order of B is 2 × 1.The number of columns of A is equal to the number of rows of B.

Hence, the product AB is defined and AB = (2 –1) 5

4c m = ( (2)(5) + (–1)4 ) = ( 6 ).

(ii) Let A = 3

5

2

1

-c m and B = 4

2

1

7c m.

The order of A is 2 × 2 and the order of B is 2 × 2.The number of columns of A is equal to the number of rows of B.

Hence, the product AB is defined and AB = 3

5

2

1

4

2

1

7

-c cm m

= ( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

3 4

5 4

2 2

1 2

3 1

5 1

2 7

1 7

+

+

- +

+

-e o = 12 4

20 2

3 14

5 7

-

+

-

+e o = 8

22

11

12

-c m

(iii) Let A = 2

4

9

1

3

0-

-e o and B =

4

6

2

2

7

1

-

-

f p.

The order of A is 2 × 3 and the order B is 3 × 2. The number of columns of A is equal to the number of rows of B. Thus, the product AB is defined and

AB = 2

4

9

1

3

0

4

6

2

2

7

1-

--

-

e fo p = 8 54 6

16 6 0

4 63 3

8 7 0

- +

+ -

+ -

- +e o = 40

22

64

1

-c m

(iv) Let A = 6

3-e o and B = (2 – 7).

The order of A is 2 × 1 and the order of B is 1 × 2.The number of columns of A is equal to the number of rows of B.Thus, the product AB is defined and

AB = 6

32 7

6 2 6 7

3 2 3 7

12 42

6 21

# #

# #-- =

-

- - -=

-

-c ^ ^

^ ^ ^e cm h h

h h ho m

3. A fruit vendor sells fruits from his shop. Selling prices of Apple, Mango and Orange are ` 20, ` 10 and ` 5 each respectively. The sales in three days are given below

Day Apples Mangoes Oranges 1 50 60 302 40 70 203 60 40 10

Write the matrix indicating the total amount collected on each day and hence find the total amount collected from selling of all three fruits combined.


Solution: Let A be the matrix that represents the selling prices of Apple, Mango and Orange in `. Prices

Then, A = 20

10

5

f p Apple

Mango

Orange

Let B be the matrix that represents the sales in three days. Apple Mango Orange

Thus, B = 50

40

60

60

70

40

30

20

10

f p Day

Day

Day

1

2

3

Thus, the total amount collected on each day is given by the matrix T = BA.

Now, T = BA & T = 50

40

60

60

70

40

30

20

10

20

10

5

f fp p

= 1000

800

1200

600

700

400

150

100

50

+

+

+

+

+

+

f p= 1750

1600

1650

f p Day

Day

Day

1

2

3

Hence, the total amount collected = ` (1750+1600+1650) = ` 5000Aliter: The total sales of three fruits can be represented by the matrix C = ( 50+40+60 60+70+40 30+20+10 ) = ( 150 170 60 ) Thus, the total amount collected from selling of all fruits is given by T = CA

= (150 170 60) 20

10

5

f p = (3000 + 1700 + 300) = (5000)

4. Find the values of x and y if x

y

x1

3

2

3 0

0

9

0

0=c c cm m m.

Solution: x

y

x1

3

2

3 0

0

9

0

0=c c cm m m x

x

y

y

0

3 0

0 2

0 3&

+

+

+

+e o = x

9

0

0c m x

x

y

y3

2

3& c m = x

9

0

0c m

Comparing the corresponding elements, we get

3x = 9 x 3& = and 2y = 0 y 0& = .

5. If ,A Xx

yC

5

7

3

5

5

11and= = =

-

-c c em m o and if AX C= , then find the values of

x and y.

Solution: Given A = 5

7

3

5c m, X = x

yc m and C = 5

11

-

-e o

Now, AX = C x

y

5

7

3

5

5

11& =

-

-c c em m o x

x

y

y

5

7

3

5&

+

+e o = 5

11

-

-e o


Comparing the corresponding elements, we get

5x + 3y = – 5 & 5x + 3y + 5 = 0 7x + 5y = –11 & 7x + 5y + 11 =0 Solving the equations by the method of cross multiplication, we get x y 1

3

5

5

11

5

7

3

5

& x y33 25 35 55 25 21

1-

=-

=-

& x y8 20 4

1=-

= 2, 5x y48

420& = = = - =-

6. If A1

2

1

3=

-c m then show that 4 5A A I O

2

2- + = .

Solution: Now, A2 = A.A

= 1

2

1

3

1

2

1

3

- -c cm m = 1 2

2 6

1 3

2 9

-

+

- -

- +e o = 1

8

4

7

- -c m

A2 – 4A + 5I2 = 5–1

8

4

741

2

1

3

1

0

0

1

- - -+c c cm m m

= 1

8

4

7

4

8

4

12

5

0

0

5

- -+

-

- -+c e cm o m

= 1 4 5

8 8 0

4 4 0

7 12 5

- - +

- +

- + +

- +e o = 0

0

0

0c m = O

7. If A B3

4

2

0

3

3

0

2and= =c cm m then find AB and BA. Are they equal?

Solution: A and B are square matrices of same order 2 × 2.

Hence, the products AB and BA are defined.

Now, AB = 3

4

2

0

3

3

0

2c cm m = 9 6

12 0

0 4

0 0

+

+

+

+e o = 15

12

4

0c m g (1)

BA = 3

3

0

2

3

4

2

0c cm m = 9 0

9 8

6 0

6 0

+

+

+

+e o = 9

17

6

6c m g (2)

From (1) and (2), we have AB ! BA.

8. If ,A B C1

1

2

2

1

3

0

1

2

2 1and=-

= =c f ^m p h, then verify ( ) ( )AB C A BC= .

Solution: A is of order 2 × 3 , B is of order 3 × 1 and C is of order 1 × 2.

Hence, AB is of order 2 × 1 and BC is of order 3 × 2.


Now, AB = 1

1

2

2

1

3

0

1

2

-c fm p = 0 2 2

0 2 6

4

8

+ +

+ +=e co m

Thus, (AB)C = 4

8c m (2 1) = 8

16

4

8c m g (1)

Now, BC = 0

1

2

f p (2 1) = 0

2

4

0

1

2

f p

Thus, A(BC) = 1

1

2

2

1

3

0

2

4

0

1

2

-c fm p = 0 4 4

0 4 12

0 2 2

0 2 6

+ +

+ +

+ +

+ +e o

= 8

16

4

8c m g (2)

From (1) and (2), we have ( ) ( )AB C A BC= .

Hence, matrix multiplication is associative.

9. If A B5

7

2

3

2

1

1

1and= =

-

-c em o verify that ( )AB B A

T T T= .

Solution: A = 5

7

2

3c m, B = 2

1

1

1-

-e o

A and B are square matrices of same order 2 × 2.

Hence, the product AB is defined.

Now, AB = 5

7

2

3

2

1

1

1-

-c em o = 10 2

14 3

5 2

7 3

-

-

- +

- +e o = 8

11

3

4

-

-e o

Thus, (AB)T = 8

3

11

4- -e o g (1)

Now, BT = 2

1

1

1-

-e o ; AT = 5

2

7

3c m

Thus, BT AT = 2

1

1

1

5

2

7

3-

-e co m = 10 2

5 2

14 3

7 3

-

- +

-

- +e o

= 8

3

11

4- -e o g (2)

From (1) and (2), we have ( )AB B AT T T= .

10. Prove that A B5

7

2

3

3

7

2

5and= =

-

-c em o are inverses to each other under

matrix multiplication.

Solution: Given A = 5

7

2

3c m and B = 3

7

2

5-

-e o


A and B are square matrices of same order 2 × 2.

Hence, the products AB and BA are defined.

AB = 57

2

3

3

7

2

5-

-c em o = 15 14

21 21

10 10

14 15

-

-

- +

- +e o = 1

0

0

1c m = I ...(1)

BA = I3 2

7 5

5 2

7 3

15 14 6 6

35 35 14 15

1 0

0 1

-

-=

- -

- + - += =c c c cm m m m ...(2)

From (1) and (2), we have AB BA I= = .Thus, the given matrices are inverses to each other under matrix multiplication.

11. Solve: xx

11

2

0

3 50

- -=^ e c ^h o m h.

Solution: (0)xx

11

2

0

3 5- -=^ e ch o m & (x 1) x

x

0

2 15

+

- -e o = (0)

& (x 1) x

x2 15- -e o = (0) & ((x)(x) + (1) (–2x–15)) = (0)

& (x2 – 2x – 15) = ( 0 )Thus, x x2 15 0

2- - = & ( )( )x x3 5 0+ - = & ,x 3 5=- .

12. If A B1

2

4

3

1

3

6

2and=

-

-=

-

-e eo o , then prove that ( ) 2A B A AB B

2 2 2!+ + + .

Solution: Now, A + B = 1

2

4

3

1

3

6

2-

-+

-

-e eo o

= 1 1

2 3

4 6

3 2

0

1

2

1

-

- +

- +

-=e co m

Thus, (A + B)2 = (A+B) (A+B)

= 0

1

2

1

0

1

2

1c cm m = 0 2

0 1

0 2

2 1

+

+

+

+e o = 2

1

2

3c m g (1)

Now, A2 = AA = 1

2

4

3

1

2

4

3-

-

-

-e eo o

= 1 8

2 6

4 12

8 9

9

8

16

17

+

- -

- -

+=

-

-e eo o

Now, AB = 1

2

4

3

1

3

6

2-

- -

-e eo o

= 1 12

2 9

6 8

12 6

- -

+

+

- -e o = 13

11

14

18

-

-e o

2AB = 2 13

11

14

18

26

22

28

36

-

-=

-

-e eo o

Now, B2 = BB = 1

3

6

2

1

3

6

2

-

-

-

-e eo o


= 1 18

3 6

6 12

18 4

+

- -

- -

+e o = 19

9

18

22-

-e o

Now, 2A AB B2 2+ + = 9

8

16

17

26

22

28

36

19

9

18

22-

-+

-

-+

-

-e e eo o o

= 9 26 19

8 22 9

16 28 18

17 36 22

- +

- + -

- + -

- +e o = 2

5

6

3

-c m ... (2)

From (1) and (2) , we get ( )A B A AB B22 2 2]+ + +

13. If ,A B C3

7

3

6

8

0

7

9

2

4

3

6and= = =

-c c cm m m, find ( )A B C AC BCand+ + .

Is ( )A B C AC BC+ = + ?

Solution: A + B = 3 8

7 0

3 7

6 9

11

7

10

15

+

+

+

+=e co m

(A + B) C = 11

7

10

15

2

4

3

6

-c cm m = 22 40

14 60

33 60

21 90

+

+

- +

- +e o

= 62

74

27

69c m g (1)

AC = 3

7

3

6

2

4

3

6

-c cm m = 6 12

14 24

9 18

21 36

+

+

- +

- +e o = 18

38

9

15c m

BC = 8

0

7

9

2

4

3

6

-c cm m = 16 28

0 36

24 42

0 54

+

+

- +

+e o = 44

36

18

54c m

AC + BC = 18

38

9

15c m+ 44

36

18

54c m = 62

74

27

69c m g (2)

From (1) and (2) , we have ( )A B C AC BC+ = + .

Exercise 4.4Choose the correct answer.

1. Which one of the following statements is not true?

(A) A scalar matrix is a square matrix (B) A diagonal matrix is a square matrix

(C) A scalar matrix is a diagonal matrix (D) A diagonal matrix is a scalar matrix.Solution: A scalar matrix is a diagonal matrix. But a diagonal matrix need not be a scalar matrix. ( Ans. (D) )

2. Matrix A aij m n

=#

6 @ is a square matrix if

(A) m n1 (B) m n2 (C) m 1= (D) m = n

Solution: In a square matrix, number of rows and number of columns are equal.

Thus, A aij m n

=#

6 @ is a square matrix if m = n. ( Ans. (D) )


3. If x

y x

y3 7

1

5

2 3

1

8

2

8

+

+ -=

-e eo o then the values of x and y are respectively

(A) –2 , 7 (B) 31- , 7 (C)

31- ,

32- (D) 2 , –7

Solution: Since the matrices are equal, the corresponding elements are equal.

Thus, 3 7 1 2 ; 1 8 7x x y y& &+ = =- + = = ( Ans. (A) )

4. If A B1 2 3

1

2

3

and= - =

-

-

^ fh p then A + B

(A) 0 0 0^ h (B) 0

0

0

f p (C) 14-^ h (D) not defined

Solution: Addition is not defined for matrices of different orders. Since A is of order 1 × 3 and B is of order 3 × 1, A + B is not defined. ( Ans. (D) )

5. If a matrix is of order ,2 3# then the number of elements in the matrix is

(A) 5 (B) 6 (C) 2 (D) 3

Solution: A matrix of order m×n has mn elements. ( Ans. (B) )

6. If 4x

8 4

8

2

1

1

2=c cm m then the value of x is

(A) 1 (B) 2 (C) 41 (D) 4

Solution: x

8 4

8

8

4

4

2=c cm m.

Equating the corresponding elements, we get 4x = ( Ans. (D) )

7. If A is of order 3 4# and B is of order 4 3# , then the order of BA is

(A) 3 3# (B) 4 4# (C) 4 3# (D) not definedSolution:

Order of B is 4 3# . Order of A is 3 4# . Thus, the order of BA is 4 4# . ( Ans. (B) )

8. If A1

0

1

21 2# =c ^m h, then the order of A is

(A) 2 1# (B) 2 2# (C) 1 2# (D) 3 2#

Solution: ( )A1

0

1

21 2

m n2 2

1 2=

##

#c m

2 1.n mand& = = Thus, A is of order 1 2# . ( Ans. (C) )

9. If A and B are square matrices such that AB = I and BA = I , then B is

(A) Unit matrix (B) Null matrix

(C) Multiplicative inverse matrix of A (D) A-

Solution: By definition, if ,AB BA I= = then B is the multiplicative inverse of A. ( Ans. (C) )


10. If x

y

1

2

2

1

2

4=c c cm m m, then the values of x and y respectively, are

(A) 2 , 0 (B) 0 , 2 (C) 0 , 2- (D) 1 , 1

Solution: x

y

1

2

2

1

2

4=c c cm m m 2

2

x y

x y

2

4&

+

+=e co m

Thus, 2 2 (1) ; 2 4 (2)x y x yg g+ = + =

Solving (1) and (2), we get ; .x y2 0= = ( Ans. (A) )

Note: By direct substitution, we see that 2 , 0x y= = satisfy both the equations.

11. If A1

3

2

4=

-

-e o and A B O+ = , then B is

(A) 1

3

2

4-

-e o (B) 1

3

2

4

-

-e o (C) 1

3

2

4

-

-

-

-e o (D) 1

0

0

1c m

Solution: B A1

3

2

4

1

3

2

4=- =-

-

-=

-

-e eo o. ( Ans. (B) )

12. If A4

6

2

3=

-

-e o, then A2 is

(A) 16

36

4

9c m (B) 8

12

4

6

-

-e o (C) 4

6

2

3

-

-e o (D) 4

6

2

3

-

-e o

Solution: A AA4

6

2

3

4

6

2

3

16 12

24 18

8 6

12 9

4

6

2

32= =

-

-

-

-=

-

-

- +

- +=

-

-e e e eo o o o

( Ans. (D) ) 13. A is of order m n# and B is of order p q# , addition of A and B is possible only if

(A) m p= (B) n = q (C) n = p (D) m = p, n = q

Solution: Addition is defined only for matrices of same order.

Thus, m p= and n = q. ( Ans. (D) )

14. If ,a

1

3

2

2

1

5

0-=c e cm o m then the value of a is

(A) 8 (B) 4 (C) 2 (D) 11

Solution: 2 3

0

5

0

a a

1

3

2

2

1

5

0&

-=

-=c e c c cm o m m m

& a a2 3 5 4&- = = ( Ans. (B) ) 15. If A

a

c

b

a=

-e o is such that A I2

= , then

(A) 1 02a bc+ + = (B) 1 0

2a bc- + =

(C) 1 02a bc- - = (D) 1 0

2a bc+ - =


Solution: A I0

0 1

0

0

12

2

2 2&

a bc

bc a=

+

+=e co m

1 1 02 2 2& &a bc a bc+ = - - = ( Ans. (C) )

16. If A aij 2 2

=#

6 @ and ,a i jij= + then A =

(A) 1

3

2

4c m (B) 2

3

3

4c m (C) 2

4

3

5c m (D) 4

6

5

7c m

Solution: A a

a

a

a

2

3

3

4

11

21

12

22

= =c cm m ( Ans. (B) )

17. a

c

b

d

1

0

0

1

1

0

0

1

-=

-c c em m o, then the values of a, b, c and d respectively are

(A) , , ,1 0 0 1- - (B) 1, 0, 0, 1 (C) , , ,1 0 1 0- (D) 1, 0, 0, 0

Solution: a

c

b

d

1

0

0

1

- -=

-c em o. Equating the corresponding elements, we get

1, 0, 0 1a b c dand=- = = =- . ( Ans. (A) )

18. If A7

1

2

3= c m and A B

1

2

0

4+ =

-

-e o, then the matrix B =

(A) 1

0

0

1c m (B) 6

3

2

1-e o (C) 8

1

2

7

- -

-e o (D) 8

1

2

7-e o

Solution: B 1

2

0

4

7

1

2

3=

-

--e co m =

8

1

2

7

- -

-e o. ( Ans. (C) )

19. If 20x5 1

2

1

3

- =^ f ^h p h, then the value of x is

(A) 7 (B) 7- (C) 71 (D) 0

Solution: x5 1

2

1

3

20- =^ f ^h p h

(10 3) (20) 13 20 7x x x& &- + = - = =- ( Ans. (B) )

20. Which one of the following is true for any two square matrices A and B of same order?

(A) ( )AB A BT T T= (B) ( )A B A B

T T T T=

(C) ( )AB BAT= (D) ( )AB B A

T T T=

Solution: By Reversal law for transpose of matrices, ( )AB B AT T T= . ( Ans. (D) )

Solution - Coordinate Geometry 133

Exercise 5.1 1. Find the midpoint of the line segment joining the points (i) ,1 1-^ h and ,5 3-^ h (ii) ,0 0^ h and ,0 4^ h.

Solution: (i) Let ( , ) (1, 1), ( , ) ( , )x y x y 5 31 1 2 2= - = - .

Midpoint is , , , ( 2,1)x x y y2 2 2

1 521 3

24

221 2 1 2+ +

= - - + = - = -` ` `j j j

Thus, the midpoint of line segment joining the points is ( 2,1)- .

(ii) Let ( , ) ( , ), ( , ) ( , )x y x y0 0 0 41 1 2 2= =

Midpoint is , (0, 2)2

0 02

0 4+ + =` j

Thus the midpoint of line segment joining the points is ( , )0 2 .

2. Find the centroid of the triangle whose vertices are (i) , , , ,1 3 2 7 12 16and -^ ^ ^h h h (ii) , , , ,3 5 7 4 10 2and- - -^ ^ ^h h h

Solution: (i) Let ( , ) (1, 3), ( , ) (2, 7) ( , ) (12, 16)x y x y x yand1 1 2 2 3 3= = = -

Centroid is , ,x x x y y y3 3 3

1 2 123

3 7 161 2 3 1 2 3+ + + += + + + -` `j j (5, 2)= - .

(ii) Let ( , ) (3, 5), ( , ) ( 7, 4) ( , ) (10, 2)x y x y x yand1 1 2 2 3 3= - = - = -

Centroid is ,3

3 7 103

5 4 2- + - + -` j , ( , )36

33 2 1= - = -8 B .

3. The centre of a circle is at (-6, 4). If one end of a diameter of the circle is at the origin, then find the other end.

Solution: Let ( , )x y be the other end of the diameter.

Centre of the circle is the midpoint of the diameter.

Thus, , ( 6, 4)x y2

02

0+ += -c m . Equating x and y co-ordinates, we get

6 12x x2

0 &+ =- =- and 4 8.y

y2

0&

+= =

Thus, the other end of the diameter is ( , )12 8-

4. If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5), then find the third vertex of the triangle.Solution: Let the third vertex be ( , )x y . Given that Centroid of the triangle is ( , )1 3

Thus, , (1, 3)x y3

7 83

6 5- + + + +=c m , (1, 3)x y

31

311

& + +=c m

Equating x and y-coordinates, we get

Co-ordinate Geometry 5


1 2x x3

1 &+ = = and 3 2.y

y3

11&

+= =-

Thus, the third vertex of the triangle is (2, )2- .

5. Using the section formula, show that the points A(1,0), B(5,3), C(2,7) and D(-2, 4) are the vertices of a parallelogram taken in order.

Solution: We know that the diagonals of a parallelogram bisect each other.

Now the midpoint of AC is , , .2

1 22

0 723

27+ + =` `j j

Also the midpoint of BD is , ,2

5 22

3 423

27- + =` `j j.

Thus the midpoints of AC and BD coincide.

Hence, ABCD is a parallelogram.

6. Find the coordinates of the point which divides the line segment joining (3, 4) and (–6, 2) in the ratio 3 : 2 externally.

Solution: Let ( , )A 3 4 and ( , )B 6 2- be the given points.

Let ( , )P x y divide AB externally in the ratio 3 : 2.

Thus, ( , )( ) ( )

,( ) ( )

( 24, 2) .P x y P P3 2

3 6 2 33 2

3 2 2 4=

-- -

--

= - -c m

7. Find the coordinates of the point which divides the line segment joining (-3, 5) and (4, -9) in the ratio 1 : 6 internally.

Solution: Let ( 3, )A 5- and ( , )B 4 9- be the given points.

Let the point P(x, y) divide AB internally in the ratio 1 : 6.

Thus, the point P(x, y) = ,Pl m

lx mxl m

ly my2 1 2 1

++

++

c m.

Hence, ( , )( ) ( )

,( ) ( )

( 2, 3) .P x y P P1 6

1 4 6 31 6

1 9 6 5=

++ -

+- +

= -c m

8. Let A (-6,-5) and B (-6, 4) be two points such that a point P on the line AB satisfies AP =

92 AB. Find the point P.

Solution: Let P(x, y) be the point on AB such that AP AB92= .

Now, AP AB92= 9 2 2( )AP AB AP PB& = = +

7 2AP PB& = . Thus, PBAP

72= . : 2 : 7.AP PB` =

Hence, P divides AB internally in the ration 2 : 7.

By section formula, we have

( , )( ) ( )

,( ) ( )

, ( 6, 3)P x y2 7

2 6 7 62 7

2 4 7 59

12 429

8 35=+

- + -++ -

= - - - = - -c `m j

Thus, the point P(x, y) = P(–6, –3).


9. Find the points of trisection of the line segment joining the points A (2,-2) and ( , )B 7 4- .

Solution: Let P and Q be the points of trisection of AB such that AP PQ QB= =

Then the point P divides AB internally in the ratio 1:2 and Q divides AB internally in the ratio 2 : 1. Thus, by section formula, we have

( ) ( ),

( ) ( ), ( 1, 0) .P

1 21 7 2 2

1 21 4 2 2

37 4

34 4=

+- +

++ -

= - + - = -c `m j

Thus, the point P is (–1, 0).

Again by section formula,

( ) ( ),

( ) ( )( , ) .Q

2 12 7 1 2

2 12 4 1 2

4 2=+

- +++ -

= -c m

Thus, the point Q is (–4, 2).

Note: Clearly, the point Q is the midpoint of PB.

Hence, the point , ( 4, 2) .Q21 7

20 4= - - + = -` j

Similarly,wecanfindP, the midpoint of AQ.

10. Find the points which divide the line segment joining A(-4 ,0) and B (0,6) into four equal parts.

Solution: Let P, Q, R be the points which divide AB into four equal parts.

Now, the point Q is the midpoint of AB.

Thus, Q = 24 0 ,

20 6 ( , )2 3- + + = -` j .

Now, P is the midpoint of AQ.

Thus, P = , , ,24 2

20 3

26

23 3

23- - + = - = -` ` `j j j.

Since R is the midpoint of QB, R is , ,22 0

23 6 1

29- + + = -` `j j.

Hence, the required points are P( 3, ), ( 2,3), ( , ) .Q R23 1

29- - -

Note: The point P divides AB internally in the ratio 1:3 and the point R divides AB internallyintheratio3:1.Usingsectionformula,onecanfindthepointsP and R.

11. Find the ratio in which the x-axis divides the line segment joining the points (6, 4) and ( , )1 7- .Solution: Let ( , )A 6 4 and ( , )B 1 7- be the given points.

Let P(x, 0) divide AB internally in the ratio l : m

Using section formula, we have ( , 0)( ) ( )

,( ) ( )

P x Pl m

l ml m

l m1 6 7 4=

++

+- +

c m

Equating the y-coordinates, we get 0 7 4 .l ml m l m

ml7 4

74& &

+- + = - =- =

Thus, the x-axis divides the line segment in the ratio 4 : 7 internally.


12. In what ratio is the line joining the points (-5, 1) and (2 , 3) divided by the y-axis? Also, find the point of intersection.

Solution: Let ( , )A 5 1- and ( , )B 2 3 be the given points.

Let P(0, y) divide AB internally in the ratio l : m.

By section formula, we have (0, )( ) ( )

,( ) ( )

P y Pl m

l ml m

l m2 5 3 1=

++ -

++

c m ... (1)

Equating the x-coordinates, we get 0 2 5 0 .l ml m l m

ml2 5

25& &

+- = - = =

Thus, the required ratio is : 5 : 2l m = .

Also from (1), we have (0, ) ,( ) ( )

( , ) .P y P P05 2

5 3 2 10

717=

++

=c m

Hence, the required point of intersection is ,0717` j.

13. Find the length of the medians of the triangle whose vertices are (1, -1) , (0, 4) and (-5, 3).

Solution: Let ( , 1)A 1 - , (0,4) ( 5,3)B Cand - be the vertices of the triangle.

Let D, E, F be the midpoints of BC, CA and AB respectively.

Then, the midpoint of BC is , ( , ) .D D2

0 52

4 325

27- + = -` j

Also, the midpoint of AC is , ( 2,1) .E E2

1 521 3- - + = -` j

The midpoint of AB is , , .F F2

1 021 4

21

23+ - + =` `j j

The length of the median 1 1AD2 2

25

27= + + - -` `j j

2 2

27

29

449

481

4130

2130= + = + = =-` `j j .

The length of the median BE 2 0 1 42 2= - - + -^ ^h h 4 9 13= + = .

The length of the median CF 5 32 2

21

23= + + -` `j j

.2 2

211

23

4121

49

2130= + = + =-` `j j

Thus, the lengths of medians of the ABCT are 2130 , 13 ,

2130 .

Exercise 5.2

1. Find the area of the triangle formed by the points (i) (0, 0), (3, 0) and (0, 2) (ii) (5, 2), (3, -5) and (-5, -1) (iii) (-4, -5), (4, 5) and (-1, -6).

Solution: (i) Let the vertices be A(0, 0), B(3, 0) and C(0, 2).


Using the formula, area = ( )x y x y x y x y x y x y21

1 2 2 3 3 1 2 1 3 2 1 3+ + - + +^ h6 @,

Thus, the area of ABCT [(0 6 0) (0 0 0)] 321= + + - + + = sq.units.

(ii) Plot the given points in a rough diagram and take them in order.

Let the vertices be A(-5, -1), B(3, -5) and C(5, 2).

The area of ABCT = 21 5

1

3

5

5

2

5

1

-

- -

-

-) 3

[( ) ( )] ( )21 25 6 5 3 25 10

21 26 38 32= + - - - - - = + = .

Thus, the area of the triangle is 32 sq.units.

(iii) Plot the given points in a rough diagram and take them in order.

Let the vertices be A(– 4, –5), B(–1, –6) and C(4, 5).


5

1

6

4

5

4

5

-

-

-

-

-

-) 3

[( ) ( )] ( )21 24 5 20 5 24 20

21 38 19= - - - - - = = .

Thus, the area of the triangle is 19 sq.units.

2. Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices Area (in sq. units) (i) ( , )0 0 , (4, a), (6, 4) 17 (ii) (a, a), (4, 5), (6,–1) 9 (iii) (a, –3), (3, a), (–1,5) 12

Solution: (i) Let the given vertices be A( , )0 0 , B(4, a), C(6, 4).

Given that the area of ABCT is 17 sq.units.

Thus, a2

1 0

0

4 6

4

0

017=' 1

(16 6 ) 17a21& - = 16 6a 34& - = .a 3& =-

Thus, the value of a is –3.

(ii) Let the vertices be A(a, a), B(4, 5) and C(6,–1).


Thus, a

a

a

a21 4

5

6

19

-=) 3

[(5 4 6 ) (4 30 )] 9a a a a21& - + - + - = 8 34 18a& &- = a

213= .

Thus, the value of a is 213 .


(iii) Let the vertices be A (a, –3), B (3, a), C (–1 , 5).


Thus, 12a

a

a

21

3

3 1

5 3-

-

-=) 3

[( 15 3) ( 9 5 )] 12a a a21 2

& + + - - - + =

a a4 3 02& - + = ( )( ) 0a a3 1& - - = 3 1a aand& = =

Hence,the values of a are 1, 3.

3. Determine if the following set of points are collinear or not.

(i) (4, 3), (1, 2) and (–2, 1) (ii) (–2, –2), (–6, –2) and (–2, 2)

(iii) 23 ,3-` j,(6, –2) and (–3, 4)

Solution: (i) Let A(4, 3), B(1, 2) and C(–2, 1) be the given points


3

1

2

2

1

4

3

-' 1

= (8 1 6) (3 4 4)21 0+ - - - + =

Hence, the given three points A,B,C are collinear.

(ii) Let the given points be A(-2, -2), B(-6, -2) and C(-2, 2).

The area of ABCT

= [(4 12 4) (12 4 4)]21 2

2

6

2

2

2

2

2 21-

-

-

-

- -

-= - + - + -) 3

[(8 12) 12]21 8 0!= - - =-

Hence, the given three points are not collinear.

(iii) Let the given points be A23 ,3-` j, B(6, -2) and C(-3, 4).

The area of ABCT

= [(3 24 9) (18 6 6)]21

3

6

2

3

4 3 212

323-

-

- -= + - - + -) 3 = 0.

Hence, the given three points are collinear.

4. In each of the following, find the value of k for which the given points are collinear. (i) ( , )k 1- , (2, 1) and (4, 5) (ii) , , , ,and k2 5 3 4 9- -^ ^ ^h h h

(iii) , , , ,andk k 2 3 4 1-^ ^ ^h h h

Solution: (i) Let the given points be A(k, -1), B(2, 1) and C(4, 5).

If the three points are collinear, then the area of ABCT is zero.

Thus, 0k k

21

1

2

1

4

5 1- -=) 3

[( 6) (2 5 )] 0k k& + - + = 1k& = .


(ii) Let the given points be , , , ,andA B C k2 5 3 4 9- -^ ^ ^h h h.

The given points are collinear.

Thus, area of ABCT = 0

0k2

1 2

5

3

4

9 2

5&

- - -=) 3

[( 8 3 45) ( 15 36 2 )] 0k k& - + - - - - + = 2.k& =

Thus, the value of k is 2.

(iii) Let the given points be , , , ,andA k k B C2 3 4 1-^ ^ ^h h h.

Given that the three points are collinear.

Thus, area of ABCT = 0

& 0k

k

k

k21 2

3

4

1-=) 3

& (3 2 4 ) (2 12 ) 0k k k k- + - + - = 6 14 0 .k k37& &- = =

Thus the value of k is 37 .

5. Find the area of the quadrilateral whose vertices are

(i) , , , , , ,and6 9 7 4 4 2 3 7^ ^ ^ ^h h h h (ii) , , , , , ,and3 4 5 6 4 1 1 2- - - -^ ^ ^ ^h h h h

(iii) , , , , , ,and4 5 0 7 5 5 4 2- - - -^ ^ ^ ^h h h h

Solution: (i) Plot the given points in a rough diagram and take the vertices in counter clockwise direction.

Let the given points be , , , , , , .andA B C D4 2 7 4 6 9 3 7^ ^ ^ ^h h h h

Area of the quadrilateral ABCD

= 21 4

2

7

4

6

9

3

7

4

2' 1

= [( ) ( )]21 16 63 42 6 14 24 27 28+ + + - + + + = [(127 93) (34) 17.

21

21- = =

Thus, the area of the quadrilateral ABCD is 17 sq.units.

(ii) Plot the given points in a rough diagram and take the vertices in counter clockwise direction.

Let the given points be , , , , , , .andA B C D5 6 4 1 1 2 3 4- - - -^ ^ ^ ^h h h h


21 5

6

4

1

1

2

3

4

5

6=

-

- -

- -

-) 3

[(5 8 4 18) ( 24 1 6 20)]21= + + + - - - - -

= [(35 51) (86) 43.21

21+ = =

Thus, the area of the quadrilateral ABCD is 43 sq.units.


(iii) Plot the given points in a rough diagram and take the vertices in counter clockwise direction. Let the given points be , , , , , ,andA B C D4 2 5 5 0 7 4 5- - - -^ ^ ^ ^h h h h.


21 4

2

5

5

0

7

4

5

4

2=

-

- -

- -

-) 3

[(20 35 0 8) ( 10 0 28 20)]21= + + + - - + - - = [( 5 ) ( ) .

21 63 8

21 121 60 5+ = =

Thus, the area of the quadrilateral ABCD is 60.5 sq.units.

6. If the three points , , ( , ) ,h a b k0 0and^ ^h h lie on a straight line, then using the area of the triangle formula, show that 1, , 0

ha

kb h kwhere !+ = .

Solution: The three points , , ( , ) ,h a b k0 0and^ ^h h lie on a straight line.

Thus, h a

b k

h

21

0

0

0' 1 = 0

& ( 0) (0 0 )hb ak kh+ + - + + = 0 & hb ak+ = kh

Since , ,h k 0! divide both sides by hk. We get, ha

kb 1+ = .

7. Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are , , , ,and0 1 2 1 0 3-^ ^ ^h h h. Find the ratio of this area to the area of the given triangle.

Solution: Let the vertices of the triangle be , , , ,andA B C0 1 2 1 0 3-^ ^ ^h h h.

Let D, E, F be the midpoints of the sides BC, CA and AB respectively.

The midpoint of BC is D2

2 0 ,2

1 3 ( , )D 1 2+ + =` j .

The midpoint of AC is E , ( , )E2

0 02

3 1 0 1+ - =` j .

The midpoint of AB is F 0 , ( , )F22

21 1 1 0+ - + =` j .

Thus, the area of TDEF = 21 1

2

0

1

1

0

0

2' 1

= [( ) ( )]21 1 0 2 0 1 0+ + - + + =1 sq.unit.

Thus, the area of the TDEF is 1 sq.unit.

Now, the area of ABCT = 21 2

1

0

3

0

1

2

1-) 3

= [( 0) ( 2)]21 6 0 0 0+ + - + - = 4 sq.units.

Thus, the area of the ABCT is 4 sq.units.Hence, the ratio of the area of DEFT to the area of ABCT is 1 : 4.


Exercise 5.3 1. Find the angle of inclination of the straight line whose slope is (i) 1 (ii) 3 (iii) 0

Solution: (i) Given that the slope of the straight line, 1 1 45tanm & &i i= = = c

Thus, the angle of inclination of the straight line is 45c.(ii) Given that the slope of the straight line, 60tanm 3 3& &i i= = = c

Thus, the angle of inclination of the straight line is 60c.(iii) Given that the slope of the straight line, 0 0 0tanm & &i i= = = c

Thus, the angle of inclination of the straight line is 0c.

2. Find the slope of the straight line whose angle of inclination is (i) 30c (ii) 60c (iii) 90c

Solution: (i) Given that 30i = c.

Thus, the slope of the straight line, 30tan tanm3

1i= = =c .(ii) Given that 06i = c. Thus, the slope of the straight line, 0tan tanm 6 3i= = =c .(iii) Given that 09i = c. Thus, the slope of the straight line, 90tan tanm i= = c,whichisnotdefined. Hence,theslopeisnotdefinedforthisstraightline.

3. Find the slope of the straight line passing through the points (i) (3 , -2) and (7 , 2), (ii) (2 , -4) and origin, (iii) ,1 3 2+^ h and ,3 3 4+^ h

Solution: (i) The slope of the line joining the points ( , ) ( , ) .x y x y mx x

y yand is

1 1 2 22 1

2 1=-

-

Thus, the slope of the line joining (3, –2) and (7, 2) is ( )1.m

7 32 2

=-

- -=

(ii) The slope of the line joining the points ( , ) ( , ) .x y x y mx x

y yand is

1 1 2 22 1

2 1=-

-

Thus, the slope of the line joining (2, –4) and (0,0) is ( )( )

2.m0 20 4

=-- -

=-

(iii) The slope of the line joining the points ( , ) ( , ) .x y x y mx x

y yand is

1 1 2 22 1

2 1=-

-

Thus, the slope ( )

m3 3 1 3

4 2

3 3 1 3

222 1=

+ - +

- =+ - -

= = .

4. Find the angle of inclination of the line passing through the points

(i) ,1 2^ h and ,2 3^ h (ii) ,3 3^ h and ,0 0^ h (iii) (a , b) and (-a , -b) Solution: (i) The slope of the line joining the points ,1 2^ h and ,2 3^ h is

1m2 13 2=-- = tan 1& i = 45 .& i = c

Thus, the angle of inclination is 45c.


(ii) The slope of the line joining the points ,3 3^ h and ,0 0^ h is

m0 30 3

33

3

1=-- = = tan

3

1& i = . Thus, 30 .i = c

Thus, the angle of inclination is 30c.(iii) The slope of the line joining the points (a , b) and (-a , -b) is

ma ab b

ab

ab

22=

- -- - =

-- = tan

ab& i =

Thus, the angle of inclination is given by the relation tanabi = .

5. Find the slope of the line which passes through the origin and the midpoint of the line segment joining the points ,0 4-^ h and (8 , 0).

Solution: Midpoint of line joining ,0 4-^ h and (8 , 0) is ,2

0 824 0+ - +` j = (4, 2)-

The slope of the line passing through (4, –2) and (0, 0) is

( ).m

0 40 2

21=

-- -

= -

6. The side AB of a square ABCD is parallel to x-axis . Find the (i) slope of AB (ii) slope of BC (iii) slope of the diagonal AC.Solution: (i) Since the side AB is parallel to x axis, the slope of AB, m 0= .

(ii) Since BC AB= , the angle made by BC with the x-axis is 90i = c.

Thus, the slope tanm 90= c,whichisnotdefined. Hence, the slope of BC isnotdefined.(iii) Now, the diagonal AC bisects the angle .DAB+

Thus, 45 . . ., 45BAC i e+ i= =c c

Hence, the slope of diagonal AC is 45tan tanm 1i= = =c .

7. The side BC of an equilateral TABC is parallel to x-axis. Find the slope of AB and the slope of BC.

Solution: Given that the side BC is parallel to x-axis and .ABC 60+ = c

Thus, the angle of inclination of the line AB is 60c.Hence, the slope of AB, 60tanm 3= =c .Now, BC is parallel to x-axis.Thus, the slope of BC, 0 0tanm = =c .

8. Using the concept of slope, show that each of the following set of points are collinear.

(i) (2 , 3), (3 , -1) and (4 , -5) (ii) (4 , 1), (-2 , -3) and (-5 , -5) (iii) (4 , 4), (-2 , 6) and (1 , 5)

Solution: (i) Let A(2 , 3), B(3 , -1) and C(4 , -5) be the given points.


Slope of the line AB is 4m3 21 3

1 =-

- - =-

Slope of the line BC is 4m4 35 1

2 =-

- + =-

Thus, slope of AB = slope of BC. Note that B is the common point.Hence, the points A, B and C are collinear.(ii) Let A(4 , 1), B(-2 , -3) and C(-5 , -5) be the given points.Slope of the line AB joining the points A(4 , 1), B(-2 , -3) is

m2 43 1

64

32

1 =- -- - =

-- = .

Slope of the line BC joining the points B(-2 , -3) and C(-5 , -5) is

m5 25 3

32

32

2 =- +- + =

-- = .

Thus, slope of AB = slope of BC.Since B is the common point, the points A, B and C are collinear.(iii) Let A (4 , 4), B (-2 , 6) and C (1 , 5) be the given points.Slope of the line AB joining the points A(4 , 4) and B(-2 , 6) is

.m2 46 4

62

31

1 =- -

- =-

=-

Slope of the line BC joining the points B(-2 , 6) and C(1 , 5) is

m1 25 6

31

2 =+- =- .

Thus, slope of AB = slope of BC.Since B is the common point, the points A, B and C are collinear.

9. If the points (a, 1), (1, 2) and (0, b + 1) are collinear, then show that a b1 1+ = 1.

Solution: Let ( ,1), (1,2) (0, 1)A a B C band + be the given points.

Now, the slope of AB is .ma a1

2 111

1 =-- =

-

Also, the slope of BC is m b b0 11 2

11

2 =-

+ - =--

Since three points are collinear, we have m m1 2= .

a

b11

11&

-=

-- a b ab& + =

Dividing by ab on both sides, we get 1a b1 1+ = .

10. The line joining the points A(-2 , 3) and B(a , 5) is parallel to the line joining the points C(0 , 5) and D(-2 , 1). Find the value of a.Solution: Since the lines AB and CD are parallel, their slopes are equal.

Thus, slope of AB = the slope of CD.


a 25 3

2 01 5&

+- =

- -- 2 1a& + = 1a& =-

Hence, the value of a is –1.

11. The line joining the points A(0, 5) and B(4, 2) is perpendicular to the line joining the points C(-1, -2) and D(5, b). Find the value of b.

Solution: Slope of AB is m1 =

4 02 5

43

-- = -

Slope of CD is m2 = b b

5 12

62

++ = +

Since the lines AB and CD are perpendicular, we have

m m1 2

= 1-

1b43

62& - + =-` `j j 2 8 6.b b& &+ = =

Thus, the value of b is 6.

12. The vertices of TABC are A(1, 8), B(-2, 4), C(8, -5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN is parallel to BC.

Solution: Midpoint M of AB is ,M2

1 22

8 4- +` j , 6M21= -` j

Midpoint N of AC is , ,N N2

8 125 8

29

23+ - + =` `j j

Now, the slope of the line MN is m6

109

129

21

23

2102

3 12

=+

-= =-

-

... (1)

Also, the slope of BC is m8 25 4

109

2 =+

- - =- ... (2)

From (1) and (2), we have m m1 2= .Hence, the straight lines BC and MN are parallel.

13. A triangle has vertices at (6 , 7), (2 , -9) and (-4 , 1). Find the slopes of its medians.

Solution: Let the vertices be A(6 , 7), B(2 , -9) and C(-4 , 1).

Let D, E, F be the midpoints of BC, CA, AB respectively.

Then AD, BE and CF are the medians of the TABC.

The midpoint of BC is D , ( 1, 4) .D2

2 429 1- - + = - -` j

The midpoint of CA is E , (1, 4) .E24 6

21 7- + + =` j

The midpoint of AB is , (4, 1) .F F2

6 22

7 9+ - = -` j


Slope of AD = 1 64 7

711

711

- -- - =

-- = . Slope of BE =

1 24 9

113 13

-+ =

-=- .

Slope of CF = 4 41 1

82

41

+- - = - = - .

Hence, the slopes of the medians are , 13 .711

41and- -

14. The vertices of a 3ABC are A(-5 , 7), B(-4 , -5) and C(4 , 5). Find the slopes of the altitudes of the triangle.Solution: Let AD, BE and CF be the altitudes of ABCD

Slope of AB4 55 7 12=

- +- - =-

Since the altitude CF is perpendicular to AB, the slope of CF121= .

Slope of BC4 45 5

810

45=

++ = = .

Since the altitude AD is perpendicular to BC, the slope of .AD54=-

Slope of AC4 55 7

92=

+- =- .

Since the altitude BE is perpendicular to CA, the slope of .BC29=

Hence, the slopes of the altitudes are , , .121

54

29-

15. Using the concept of slope, show that the vertices (1 , 2), (-2 , 2), (-4 , -3) and (-1, -3) taken in order form a parallelogram.Solution: Plot the given points in a rough diagram and take the vertices in order.

Let the vertices be A (-4 , -3), B (-1, -3), C (1 , 2) and D (-2 , 2).

Slope of AB1 43 3 0=

- +- + = ; Slope of 0CD

2 12 2=

- -- = .

Since the slopes of AB and CD are equal, AB is parallel to CD. ... (1)

Slope of BC1 12 3

25=

++ = ; Slope of AD

2 42 3

25=

- ++ = .

Since the slopes of BC and AD are equal, BC is parallel to AD. ... (2)From (1) and (2), we get the opposite sides of the quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.

16. Show that the opposite sides of a quadrilateral with vertices AA(-2 ,-4), B(5 , -1), C(6 , 4) and D(-1, 1) taken in order are parallel.

Solution: Given the poinst are A(-2 ,-4), B(5 , -1), C(6 , 4) and D(-1, 1).

Slope of AB5 21 4

73=

+- + = ; Slope of CD

1 61 4

73

73=

- -- =

-- = .


Thus, the slopes of AB and CD are equal. Hence, AB is parallel to CD ...(1)

Now, slope of 5;AD1 21 4=

- ++ = slope of 5.BC

6 54 1=-+ =

Thus, the slopes of BC and AD are equal. Hence, BC is parallel to AD ...(2)

From (1) and (2), we see that the opposite sides of the quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.

Exercise 5.4

1. Write the equations of the straight lines parallel to x-axis which are at a distance of 5 units from the x-axis.

Solution: Equation of a straight line parallel to x-axis is y k= .

Thus, the equations of the straight lines parallel to x-axis which are at a distance of 5 units from the x-axis are given by , .y y5 5= =-

2. Find the equations of the straight lines parallel to the coordinate axes and passing through the point (-5,-2).

Solution: Equation of the straight line passing through (–5, –2) and parallel to x-axis is 2y =- .

Equation of the straight line passing through (–5, –2) and parallel to y-axis is x 5=- .

3. Find the equation of a straight line whose (i) slope is -3 and y-intercept is 4, (ii) angle of inclination is 60c and y-intercept is 3.

Solution: (i) Given that slope 3 andm =- the y-intercept .c 4=

Equation of a straight line in the slope-intercept form is .y mx c= +

Thus, the required equation is 3 4y x=- + or x y3 4 0+ - = .(ii) Given that the angle of inclination is 60cand y-intercept is 3. Thus, the slope 60tanm 3= =c and the y-intercept .c 3=

Equation of the straight line in slope-intercept form is y mx c= + . Thus, the required equation is 3y x3= + or 3 0x y 3- + = .

4. Find the equation of the line intersecting the y- axis at a distance of 3 units above the origin and tan

21i = , where i is the angle of inclination.

Solution: Given that the slope, tanm21i= = ; y-intercept, c 3= .

Equation of the straight line in slope-intercept form is .y mx c= +

Thus, the equation of the straight line is 3y x21= + & x y2 6 0- + = .


5. Find the slope and y-intercept of the line whose equation is (i) 1y x= + , (ii) 5 3x y= (iii) 4 2 1 0x y- + = , (iv) 10 15 6 0x y+ + = .

Solution: (i) Comparing y x 1= + with y mx c= + , we get slope m 1= and y-intercept .c 1=

(ii) Given equation x y5 3= is rewritten as y x35= .

Comparing the equation y x35= with y mx c= + , we have

slope m35= and y-intercept .c 0=

(iii) x y4 2 1 0- + = 2 4 1 2 .y x y x21& &= + = +

Comparing the equation 2y x21= + with the equation y mx c= + , we get,

slope m 2= and y-intercept .c21=

(iv) x y10 15 6 0+ + = .y x32

52& = - -

Comparing the equation y x32

52= - - with y mx c= + we get,

slope m32= - and y-intercept .c

52= -

6. Find the equation of the straight line whose (i) slope is -4 and passing through ( , )1 2 (ii) slope is

32 and passing through (5, -4).

Solution: (i) Given that slope m 4=- and a point ( , ) ( , )x y 1 21 1 = .

Equation of the straight line in slope-point form is ( )y y m x x1 1- = -

2 4( 1) 4 6 0.y x x y& &- =- - + - =

(ii) Slope m32= and passing through the point ( , ) ( , )x y 5 41 1 = - .

Equation of the straight line in slope-point form is ( )y y m x x1 1- = - . 4 ( 5) 3( 4) 2( 5) 2 3 22 0y x y x x y

32& & &+ = - + = - - - = .

7. Find the equation of the straight line which passes through the midpoint of the line segment joining (4, 2) and (3, 1) whose angle of inclination is 30c.

Solution: Given that the angle of inclination 30i = c.

Thus, the slope 3

tanm 30 1= =c .

The midpoint of the straight line joining (4, 2) and (3, 1) is

( , ) , , .x y2

4 32

2 127

23

1 1 = + + =` `j j

Now, the equation of the straight line in slope-point form is ( )y y m x x1 1- = - . & ( )y x y x

23

3

127 3 2 3 2 7&- = - - = -^ h

2 2 (3 7) 0.x y3 3& - + - =


8. Find the equation of the straight line passing through the points (i) (– 2, 5) and (3, 6) (ii) (0, – 6) and (– 8, 2).

Solution: (i) Equation of the straight line in two-points form is y yy y

x xx x

2 1

1

1

1

2--

=-- .

Here, ( , ) ( 2, 5), ( , ) (3, 6) .x y x y1 1 2 2= - =

The required equation is y x6 5

53 2

2--

=++

y x15

52&

-= + x y5 27 0& - + = .

(ii) Equation of the straight line in two-points form is y yy y

x xx x

2 1

1

1

1

2--

=-- .

Here, ( , ) (0, 6), ( , ) ( 8, 2) .x y x y1 1 2 2= - = -

The required equation is y x2 6

68 0

0++

=- -

- 6 0.x y& + + =

9. Find the equation of the median from the vertex R in a TPQR with vertices at P(1, -3), Q(-2, 5) and R(-3, 4).

Solution : Let M be the midpoint of PQ.

Then , ( ,1) .M M2

1 223 5

21- - + = -` j

The median RM is the straight line joining the points R(-3, 4) and , .M21 1-` j

Thus, the equation of RM is

( )y x1 4

4

3

3

21-

-=

- -

- -- ^ h

( )y x34

52 3

&--

=+ 6 5 2 0.x y& + - =

10. By using the concept of the equation of the straight line, prove that the given three points are collinear. (i) (4, 2), (7, 5) and (9, 7) (ii) (1, 4), (3, -2) and (-3, 16).

Solution: (i)Letusfindthe equation of the straight line joining the points (4, 2), (7, 5).

The equation is y x5 2

27 4

4--

=-- 2 0.x y& - - = g (1)

Put ,x y9 7= = in (1). We have, 9 7 2 0- - = .

Thus, the point (9, 7) lies on the line joining the points (4, 2) and (7, 5).

Hence, the three points (4, 2), (7, 5) and (9, 7) are collinear.

(ii) Letusfindthe equation of the straight line joining the points (1, 4), (3, -2).

The equation of the straight line joining the points (1, 4), (3, -2) is

y x2 4

43 1

1 &- -

-=

-- 6 2 14 0.x y+ - = 3 7 0.x y& + - = g (1)

Put ,x y3 16=- = in (1). We get , ( ) .3 3 16 7 0- + - =

Thus, the point (-3, 16) lies on the line joining the points (1, 4), (3, –2).

Hence, the points (1, 4), (3, -2) and (-3, 16) are collinear.


11. Find the equation of the straight line whose x and y-intercepts on the axes are given by

(i) 2 and 3 (ii) 31- and

23 (iii)

52 and

43- .

Solution: (i) Given that x-intercept, a 2= and y-intercept, b 3= .

The equation of the straight line in intercepts form is 1.ax

by

+ =

Thus, the required equation is 1x y2 3+ = 3 2 6 0.x y& + - =

(ii) Given that x-intercept, a31=- and y-intercept, b

23= .

The equation of the straight line in intercepts form is

1.ax

by

+ =

1x y

31

23

&-

+ = 9 2 3 0x y& - + =

(iii) Given that x-intercept, a52= and y-intercept, b

43=- .

The equation of the straight line in intercepts form is 1.ax

by

+ =

1x y

52

43

& +-

= 15 8 6 0.x y& - - =

12. Find the x and y-intercepts of the straight line

(i) x y5 3 15 0+ - = (ii) 2 1 0x y 6- + = (iii) x y3 10 4 0+ + =

Solution: (i) Given equation is x y5 3 15 0+ - = 5 3 15.x y& + =

Dividing both sides by 15, we get 1x y3 5+ = g (1)

Comparing (1) with intercepts form of the straight line 1ax

by

+ = , we have

x-intercept, a 3= . y-intercept, b 5= .

(ii) Given equation is 2 1 0x y 6- + = 2 16x y& - =-

Dividing both sides by –16, we get 1x y8 16-+ = g (1)

Comparing (1) with intercepts form of the straight line 1ax

by

+ = ,we have

x-intercept, a 8=- . y-intercept, b 16= .

(iii) Given equation is x y3 10 4 0+ + = 3 10 4x y& + =-

Dividing both sides by –4, we get 1 1x y x y43

410

34

52

&-

+-

=-

+-

=^ ^h h

g (1)

Comapring (1) with intercepts form of the straight line 1ax

by

+ = , we have

x-intercept, a34=- . y-intercept,

5b 2=- .

Note: One can get x-intercept by substituting y 0= and the y-intercept by substituting 0x = in the equation of the straight line.


13. Find the equation of the straight line passing through the point (3, 4) and has intercepts which are in the ratio 3 : 2.

Solution: Let x and y-intercepts of the straight line be a and b respectively.

Then, a : b = 3 : 2. Let a = 3k and b = 2k, where k is a non-zero constant.

Thus, the equation of the straight line in intercepts form is 1kx

ky

3 2+ = g (1)

Since this line passes through (3, 4), we have 1 1 3.k k k k

k33

24 1 2& &+ = + = =

Substituting k 3= in (1), we get 1x y9 6+ = 2 3 18 0.x y& + - =

14. Find the equation of the straight lines passing through the point (2, 2) and the sum of the intercepts is 9.

Solution: Let x and y-intercepts of the straight line be a and b respectively.

Then, a+b = 9 or .b a9= -

Now, the equation of the straight line in intercepts form is 1ax

ay

9+

-= g (1)

Since (1) passes through (2, 2), we have 1a a2

92+-

= 9 18 0a a2& - + =

( 3)( 6) 0. , 3 ( ) 6.Thusa a a aor& - - = = =

When a 3= , from equation (1) we have 1 2 6 0.x yx y

3 6&+ = + - =

When a 6= , from equation (1) we have 1 2 6 0.x yx y

6 3&+ = + - =

( Here, we have two straight lines satisfying the given conditions. )

15. Find the equation of the straight line passing through the point (5, -3) and whose intercepts on the axes are equal in magnitude but opposite in sign.

Solution: Let a be the x-intercept of the straight line.

Then, y-intercept is a- .Now, the equation of the line in intercepts form is 1

ax

ay

x y a&+-

= - = g (1)

Since (1) passes through (5, –3), we get 5 3 8.a = + =

Substituting 8a = in (1), the equation of the straight line is 8 0.x y- - =

16. Find the equation of the line passing through the point (9, -1) and having its x-intercept thrice as its y-intercept.

Solution: Equation of the straight line in intercepts form is 1ax

by

+ = .

It is given that a b3= .

The equation of the straight line is 1 3 3 .bx

by

x y b3

&+ = + = g (1)

Since (1) passes through (9, –1), we have 3 2.b b9 3 &= - =

Hence, the eqation of the required straight line is 3 6 0.x y+ - =


17. A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2), then find the equation of AB.

Solution: Equation of the straight line in intercepts form is 1ax

by

+ = g (1)

Now, (1) cuts the x-axis at A. At A, we have y = 0. So, (1) 1 .ax x a& &= =

Hence, A is A(a, 0). Also, (1) cuts the y-axis at B. At B, we have x = 0.

So, (1) 1 .by

y b& &= =

Thus, B is B (0, b).

Now, the midpoint of , (3, 2) .isAB a b20

20+ + =` j

Thus, 3, 6 2 4.a a b b20

20and& &+ = = + = =

Hence, the equation of the straight line is x y6 4

1+ =

& 2 3 12 2 3 12 0orx y x y+ = + - = .

18. Find the equation of the line passing through (22, -6) and having intercept on x-axis exceeds the intercept on y-axis by 5.

Solution: Let the y-intercept be a.Then, x-intercept is a 5+ .

Thus, the equation of the straight line in intercepts form is 1ax

ay

5++ = g (1)

Since the line passes through (22, –6), we have a a522 6 1+

- =

( )( )

1a a

a a5

22 6 5&

+- +

= 11 30a a 02& - + = 5a a 6or& = =

Thus, we get two straight lines satisfying the given conditions.

When 5a = , (1) & 1 2 10 0x yx y

10 5&+ = + - = .

When a 6= , (1) & 1 6 11 66 0x yx y

11 6&+ = + - = .

Hence, the equations of lines are 2 10 0, 6 11 66 0.x y x y+ - = + - =

19. If A(3, 6) and C(-1, 2) are two vertices of a rhombus ABCD, then find the equation of straight line that lies along the diagonal BD.

Solution: The diagonals AC and BD of the rhombus ABCD bisect each other at right angle. The slope of AC =

1 32 6 1

- -- = . Thus, the slope of BD 1=- .

Now, the midpoint of AC = , (1, 4) .2

3 12

6 2- + =` j

The equation of the line along BD and passing through (1, 4) with slope –1 is 4 1( 1) 4 1y x y x x y 5 0& &- =- - - =- + + - = .


20. Find the equation of the line whose gradient is 23 and which passes through P,

where P divides the line segment joining A(-2, 6) and B (3, -4) in the ratio 2 : 3 internally.

Solution: The point P divides AB in the ratio 2:3 internally.

Thus, the point P is ( ) ( ),

( ) ( )2 3

2 3 3 22 3

2 4 3 6+

+ -+

- +c m (0, 2)= .

Hence, the equation of the straight line passing through (0, 2) with slope 23 is

2 ( 0) 2 4 3 3 2 4 0y x y x x y23 & &- = - - = - + = .

Exercise 5.5

1. Find the slope of the straight line (i) 3 4 6 0x y+ - = , (ii) 7 6y x= + , (iii) 4 5 3x y= + .

Solution: (i) The slope of the straight line 0 .isax by cba+ + = -

Thus, the slope of the straight line x y3 4 6 0+ - = is ba

43- =- .

(ii) The slope of the straight line 0 .isax by cba+ + = -

Thus, the slope of the straight line 7 6 0x y- + = is ba

17 7- =

-- = .

(iii) The slope of the straight line 0 .isax by cba+ + = -

Thus, the slope of the straight line 4 5 3 0x y- - = is ba

54

54- =

-- = .

2. Show that the straight lines 2 1 0x y+ + = and 3 6 2 0x y+ + = are parallel.

Solution: Slope of x y2 1 0+ + = is .m21

1 =-

Slope of x y3 6 2 0+ + = is .m21

2 =-

Since m m1 2= , the given two straight lines are parallel.

Aliter : ;aa

bb

aa

bb

31

62

31

2

1

2 2

1

2

1 1&= = = = . Thus, the two lines are parallel.

3. Show that the straight lines 3 5 7 0x y- + = and 15 9 4 0x y+ + = are perpendicular.

Solution: Slope of x y3 5 7 0- + = is .m53

53

1 =-- =

Slope of x y15 9 4 0+ + = is .m915

35

2 = - = -

Now, 1.m m53

35

1 2# = - =-` `j j Thus, the straight lines are perpendicular.

Note: 3(15) ( 5)(9) 0a a b b1 2 1 2+ = + - = . Thus, the lines are perpendicular.

4. If the straight lines 5 3y

x p ax y2

and= - + = are parallel, then find a .

Solution: Given straight lines are 5 3y

x p ax y2

and= - + =


0 3 5 0.andxy

p ax y2

& - - = - + =

Since the given lines are parallel, we have 6.aa

bb

aa1

32

1

2

21

1 & &= =-

-=

5. Find the value of a if the straight lines 5 2 9 0x y- - = and 2 11 0ay x+ - = are perpendicular to each other.

Solution: Slope of x y5 2 9 0- - = is .m25

25

1 =-- =

Slope of 11 0ay x2+ - = is .ma2

2 = -

Since the straight lines are perpendicular to each other, we have

1 1 5.m ma

a25 2

1 2 & &=- - =- =` `j j

Thus, the value of a is 5.

6. Find the values of p for which the straight lines 8 1 0px p y2 3+ - + =^ h and 8 7 0px y+ - = are perpendicular to each other.

Solution: Slope of px p y8 2 3 1 0+ - + =^ h is .mpp

2 38

1 =--

Slope of px y8 7 0+ - = is m p8

2 =- .


1 1m mpp p

2 38

81 2 &=-

-- -

=-c `m j

3 2 0 ( 1)( 2) 0 1,2.p p p p p2& & &- + = - - = =

7. If the straight line passing through the points ,h 3^ h and (4, 1) intersects the line 7 9 19 0x y- - = at right angle, then find the value of h .

Solution: Slope of straight line joining ,h 3^ h and (4, 1) is

.mh h4

1 34

21 =

-- =

--

Slope of the straight line x y7 9 19 0- - = is .m97

2 =


1 1 1m mh h42

97

36 914

1 2 & & &=--- =-

-- =-` `j j h

922= .

8. Find the equation of the straight line parallel to the line 3 7 0x y- + = and passing through the point (1, -2).

Solution: Equation of the straight line parallel to x y3 7 0- + = is 3 0.x y k- + =

Since it passes through (1, –2), we have

3(1) ( 2) 0 5.k k&- - + = =-

Thus, the equation of the required straight line is 3 0x y 5- - =


9. Find the equation of the straight line perpendicular to the straight line 2 3 0x y- + = and passing through the point (1, -2).

Solution: Equation of the line perpendicular to x y2 3 0- + = is

of the form 0x y k2 + + = .

Since it passes through (1, –2), we have

2(1) (2) 0 0.k k&- + = =

Thus, the equation of the required straight line is 0x y2 + = .

10. Find the equation of the perpendicular bisector of the straight line segment joining the points (3, 4) and (-1, 2).

Solution: Midpoint of the straight line joining the points A(3, 4) and B(-1, 2)

, , .2

3 12

4 2 1 3- + =` ^j h

Equation of the straight line joining (3, 4) and (–1, 2) is

4 16 2 6y x y x2 4

41 3

3 & &--

=- -

- - + =- + 2 5 0.x y- + =

Thus, the equation of the line perpendicular to 2 0x y 5- + = is of the form

0x y k2 + + = .Since it passes through (1, 3), we have 2(1) 3 0 5.k k&+ + = =-

Thus, the equation of the required straight line is 0x y2 5+ - = .Note: One can get the required equation using slope-point formula.

11. Find the equation of the straight line passing through the point of intersection of the lines 2 3 0x y+ - = and 5 6 0x y+ - = and parallel to the line joining the points (1, 2) and (2, 1).Solution: Let the given points be A(1, 2) and B(2, 1).The given equations are 2 3 (1) ; 5 6 (2) .x y x yg g+ = + =

Subtracting (1) from (2), we have 3 3 1x x&= = and hence .y 1=

The point of intersection is (1, 1).

The slope of the line joining (1, 2) and (2, 1) is 1.m1 22 1=-- =-

Hence, the slope of the required line( parallel line) is –1Now, the equation of the straight line passing through (1, 1) with slope –1 is 1 1( 1) 2 0y x x y&- =- - + - = .

12. Find the equation of the straight line which passes through the point of intersection of the straight lines 5 6 1x y- = and 3 2 5 0x y+ + = and is perpendicular to the straight line 3 5 11 0x y- + = .Solution:The given equations are 5 6 1 (1) ; 3 2 5 (2)x y x yg g- = + =-


Solving (1) and (2), we get the point of intersection as (–1, –1).

Slope of the line x y3 5 11 0- + = is .m53

53=

-- =

Thus, the slope of the required line (perpendicular line) is 35- .

Hence, the equation of the line passing through (–1, –1) with slope

35- is

1 ( 1) 5 3 8 0.y x x y35 &+ = - + + + =

13. Find the equation of the straight line joining the point of intersection of the lines 3 9 0x y- + = and 2 4x y+ = and the point of intersection of the lines 2 4 0x y+ - = and 2 3 0x y- + = .

Solution: Given equations can be rewritten as 3 9 (1)x y g- =- ; 2 4 (2) .x y g+ =

2 4 (3)x y g+ = ; 2 3 ( )x y 4g- =- .

Solving (1) and (2), the point of intersection is (–2, 3).

Solving (3) and (4), the point of intersection is (1, 2).

The equation of the straight line joining the points (–2, 3) and (1, 2) is

y x y x2 3

31 2

213

32&

--

=++

--

= +

3 7 0x y& + - =

Thus, the required equation is 3 7 0x y+ - = .

14. If the vertices of a 3ABC are A(2, -4), B(3, 3) and C(-1, 5), then find the equation of the straight line along the altitude from the vertex B.

Solution: Let BD be the altitude from the vertex B.

Now, the slope of AC is 3.1 25 4

39

- -+ =

-=-

Thus, the slope of the straight line along the altitude BD is 31 . ( .AC BD= )

Now, the required line is passing through (3, 3) with slope 31 .

Hence, the required equation is 3 ( 3) 3 9 3 3y x y x x y

31 6 0& &- = - - = - - + = .

15. If the vertices of a 3ABC are A(-4,4 ), B(8 ,4) and C(8,10), then find the equation of the straight line along the median from the vertex A.

Solution: Let AD be the median through the vertex.

Midpoint D of BC is , (8, 7) .D D2

8 82

4 10+ + =` j


The equation of the median AD joining the points A(-4,4 ), D(8, 7) is

y x7 4

48 4

4--

=++ 4 16 4 4 20 0.y x x y& &- = + - + =

16. Find the coordinates of the foot of the perpendicular from the origin on the straight line 3 2 13x y+ = .

Solution: Let P be the foot of the perpendicular OP on the line x y3 2 13+ = .

Thus, the equation of straight line OP is of the form x y k2 3 0- + = .

It passes through the origin O (0, 0). Thus, .k 0=

Hence, the equation of the straight line OP is 2 3 0x y- = .

Now, P is the point of intersection of the straight lines

3 2 13 (1)x y g+ = and 2 3 0 (2)x y g- =

Solving (1) and (2), we get , .x y3 2= =

Hence, the foot of the perpendicular is P (3, 2).

17. If 2 7x y+ = and 2 8x y+ = are the equations of the lines of two diameters of a circle, find the radius of the circle if the point (0, -2) lies on the circle.

Solution: The equations of the diameters are 2 7 (1) ; 2 8 (2)x y x yg g+ = + =

Their point of intersection is the centre of the circle.Solving (1) and (2), we get the centre C(3, 2).

Let the given point on the circle be P(0, –2).Thus, the radius ( ) ( ) 5.r CP 3 0 2 2 252 2

= = - + + = = units.

18. Find the equation of the straight line segment whose end points are the point of intersection of the straight lines 2 3 4 0x y- + = , 2 3 0x y- + = and the midpoint of the line joining the points (3, -2) and (-5, 8).Solution: Let P be the point of intersection of the straight lines 2 3 4 0 (1)x y g- + = and 2 3 0 (2) .x y g- + =

Thus, solving (1) and (2), we get the point of intersection P(1, 2). The midpoint M of the line joining (3, –2) and (–5, 8) is

, ( 1, 3) .M M2

3 522 8- - + = -` j

Thus, the equation of the required line MP is

y x3 2

21 1

1--

=- -

- 1 2 4 2 5 0.x y x y& &- =- + + - =


19. In an isosceles 3PQR, PQ = PR. The base QR lies on the x-axis, P lies on the y- axis and 2 3 9 0x y- + = is the equation of PQ. Find the equation of the straight line along PR.

Solution: Equation of PQ is 2 3 9 0 ( )x y 1g- + =

At P, we have x 0= Now, (1) 2(0) 3 9 0 3.y y& &- + = =

Thus, the point P is P(0, 3).

At Q, we have 0y = . Now, (1) 2 0 9 0 .x x29& &- + = =-

Thus, the point Q is 29 , 0-` j.

Since PQ PR= and QR lie on the x-axis, the point is R , .29 0` j

Thus, the equation of the line along PR is

y x0 3

3

00

29-

-=

- -- 9 27 6 2 3 9 0.y x x y& &- =- + - =

Exercise 5.6

Choose the correct answer.

1. The midpoint of the line joining ,a b-^ h and ,a b3 5^ h is

(A) ,a b2-^ h (B) ,a b2 4^ h

(C) ,a b2 2^ h (D) ,a b3- -^ h

Solution: Midpoint is , ( , 2 )a a b b a b23

25 2+ - + =` j . ( Ans. (C) )

2. The point P which divides the line segment joining the points ,A 1 3-^ h and ,B 3 9-^ h internally in the ratio 1:3 is

(A) ,2 1^ h (B) ,0 0^ h

(C) ,35 2` j (D) ,1 2-^ h

Solution: Point P is ( ) ( ),

( ) ( ), (0, 0)

1 31 3 3 1

1 31 9 3 3

43 3

49 9

+- +

++ -

= - + - =c `m j .

( Ans. (B) )

3. If the line segment joining the points ,A 3 4^ h and ,B 14 3-^ h meets the x-axis at P, then the ratio in which P divides the segment AB is

(A) 4 : 3 (B) 3 : 4 (C) 2 : 3 (D) 4 : 1

Solution: If a line intersects x-axis, then y 0= .

( ) ( )0 3 4 0 3 4

l ml m

l m l mml3 4

34& & & &

+- +

= - + = = = ( Ans. (A) )


4. The centroid of the triangle with vertices at ,2 5- -^ h, ,2 12-^ h and ,10 1-^ h is

(A) ,6 6^ h (B) ,4 4^ h (C) ,3 3^ h (D) ,2 2^ h

Solution: Centroid is 3

2 2 10 ,3

5 12 1 ( , )2 2- - + - + - =` j ( Ans. (D) )

5. If ,1 2^ h, ,4 6^ h, ,x 6^ h and ,3 2^ h are the vertices of a parallelogram taken in order, then the value of x is

(A) 6 (B) 2 (C) 1 (D) 3

Solution: In a parallelogram the diagonals bisect each other.

Midpoint of AC = Midpoint of BD is

, , 6x x x2

12

2 62

4 32

6 22

12

4 3& &+ + = + + + = + =` `j j ( Ans. (A) )

6. Area of the triangle formed by the points (0,0), ,2 0^ h and ,0 2^ h is

(A) 1 sq. units (B) 2 sq. units (C) 4 sq. units (D) 8 sq. units

Solution : Area is (4) 221 0

0

2

0

0

2

0

0 21= =' 1

( ) (2)(2) 2or ab21

21= = units. ( Ans. (B) )

7. Area of the quadrilateral formed by the points ,1 1^ h, ,0 1^ h, ,0 0^ h and ,1 0^ h is

(A) 3 sq. units (B) 2 sq. units (C) 4 sq. units (D) 1 sq. units

Solution: Area is (2) 121 1

1

0

1

0

0

1

0

1

1 21= =' 1

( ) (1) 1Area of the squareor a2 2= = unit. ( Ans. (D) )

8. The angle of inclination of a straight line parallel to x-axis is equal to

(A) 0c (B) 60c (C) 45c (D) 90c

Solution: The angle of inclination of line parallel to x-axis is 0c. ( Ans. (A) ) 9. Slope of the line joining the points ,3 2-^ h and , a1-^ h is

23- , then the value of a is

equal to

(A) 1 (B) 2 (C) 3 (D) 4

Solution: Slope 2 4 12 4.a a a1 3

223 & &

- -+ = - + = = ( Ans. (D) )

10. Slope of the straight line which is perpendicular to the straight line joining the points ,2 6-^ h and ,4 8^ h is equal to

(A) 31 (B) 3 (C) -3 (D)

31-

Solution: Slope of the straight line joining the points ,2 6-^ h and ,4 8^ h is .4 28 6

31

+- =

Thus, the slope of the perpendicular line is –3. ( Ans. (C) )


11. The point of intersection of the straight lines x y9 2 0- - = and x y2 9 0+ - = is (A) ,1 7-^ h (B) ,7 1^ h

(C) ,1 7^ h (D) ,1 7- -^ h

Solution: 9 2 0 (1) 2 9 0 (2)x y x yg g- - = + - =

On solving the equations, the point is (1, 7). ( Ans. (C) ) 12. The straight line x y4 3 12 0+ - = intersects the y- axis at

(A) ,3 0^ h (B) ,0 4^ h

(C) ,3 4^ h (D) ,0 4-^ h

Solution: On the y-axis, 0x = . So, the point is (0, 4) ( Ans. (B) )

13. The slope of the straight line y x7 2 11- = is equal to

(A) 27- (B)

27 (C)

72 (D)

72-

Solution: Slope .mba

72

72= - =- - =` j ( Ans. (C) )

14. The equation of a straight line passing through the point (2 , –7) and parallel to x-axis is

(A) x 2= (B) x 7=- (C) y 7=- (D) y 2=

Solution: Equation of the line parallel to x-axis is y k= .

This line passes through the point (2, –7). Thus, the line is y 7=- ( Ans. (C) )

15. The x and y-intercepts of the line x y2 3 6 0- + = , respectively are

(A) 2, 3 (B) 3, 2 (C) -3, 2 (D) 3, -2

Solution: Tofindthex-intercept, put y 0= in the equation. Thus, 3.x =-

Tofindthey-intercept, put 0x = in the equation. Thus, 2.y = ( Ans. (C) )

16. The centre of a circle is (-6, 4). If one end of the diameter of the circle is at (-12, 8), then the other end is at

(A) (-18, 12) (B) (-9, 6) (C) (-3, 2) (D) (0, 0)

Solution: Midpoint of the line joining (-12, 8) and (x, y) is

, ( 6, 4) , .x yx y

212

28

0 0&- + += - = =c m ( Ans. (D) )

17. The equation of the straight line passing through the origin and perpendicular to the straight line x y2 3 7 0+ - = is

(A) x y2 3 0+ = (B) x y3 2 0- =

(C) y 5 0+ = (D) y 5 0- =

Solution: The required line is 3 0.x y k2- + = As it passes through the origin, k 0= ( Ans. (B) )


18. The equation of a straight line parallel to y-axis and passing through the point ,2 5-^ h is

(A) x 2 0- = (B) x 2 0+ =

(C) y 5 0+ = (D) y 5 0- =

Solution: Equation of the line parallel to y-axis is x k=

The line passes through (–2, 5). Thus, 2 2 0.x x&=- + = ( Ans. (B) )

19. If the points (2, 5), (4, 6) and ,a a^ h are collinear, then the value of a is equal to

(A) -8 (B) 4 (C) -4 (D) 8

Solution: Using the concept of slope, .aa a

4 26 5

25 8&

-- =

-- = ( Ans. (D) )

20. If a straight line y x k2= + passes through the point (1, 2), then the value of k is equal to

(A) 0 (B) 4 (C) 5 (D) -3

Solution: The line passes through the point (1, 2). So, 2(1) 0k k2&+ = =

( Ans. (A) ) 21. The equation of a straight line having slope 3 and y-intercept -4 is

(A) x y3 4 0- - = (B) x y3 4 0+ - =

(C) x y3 4 0- + = (D) x y3 4 0+ + =

Solution: 3, 4. , 3 4 3 4 0.Som c y mx c y x x y& &= =- = + = - - - =

( Ans. (A) ) 22. The point of intersection of the straight lines y 0= and x 4=- is

(A) ,0 4-^ h (B) ,4 0-^ h (C) ,0 4^ h (D) ,4 0^ h

Solution: The point intersection is ,4 0-^ h ( Ans. (B) )

23. The value of k if the straight lines 3x + 6y + 7 = 0 and 2x + ky = 5 are perpendicular is

(A) 1 (B) –1 (C) 2 (D) 21

Solution: 1m m1 2 &=- 1 1.k

k63 2 &- - =- =-` `j j

or using a a b b k k0 3 2 6 0 11 2 1 2

& &+ = + = =-^ h ( Ans. (B) )

Solution - Geometry 161

Exercise 6.1

1. In a ABCT , D and E are points on the sides AB and AC respectively such that DE BC< .

(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, then find AC. (ii) If AD = 8 cm, AB = 12 cm and AE = 12 cm, then find CE. (iii) If AD = 4x – 3, BD = 3x – 1 , AE = 8x – 7 and EC = 5x – 3, then find the value of x.

Solution: (i) In ,ABC DE BCT < . By Thales theorem (BPT), we have

DBAD

ECAE= 12 .cmEC

ADAE DB EC

68 9& &# #= = =

AC AE EC` = + = 8 + 12 = 20 cm (ii) Given that AD = 8 cm, AB = 12 cm and AE = 12 cm So, 12 8 4cmBD AB AD= - = - =

In ,ABC DE BCT < . By Thales theorem (BPT), we have

DBAD

ECAE= 6 .cmEC

ADAE DB EC

812 4& &# #= = =

` 6cmCE =

(iii) Given AD = 4x – 3, BD = 3x – 1 , AE = 8x – 7 and EC = 5x – 3. In ,ABC DE BCT < . By Thales theorem (BPT), we get

DBAD

ECAE= &

xx

xx

3 14 3

5 38 7

-- =

--

& (4 3)(5 3)x x- - = (8 7)(3 1)x x- -

& 4 2 2x x2- - = 0

& 2 1x x2- - = 0

& ( 1)(2 1)x x- + = 0

Thus, , 1x x21=- = . Since x

21!- (distance), we have 1x = .

2. In the figure, AP = 3 cm, AR = 4.5cm, AQ = 6cm, AB = 5 cm, and AC = 10 cm. Find the length of AD.

Solution: From the given data, we have APAB

35= and

AQAC

610

35= = .

Now, in ,ABCT APAB

AQAC=

Hence, by the converse of Thale’s theorem, PQ || BC

Let RD = x. In ,ABD PR BD3 < . So, APAB

ARAD=

.. x

35

4 54 5& = +

& 13.5 3 .x 22 5+ = or 3x = 9. & x = 3. Thus, 4.5 3 7.5AD AR RD cm= + = + = .

Geometry 6


Aliter : We have PB AB AP 5 3 2= - = - = cm, QC AC AQ 10 6 4= - = - = cm.

Now, PBAP

23= and

QCAQ

46

23= = . So, in ,ABCT

PBAP

QCAQ

=

Hence, by the converse of Thale’s theorem, PQ || BC

In ,ABDPBAP

RDART = . .

RD23 4 5& =

` RD = . 33

4 5 2# =

Hence, 4.5 3 7.5AD AR RD cm= + = + =

3. E and F are points on the sides PQ and PR respectively, of a PQRT . For each of the following cases, verify EF QR< .

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.

Solution: (i) Given that PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.

In , . 1.3PQREQPE

33 9T = = and

.

. 1.FRPF

2 43 6 5= =

EQPE

FRPF` !

Hence, EF is not parallel to QR.

(ii) Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.

So, .EQ

PE4 54

98= = and

FRPF

98=

EQPE

FRPF` =

Hence, by the converse of BPT, EF QR<

4. In the figure, AC BD< and CE DF< . If OA =12cm, AB = 9 cm, OC = 8 cm and EF = 4.5 cm , then find FO.

Solution: In ,OBD AC BDT < .

By Thales theorem (BPT), we have

ABOA =

CDOC 6 .

CDCD

912 8

128 9 cm& & #= = =

In ,ODF CE DFT < . By Thales theorem (BPT), we have

CDOC =

EFOE

.. 6 .OE OE

68

4 5 68 4 5 cm& & #= = =

Hence, 6 4.5OF OE EF= + = + = 10.5 cm.

5. ABCD is a quadrilateral with AB parallel to CD. A line drawn parallel to AB meets

AD at P and BC at Q. Prove that PDAP

QCBQ

= .

Solution: Construction: Join BD and let it meet PQ at E


In ,DAB PE ABT < . By Thales theorem (BPT), we have

PDAP

EDBE= g (1)

In ,BCD EQ DCT < . By Thales theorem (BPT), we have

EDBE

QCBQ

= g (2)

From (1) and (2), we get PDAP

QCBQ

= .

6. In the figure, PC QK< and BC HK< . If AQ = 6 cm, QH = 4 cm, HP = 5cm, KC = 18cm, then find AK and PB.Solution: In ,APC PC QKT < .

By Thales theorem, we have QPAQ

KCAK= .

AKQPAQ

KC& #= [aQP QH HP 4 5 9= + = + = ]

AK& = 12cm96 18# =

Also, given that in ,ABC BC HKT < . By Thales theorem, we have

HBAH

KCAK=

HB10& =

1812 [a AH = AQ + QH = 6+ 4 = 10]

& 15 .cmHB12

10 18#= =

Hence, PB = 15 5 10cmHB HP- = - =

7. In the figure, DE AQ< and DF AR< . Prove that EF QR< .Solution: In ,PQA DE AQT < . By Thales theorem, we have

EQPE =

DAPD g (1)

Also, given that in ,PAR DF ART < . By Thales theorem, we have

DAPD =

FRPF g (2)

From (1) and (2), we get EQPE

FRPF=

Hence, by the converse of Thale’s theorem, we have EF QR< .

8. In the figure,DE AB< and DF AC< . Prove that EF BC< .Solution: In ,ABP DE ABT < . By Thales theorem, we have

DAPD =

EBPE g (1)

Also, given that in ,CAP DF ACT < . By Thales theorem, we have

DAPD =

FCPF g (2)

From (1) and (2), we get EBPE

FCPF=

Hence, by the converse of Thale’s theorem, we have EF BC< .


9. In a ABCT , AD is the internal bisector of A+ , meeting BC at D. (i) If BD = 2 cm, AB = 5 cm, DC = 3 cm find AC. (ii) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm find BC. (iii) If AB = x, AC = x – 2, BD = x + 2 and DC = x – 1 find the value of x.

Solution: (i) Given that AD is the internal bisector of A+ .

In ABCT , by Angle bisector theorem, we have

ACAB =

DCBD

& AC5 =

32

& AC = 7.52

5 3 cm# =

(ii) Given that AD is the internal bisector of A+ . In ,ABCT by ABT, we have

ACAB =

DCBD

& .65 6 = . 2.8BD BD

3 65 6 3 cm& #= =

Hence, BC = 2.8 3 5.8cmBD DC+ = + =

(iii) Given that AD is the internal bisector of A+ . In ,ABCT by ABT, we have

ACAB =

DCBD

& x

x2-

= xx

12

-+

& ( 1)x x - = ( 2)( 2)x x+ -

& x x2- = x 42

- x` = 4

10. Check whether AD is the bisector of A+ of ABCT in each of the following. (i) AB = 4 cm, AC = 6 cm, BD = 1.6 cm, and CD = 2.4 cm. (ii) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 3 cm.

Solution: (i) We have, DCBD =

.

.2 41 6

32= g (1)

ACAB =

64

32= g (2)

From (1) and (2), we get DCBD

ACAB=

Hence, by the converse of ABT, AD is the bisector of A+ .

(ii) We have, DCBD = .

31 5

21= g (1)

ACAB =

86

43= g (2)

From (1) and (2) it follows DCBD

ACAB!

Hence, AD is not the bisector of A+ .


11. In a MNOT , MP is the external bisector of M+ meeting NO produced at P. If MN cm10= , MO = 6 cm, NO = 12 cm, then find OP.

Solution: Given that MP is the external bisector of M+ .

In ,MNOT by ABT, we have

OPNP =

MOMN

& OP

OP12 + = MOMN [ 12NP NO OP OPa = + = + ]

& 12OP

OP+ = 610

& 72 6 OP#+ = 10 4OP OP 72&# # =

Thus, OP = 18 cm

12. In a quadrilateral ABCD, the bisectors of B+ and D+ intersect on AC at E. Prove that BC

ABDCAD= .

Solution: DE is the internal bisector of D+ .

In ,ADCT by ABT, we have ECAE =

DCAD g (1)

BE is the internal bisector of B+ .

In ,ABCT by ABT, we have ECAE =

BCAB g (2)

From (1) and (2), we get BCAB

DCAD= .

13. The internal bisector of A+ of ABCT meets BC at D and the external bisector

of A+ meets BC produced at E. Prove that BEBD

CECD= .

Solution: In ABCT , AD is the internal bisector of A+ .

In ,ABCT by ABT, we have

DCBD =

ACAB g (1)

In ,ABCT AE is the external bisector of A+ .

By ABT, we have CEBE =

ACAB g (2)

From (1) and (2), we get

DCBD

CEBE= &

BEBD

CEDC

BEBD

CECD(or)= =

14. ABCD is a quadrilateral with AB =AD. If AE and AF are internal bisectors of BAC+ and DAC+ respectively, then prove that EF BD< .

Solution: In ,ABCT AE is the internal bisector of BAC+ .

So, by ABT, we have

ACAB =

ECBE g (1)


In ,ADCT AF is the internal bisector of DAC+ . So, by ABT, we have

ACAD

FCDF= or

ACAB

FCDF= g (2) [ AB = AD ]

From (1) and (2), we get ECBE

FCDF=

Hence, in CDBT , using the converse of Thale’s Theorem, we have EF BD< .

Aliter : In ,ADCT AF is the internal bisector of DAC+ .

By ABT, we have ADAC =

FDCF g (1)

In ,ABCT AE is the internal bisector of BAC+ . By ABT, we have ABAC =

EBCE .

Since AB = AD, we getADAC =

EBCE g (2)

From (1) and (2), we get FDCF

EBCE=

Hence, in CDBT , using the converse of BPT, we have EF || BD.

Exercise 6.2

1. Find the unknown values in each of the following figures. All lengths are given in centimetres. (All measures are not in scale)

(i) (ii) (iii)

Solution: (i) In ABCT and ,ADET we have ABC+ = ADE+ (corresponding angles)and A+ = A+ (common angle)

So, by AA criterion, we have ABC ADET T+ . Hence, AEAC

DEBC= .

& x

x8+

= 248 &

xx8 3

1+

= & 4x cm= .

Also, EAGT and ECFT are similar.

Thus, EAEC

AGCF= & AG

ECCF EA#= [a EA = EC + CA = 8 + 4 = 12]

& y = 12 9. , 9cmy86 Thus# = = .

(ii) In HBCT , FG < BC.

HFG HBCHBHF

BCFG&` T T+ = 3.6 cmx x

104

9 104 9& & #= = =

Consider .FBD FHGandT T Here BD GH<

FBD FHG`+ += [Alternate angles] BFD HFG+ += [Vertically opposite angles] Thus, by AA criterion of similarity, we have FBD FHGT T+

a


& FDFG

FBFH=

yx3 6

4&+

= .y 33 6

32&

+=

& 2 6 .y 10 8 &+ = 2.4cmy = .

Again, consider AEGT and ABCT . Here EG || BC

Thus, by AA criterion of similarity, we have AEG ABCT T+ . A+ is common

& ABAE =

BCEG AEG+ = ABC+

corresponding angles

& z

z5+

= x y9+

zz5 9

6&+

=

& 3z = 2 10z + & 0cmz 1=

(iii) EFCD is a parallelogram.

So, 7cmEF DC= = , 6cmDE CF= = . Consider AEFT and ABCT .

From the figure, it is clear that AEF ABCT T+

ACAF =

BCEF

& x

x6+

= 127 & x x12 7 42= +

& x = 8.4 cm Consider BDGT and BCFT . From the figure, it is clear that BDG BCFT T+

BCBD` =

CFDG

& DGBCBD CF#= & 2.5cmy

125 6#= =

2. The image of a man of height 1.8 m, is of length 1.5 cm on the film of a camera. If the film is 3 cm from the lens of the camera, how far is the man from the camera?Solution: Let AB be the height of the man,

CD be the height of the image of the man of height 1.8m,L be the position of the lens of the camera,LM be the distance between man and lens,LN be the distance between lens and film.

Then, AB || CD, AB = 1.8m, CD = 1.5cm, LN = 3cm,Consider LABT and LCDT , we have LAB+ = LCD+ [Alternate angles] BLA+ = DLC+ [Vertically opposite angles]Thus, by AA criterion of similarity, we have LAB LCDT T+

& CDAB =

LNLM &

.1 5180 = LM

3

& LM = .

360cmCDAB LN

1 5180 3# #= = = 3.6 m

Hence, the distance between the man and camera is 3.6m.

AEF ABC+ +=

corresponding angles

A+ is common` By AA criterionAEF ABCT T+


3. A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 0.6 m/sec. If the lamp is 3.6m above the ground level, then find the length of her shadow after 4 seconds.Solution: Let AB be the height of the Lamp post CD be the height of the girl CE be the length of the shadow of the girlThen, AB = 3.6 m, CD = 120 cm = 1.2 mGiven that the walking speed of the girl is 0.6 m/sec.The distance AC travelled by the girl in 4 seconds 4 0.6 2.4m#= =

Now, consider ECDT and EABT . Clearly, CD AB< . ECD+ = EAB+ [corresponding angles] E+ = E+ [common angle]` By AA criterion of similarity, we have ECD EABT T+

Thus, EAEC =

ABCD

& . ECEC

2 4 + =

.

.3 61 2

31= & .EC EC3 2 4= +

& EC = 1.2 mHence, the length of the shadow of the girl after 4 seconds is 1.2m.

4. A girl is in the beach with her father. She spots a swimmer drowning. She shouts to her father who is 50 m due west of her. Her father is 10 m nearer to a boat than the girl. If her father uses the boat to reach the swimmer, he has to travel a distance 126 m from that boat. At the same time, the girl spots a man riding a water craft who is 98 m away from the boat. The man on the water craft is due east of the swimmer. How far must the man travel to rescue the swimmer?

( This problem is not for examination )Solution: Let A be the position of the father C be the position of the girl B be the position of the boat D be the position of the water craft and E be the position of the swimmer. Let BC = x m. Then, AB = ( 10)mx -

Consider ABCT and DBET . We have, ABC+ = DBE+ [vertically opposite angles] BAC+ = BDE+ [alternate angles]Thus, by AA criterion of similarity, we have ABC DBET T+ .

So, DBAB =

BEBC

DEAC=

Now, DBAB =

BEBC x x

9810

126& - =


& x = 28

1260 = 45 ` BC = 45 m

Now, BEBC

DEAC= DE& =

BCAC BE#

& DE = 45

50 126# = 140

So, DE = 140 m

Hence, the man on the water craft must travel 140 m distance to rescue the swimmer.

5. P and Q are points on sides AB and AC respectively, of ABCT . If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.

Solution: Now, ABAP

93

31= = ,

ACAQ

155

31= =

From APQT and ABCT , we have

ABAP =

ACAQ

and A+ = A+ [common angle]

Thus, by SAS criterion of similarity, we have APQ ABCT T+

& ABAP =

ACAQ

BCPQ

=

Now, ABAP =

BCPQ

BCPQ

93& = & BC = 3PQ

6. In ABCT , AB = AC and BC = 6 cm. D is a point on the side AC such that cmAD 5= and CD = 4 cm. Show that BCD ACBT T+ and hence find BD.

Solution: Given that in ABCT , AB = AC

Now, ACBC =

96

32= ;

CBCD

64

32= =

Thus, in BCDT and ACBT , we have

ACBC =

CBCD

and C+ = C+ [common angle]

Thus, by SAS criterion of similarity, we have BCD ACBT T+

& ABBD =

ACBC

ACBD

96& = [ ]AB ACa =

& BD9

= 96

Hence, BD = 6 cm. 7. The points D and E are on the sides AB and AC of ABCD respectively, such that

DE || BC. If AB = 3 AD and the area of ABCT is 72 cm2, then find the area of the quadrilateral DBCE.

Solution: Given that, in figure, DE BC< and AB = 3AD

& ABAD =

31

Consider ADET and ABCT .


ADE+ = ABC+ [corresponding angles]

and A+ = A+ [common angle]

Thus, by AA criterion of similarity, we have ADE ABCT T+

Now, ABCADE

Area ofArea of

TT =

AB

AD2

2

ADE72 9

1area of& T =

Thus, ADEArea of T = 8cm2

Area of the quadrilateral DBCE = Area of ABCT - Area of ADET

= 72 8 64cm2- =

8. The lengths of three sides of a triangle ABC are 6cm, 4 cm and 9cm. .PQR ABCT T+ One of the lengths of sides of PQRT is 35cm. What is the greatest perimeter possible for PQRT ?

Solution: Given, PQR ABCT T+

& ABPQ =

Perimeter ofPerimeter of

BCQR

ACPR

ABCPQR

TT

= = g (1)

Let QR = 35.From (1), we see that the perimeter of PQRT is the greatest only when the corresponding side to QR must be BC.


ABCPQR

TT =

BCQR

435=

Thus, the greatest perimeter of PQRT = 19435

# = 166.25 cm.

Aliter: Since PQR ABCT T+ ,

we have PQ6

= PR435

9= (QR is the corresponding side to BC)

Consider PQ6

= 6 52.5PQ435

435

2105& #= = =

Again consider PR9

= 9 78.75PR435

435

4315& #= = =

The greatest possible perimeter of PQRT

= PQ QR PR+ + = 52.5 35 78.75 166.25cm+ + =

9. In the figure, DE || BC and BDAD

53= , calculate the value of

(i) ABCADE

area ofarea of

TT , (ii)

ABCBCED

area ofarea of trapezium

T.

Solution: (i) In ABC, DE || BC. ` ADE ABCT T+

Now, BDAD = 3 , 5AD k BD k

53 & = =

Also, ABCADE

Area ofArea of

TT =

AB

AD2

2

( )

( )

k

k

8

3649

2

2

& =


(ii) Area of 9ADE kT = and Area of 64ABC kT =

Now, Area of trapezium BCED = Area of ABCT - Area of ADET

= 64 9 55k k k- =

ABC

BCED

Area ofArea of trapezium

T =

kk

6455

= 6455 .

10. The government plans to develop a new industrial zone in an unused portion of land in a city. The shaded portion of the map shown on the right, indicates the area of the new industrial zone. Find the area of the new industrial zone.

Solution: Consider ,EAB EDCT T . Clearly AB CD<

Further, AEB+ = DEC+ [Vertically opposite angles]

EAB+ = EDC+ [Alternate angles]

By AA criterion of similarity, we have EAB EDCT T+

DCAB =

EGEF

& EF = ( . )DCAB EG

13 1 4# #= = 4.2km .

The area of the new industrial zone

= EABArea of T

= AB EF21# # 3 4.2 6.3km

21 2# #= =

11. A boy is designing a diamond shaped kite, as shown in the figure where,cmAE 16= EC = 81cm. He wants to use a straight cross bar BD. How long

should it be?

Solution: We know that if a perpendicular is drawn from the vertex of a right angled triangle to its hypotenuse. then the triangles on each side of the perpendicular are similar. So, EAD EDC+D D . Thus, we have

EDEA

ECED=

& ED2 = 16 81EA EC# #=

& ED = 4 9 3616 81# #= = .

Now, ABDT is an isosceles triangel and AE BD=

So, BE = ED

Thus, BD = ED2 = 2 × 36 = 72 cm.

(OR) Area of

Area of trapezium

ABC

BCED

T

= Area

ABC

ABC ADE

Area ofArea of of

TT T-

= ABC

ADE1

Area ofArea of

TT

- = 164

9

64

55- =


12. A student wants to determine the height of a flagpole. He placed a small mirror on the ground so that he can see the reflection of the top of the flagpole. The distance of the mirror from him is 0.5 m and the distance of the flagpole from the mirror is 3 m. If his eyes are 1.5 m above the ground level, then find the height of the flagpole.(The foot of student, mirror and the foot of flagpole lie along a straight line).

Solution: Let AB and ED be the heights of the man and the tower respectively. Let C be the point of incidence of the flagpole in the mirror. In ,ABC EDCandT T 90 ,ABC EDC BCA DCE+ + + += = =c

Thus, by AA criterion of similarity, we have ABC EDCT T+ .

So, EDAB =

DCBC . .

ED1 5

30 5& =

& . ED0 5 = 4.5 & 9mED = .

Thus, the height of the flagpole is 9m.

13. A roof has a cross section as shown in the diagram, (i) Identify the similar triangles, (ii) Find the height h of the roof.

Solution: (i) We know that if a perpendicular is drawn from the vertex of a right angled triangle to its hypotenuse, then the triangles on each side of the perpendicular are similar to the whole triangle.Thus from the given figure, we have the similar triangles(i) WZY YZXT T+ , (ii) WYX YZXT T+ and (iii) WZY WYXT T+ (or)(i) XWY YWZ+D D , (ii) YWZ XYZT T+ and (iii) XWY XYZT T+

(ii) In XWY XYZT T+ In YWZ XYZ+D D

YZWY

XZXY= & h

8 106= (or)

XYYW

XZYZ= & h

6 108=

4.8mh` = 4.8mh` =

Exercise 6.3

1. In the figure TP is a tangent to a circle. A and B are two points on the circle. If BTP 72+ = c and +ATB = 43c find +ABT.

Solution: BAT+ = 72PTB+ = c [ Angles in alternate segment]Now, in ABTT , we have ATB BAT ABT+ + ++ + = 180c

43 72 ABT++ +c c = 180c

ABT` + = 65c

[Angular elevation is same at the same instant. i.e. The angle of incidence and the angle of reflection are the same]


2. AB and CD are two chords of a circle which intersect each other internally at P. (i) If CP = 4 cm, AP = 8 cm, PB = 2 cm, then find PD. (ii) If AP = 12 cm,

AB = 15 cm, CP = PD, then find CD.Solution: (i) The chords AB and CD interset at a point P, inside the circle. we have PA PB# = PC PD#

& PD = PC

PA PB# = 4cm4

8 2# = .

(ii) Given that CP = PD, AP = 12cm. AB = 15 cm.

Now, 15 15 12 3 .cmAP PB PB&+ = = - =

Now, PA PB# = PC PD#

& PD2 = PA PB# PC PDa =6 @& PD2 = 36 & 6cmPD = .Hence, CD = 2 12cmPD = .

3. AB and CD are two chords of a circle which intersect each other externally at P (i) If AB = 4 cm BP = 5 cm and PD = 3 cm, then find CD. (ii) If BP = 3 cm, CP = 6 cm and CD = 2 cm, then find AB.

Solution: (i) Since the chords AB and CD meet externally at P, we have PA PB# = PC PD#

& 9 5# = [3 ] (3)CD+

& 3 CD+ = 15cm3

9 5# =

Thus, CD = 12 cm(ii) Given that CP = 6 cm& CD PD+ = 6 4cmPD` = .We have PA PB# = PC PD#

i.e., ( )AB PB PB#+ = PC PD#

& (3 ) 3AB #+ = 6 4#

& 3 AB+ = 83

6 4# =

Thus, AB = 5 cm. 4. A circle touches the side BC of TABC at P, AB and AC produced at Q and R

respectively, prove that AQ = AR = 21 ( perimeter of TABC)

Solution: We know that the lengths of two tangents drawn to a circle from an external point are equal. Thus, we have BQ = BP g (1) [tangents from the point B] CP = CR g (2) [tangents from the point C] AQ = AR g (3) [tangents from the point A]


Perimeter of ABCT = AB BC CA+ +

= AB BP PC CA+ + +

= ( ) ( )AB BP PC CA+ + +

= ( ) ( )AB BQ CR CA+ + + Using (1) and (2) = 2AQ AR AR AR AR+ = + = , Using (3)

& AR = ( )ABC21 Perimeter of T

Thus, from (3) we have AR = AQ ( )ABC21 Perimeter of T=

5. If all sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.

Solution: Let ABCD be a parallelogram. The sides AB, BC, CD and DA touch the circle at the points P,Q,R, and S respectively.

We know that the lengths of two tangents drawn to a circle from an external point are equal. Thus, (1) ; (2)AP AS BP BQg g= =

(3) ; (4)CR CQ DR DSg g= =

Adding (1), (2), (3) and (4), we get

AP BP CR DR+ + + = AS BQ CQ DS+ + +

& ( ) ( )AP BP CR DR+ + + = ( ) ( )AS DS BQ CQ+ + +

& AB CD+ = AD BC+

& 2 2AB AD= or AB AD=

Thus, we have AB BC CD AD= = = ABCD` is a rhombus.

6. A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water 40 cm away from the original position of the stem. How much of the stem was below the water surface originally?

Solution: Let O be the bottom of the stem immersed in water. Let B be the lotus. Then AB be the length of the stem above the water surface and OA be the length of the stem below the water surface.

Let OA = cmx

Let C be the point where the lotus touches the water surface when the wind blow.

Now, OC = OA + AB 20cmx= +

By Pythagoras theorem, we have OC OA AC2 2 2= +

( 20)x 2+ = 40x2 2

+

40 400x x2+ + = 1600x2 + & x40 = 1200 & x = 30 cm

Thus, the stem is 30 cm below the water surface.

[ABCD is a parallelogram &AB CD BC AD= = ]


7. A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that OA OC OB OD2 2 2 2

+ = + .

Solution: Through O, draw EOF AB< .

Now, ABFE and EFCD are rectangles In right OEAT , by Pythagoras Theorem OA OE EA2 2 2

= + g (1) In right OFCT , by Pythagoras Theorem OC OF FC2 2 2

= + g (2) In right OFBT , by Pythagoras Theorem OB OF FB2 2 2

= + g (3) In right OEDT , by Pythagoras Theorem OD OE ED2 2 2

= + g (4)

Adding (3) and (4),

OB OD2 2+ = OF FB OE ED2 2 2 2

+ + +

= ( ) ( )OE FB OF ED2 2 2 2+ + +

= ( ) ( )OE EA OF FC2 2 2 2+ + +

[ ABCDa and EFCD are rectangles, FB = EA and ED = FC

= OA OC2 2+ . using (1) and (2)

Exercise 6.4

1. If a straight line intersects the sides AB and AC of a ABCT at D and E respectively and is parallel to BC, then

ACAE =

(A) DBAD (B)

ABAD (C)

BCDE (D)

ECAD

Solution: By Thales theorem, ACAE

ABAD= ( Ans. (B) )

2. In ABCT , DE is < to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm , then AC is equal to

(A) 6.5 cm (B) 4.5 cm (C) 3.5 cm (D) 5.5 cm

Solution: By Thales Theorem

BDAD =

ECAE & EC =

ADAE BD# = . 1.8cm

32 7 2# =

Thus, AC = 2.7 1.8 4.5cmAE EC+ = + = ( Ans. (B) )

3. In PQRT , RS is the bisector of R+ . If PQ = 6 cm, QR = 8 cm, RP = 4 cm then PS is equal to

(A) 2 cm (B) 4 cm (C) 3 cm (D) 6 cm

Solution: Let cmPS x= . 6 cmSQ x= -

RS is the bisector of PRQ+ , we have

QRPR =

SQPS

84& = 2 6 2

xx x x x

6& &

-= - = ( Ans. (A) )


4. In figure, if ACAB

DCBD= , ,B 40c+ = and ,C 60c+ = then BAD+ =

(A) 30c (B) 50c

(C) 80c (D) 40c

Solution: ACAB

DCBD AD&= is the angle bisector of BAC+

But ABC BCA CAB+ + ++ + = 180c 40 60 2 BAD++ + = 180c & 40BAD+ = c ( Ans. (D) )

5. In the figure, the value x is equal to

(A) 4 2$ (B) 3 2$

(C) 0 8$ (D) 0 4$

Solution: By Thales theorem, .BDAD

ECAE x x

8 104 3 2& &= = = ( Ans. (B) )

6. In triangles ABC and DEF, ,B E C F+ + + += = , then

(A) DEAB

EFCA= (B)

EFBC

FDAB= (C)

DEAB

EFBC= (D)

FDCA

EFAB=

Solution: By AA - Criterion ~ABC DEFT T . Thus DEAB

EFBC= ( Ans. (C) )

7. From the given figure, identify the wrong statement.

(A) ADBT + ABCT (B) ABDT + ABCT

(C) BDCT + ABCT (D) ADBT + BDCT

Solution: ~ABD ABCT T is wrong statement ( Ans. (B) ) 8. If a vertical stick 12 m long casts a shadow 8 m long on the ground and at the same time

a tower casts a shadow 40 m long on the ground, then the height of the tower is

(A) 40 m (B) 50 m (C) 75 m (D) 60 m

Solution: Shadow of towerHeight of tower = Shadow of stick

Height of stick

& Height of tower = 40812

# = 60m. ( Ans. (D) )

9. The sides of two similar triangles are in the ratio 2:3, then their areas are in the ratio

(A) 9:4 (B) 4:9 (C) 2:3 (D) 3:2

Solution: 2 : 3 4 : 92 2= . ( Ans. (B) )

10. Triangles ABC and DEF are similar. If their areas are 100 cm2 and 49 cm2 respectively and BC is 8.2 cm then EF =

(A) 5.47 cm (B) 5.74 cm (C) 6.47 cm (D) 6.74 cm

Solution: DEF

ABC

EF

BC

EF

BC49100

AreaArea

2

2

2

2&

TT

= =^^

hh

Thus, . 5.74 cmEFBC EF

710

107 8 2& #= = = ( Ans. (B) )


11. The perimeters of two similar triangles are 24 cm and 18 cm respectively. If one side of the first triangle is 8 cm, then the corresponding side of the other triangle is

(A) 4 cm (B) 3 cm (C) 9 cm (D) 6 cm

Solution:

the

of the the

Perimeter of second trianglePerimeter first triangle

corresponding side of the second trianglea side of first triangle

=

Corresponding side of the second triangle = 6 cm24

8 18# = ( Ans. (D) )

12. AB and CD are two chords of a circle which when produced to meet at a point P such that AB = 5 cm, AP = 8 cm, and CD = 2 cm then PD =

(A) 12 cm (B) 5 cm (C) 6 cm (D) 4 cmSolution: Let PD = x cm & PA × PB = PC × PD

8 × 3 = (x + 2) x & x x2 24 02+ - = & x = 4 ( Ans. (D) )

13. In the adjoining figure, chords AB and CD intersect at P. If AB = 16 cm, PD = 8 cm, PC = 6 and AP > PB, then AP =

(A) 8 cm (B) 4 cm (C) 12 cm (D) 6 cm

Solution: Let PA = x cm & PA × PB = PC × PD

& x (16 – x) = 6 × 8 & x x16 48 02- + = & (x – 4) (x – 12) = 0

& x = 4 or x = 12. But AP > PB AP = 12 cm ( Ans. (C) ) 14. A point P is 26 cm away from the centre O of a circle and PT is the tangent

drawn from P to the circle is 10 cm, then OT is equal to

(A) 36 cm (B) 20 cm (C) 18 cm (D) 24 cm

Solution: 26 10 24OP OT TP OT2 2 2 2 2 2&= + = - = ( Ans. (D) )

15. In the figure, if PAB 120+ = c then BPT+ =

(A) 120o (B) 30o

(C) 40o (D) 60o

Solution: 120BCP+ + c = 180° (a ABCP is a cyclic quadrilateral)

` BCP+ = 60c

But BPT BCP+ += = 60c ( Ans. (D) ) 16. If the tangents PA and PB from an external point P to circle with centre O are

inclined to each other at an angle of 40o, then POA+ =

(A) 70o (B) 80o (C) 50o (D) 60o

Solution: Since , 20OAP OBP APOT T +, = c

In , 180OAP POA 90 20T + + + =c c c ` POA 70+ = c ( Ans. (A) )


17. In the figure, PA and PB are tangents to the circle drawn from an external point P. Also CD is a tangent to the circle at Q. If PA = 8 cm and CQ = 3 cm, then PC is equal to

(A) 11 cm (B) 5 cm

(C) 24 cm (D) 38 cm

Solution: PB = PA & PC + BC = 8 & PC + PQ = 8

& PC = 5 cm ( Ans. (B) )

18. ABCT is a right angled triangle where B 90+ = c and BD AC= . If BD = 8 cm, 4cmAD = , then CD is

(A) 24 cm (B) 16 cm (C) 32 cm (D) 8 cm

Solution: DBDC

DADB= & DCB DBA

DBDC

DADB&T T+ =

& DC DA# = DB2

& 4DC = 82 & DC = 16 cm ( Ans. (B) )

19. The areas of two similar triangles are 16 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 3 cm, then the corresponding altitude of the other triangle is

(A) 6.5 cm (B) 6 cm (C) 4 cm (D) 4.5 cm

Solution:

Corresponding altitude of the second trianglethe

the the

Area of second triangleArea of first triangle Altitude of first triangle

2

2

=^

^

h

h

& Altitude of second triangle = 16

36 9# = 4.5 cm ( Ans. (D) )

20. The perimeter of two similar triangles ABCT and DEFT are 36 cm and 24 cm respectively. If DE = 10 cm, then AB is

(A) 12 cm (B) 20 cm (C) 15 cm (D) 18 cm

Solution: DEFABC


TT =

DCAB

& AB = 24

36 10# = 15 cm ( Ans. (C) )

Solution - Trigonometry 179

Exercise 7.1 1. Determine whether each of the following is an identity or not. (i) cos sec sin2

2 2i i i+ = + , (ii) cot cos sin

2 2i i i+ = .

Solution: (i) When 45oi= , 2 2.5cos sec2

1 2212 2 2 2

i i+ = + = + =c ^m h

and 2+sini =2 + 2

1

Thus, if 45oi= , 2cos sec sin2 2

!i i i+ +

Hence, 2cos sec sin2 2i i i+ = + is not an identity.

(ii) When 30oi= , 3cot cos23 3

2322

i i+ = + = +^ h and 21sin

412

2

i= =` j Thus, if 30oi= , cot cos sin

22!i i i+ .

Hence, cot cos sin22

i i i+ = is not an identity.

2. Prove the following identities (i) sec cosec sec cosec2 2 22

i i i i+ =

Solution: We consider sec cosec22

i i+

= cos sin1 1

2 2i i+ =

cos sinsin cos

2 2

2 2

i ii i+ =

cos sin1

2 2i i

= .sec cosec22

i i

Note: In this problem, sum and product of two terms are equal.

(ii) cos

sin cosec cot1 i

i i i-

= + .

Solution: We consider cos

sin1 i

i-

= cos

sincoscos

1 11

#ii

ii

- ++c m = (1 )

cossin cos

1 2 #i

i i-

+

= (1 )sinsin cos

2ii i+ = 1

sincosii+

= sin sin

cos1i i

i+ = cosec coti i+

(iii) sinsin sec tan

11

ii i i

+- = - .

Solution: We consider sinsin

sinsin

sinsin

11

11

11

#ii

ii

ii

+- =

+-

--

= 1

sin

sin

12

2

i

i

-

-^ h

= cos

sin12

2

i

i-^ h = cossin1ii-

= sec tani i-

Aliter:

( )( )cos cosec cot1 i i i- +

= cosec cot cotsincos2i i iii+ - -

= sin i

Note: One can use90i = c also.

Trigonometry 7


(iv) 1sec tan

cos sini ii i

-= + .

Solution: We consider sec tan

cosi ii

-

= sec tan sec tan

cos sec tan

i i i i

i i i

- +

+

^ ^^

h hh

= 1coscos cos

sinii i

i+c m sec tan 12 2a i i- =^ h

= sin1 i+ .

Note: ( )( )sec tan sin sec tan tancossin1

2i i i i i i

ii- + = + - -

= cossin cos1 2

ii i- =

(v) sec cosec tan cot2 2i i i i+ = + .

Solution: We consider sec cosec2 2i i+

= 1 1tan cot2 2i i+ + +^ ^h h

= tan cot 22 2i i+ +

= 2tan cot tan cot2 2i i i i+ + tan cot 1a i i =^ h

= tan cot 2i i+^ h = tan coti i+

(vi) sin cos

cos sin cot1

12

i ii i i+

+ - =^ h

.

Solution: We consider sin cos

cos sin1

1 2

i ii i+

+ -^ h

= 1sin cos

sin cos1

2

i ii i+

- +^ h

= sin coscos cos

1

2

i ii i++

^ h

= ( )sin coscos cos

11

i ii i

++

^ h =

sincosii = coti

(vii) 1sec sin sec tan1i i i i- + =^ ^h h .

Solution: We consider sec sin sec tan1i i i i- +^ ^h h

= cos

sin sec tan1 1i

i i i- +^ ^h h

= 1cos cos

sin sec tani i

i i i- +c ^m h

= sec tan sec tani i i i- +^ ^h h

= sec tan2 2i i-^ h = 1

(viii) cosec cot

sini ii+

= 1 cosi- .

Solution: We consider cosec cot

sini ii+

= cosec cot

sincosec cotcosec cot

#i ii

i ii i

+ --c m

Aliter: ( ) { (1 )}cot sin cosi i i+

= ( )( )cos cos1i i+

= cos cos2i i+

= cos sin1 2i i+ -

Aliter: [ ( )] ( )sec sin sec tan1i i i i- +

= ( )( )sec tan sec tani i i i- +

= sec tan 12 2i i- =


= cosec cot

sin cossin sin2 2

1

i i

i

-

-i i

ic m = sin

sinsin

sincos1

#ii

iii- c m

cosec cot 12 2a i i- =^ h

= cos1 i-

Aliter-I Aliter-II

sin sincos

sin1i i

ii

+ =

cossin

1

2

ii

+ ( )( )cos cosec cot1 i i i- +

= coscos

11 2

ii

+- = cosec cot cot

sincos2i i iii+ - -

= 1

1 1

cos

cos cos

i

i i

+

- +^ ^h h = sin sin

cos sin1 2

i ii i- =

= cos1 i-

3. Prove the following identities.

(i) sin

sin

coscos sec

1

90

1 902

i

i

ii i

+

-+

- -=

c

c^

^h

h.

Solution: We consider sin

sin

coscos

1

90

1 90i

i

ii

+

-+

- -

c

c^

^h

h

= 1sin

cossin

cos1 i

iii

++

-

= sin sin

cos sin cos sin

1 1

1 1

i i

i i i i

+ -

- + +

^ ^^ ^

h hh h

= sin

cos cos sin cos cos sin

12i

i i i i i i

-

- + +

= coscos2

2ii =

cos2i

= sec2 i sin cos1 2 2a i i- =^ h

(ii) 1cot

tantan

cot sec cosec1 1i

iii i i

-+

-= + .

Solution: We consider cot

tantan

cot1 1i

iii

-+

-

= tantan

tancot

1 1

2

ii

ii

--

-

= tan

tantan1

1 12

ii

i--

^`

hj

= ( )tan tan

tan1

13

i ii-

-

= tan tan

tan tan tan

1

1 12

i i

i i i

-

- + +

^^ ^

hh h

= tan

tan tan12

ii i+ +

Aliter: We consider

sin

cos

cos

sin

cos

sin

sin

cos

1 1i

i

i

i

i

i

i

i

-

+

-

= ( ) ( )cos sin cos

sin

sin cos sin

cos2 2

i i i

i

i i i

i

-+

-

= ( )sin cos sin cos

sin cos3 3

i i i i

i i

-

-

= ( )

( ) ( )

sin cos sin cos

sin cos sin cos sin cos2 2

i i i i

i i i i i

-

- + +

= 1sin cos

sin coscosec sec

1

i i

i ii i

+= +


= tansec

tantan

2

ii

ii+ =

cos sincos1 1

2 #i i

i +

= 1cos sin1 1

#i i

+ = sec cosec 1i i+

(iii) tan

sincot

coscos sin

190

190

0 0

ii

ii

i i-

-+

--

= +^ ^h h .

Solution: We consider tan

sincot

cos190

190

0 0

ii

ii

--

+-

-^ ^h h

=

cossin

cos

sincos

sin

1 1iii

iii

-+

-

= cos sin

cossin cos

sin2 2

i ii

i ii

-+

-

= cos sin

cos sin1 2 2

i ii i

--^ h

= cos sin

cos sin cos sin

i i

i i i i

-

+ -^ ^h h

= cos sini i+

(iv) 2 .cosectan

cotcosec sec

190 1

0

ii

ii i

+-

+ + =^ h

Solution: We consider cosectan

cotcosec

190 1

0

ii

ii

+-

+ +^ h

= cosec

cotcot

cosec1

1ii

ii

++ +

= sin

coscossin

11

ii

ii

++ +

= cos sin

cos sin

1

12 2

i i

i i

+

+ +

^^

hh

= 1

2cos sin

cos sin sin12 2

i ii i i

++ + +

^ h

= 1

1

cos sin

sin2

i i

i

+

+

^^

hh = sec2 i

(v) .cot coseccot cosec cosec cot

11

i ii i i i- ++ - = +

Solution: We consider cot coseccot cosec

11

i ii i- ++ -

= ( )cot cosec

cot cosec cosec cot1

2 2

i ii i i i

- ++ - - ( )cosec cot 12 2a i i- =

Aliter:

= tan

costansin tan

1 1ii

ii i

-+

-

= tan

cos sincossin

1 i

i iii

-

- c m

= cos sincos sin cos sin

2 2

i ii i i i-- = +

Aliter:

= ( )( )

cot coseccot cosec

112 2

i ii i

++ +

= ( )

( ) ( )cot cosec

cosec cosec cosec1

1 1 22 2

i ii i i

+- + + +

= ( )( )

cot coseccosec cosec

sec1

2 12

i ii i

i++

=


= 1

( )( )cot cosec

cot cosec cosec cot cosec coti i

i i i i i i

- ++ - + -

= ( )( ( ))cot cosec


1i i

i i i i

- ++ - -

= ( )

( )( )cot cosec


1i i

i i i i

- ++ - +

= cot coseci i+

(vi) 2cot cosec tan sec1 1i i i i+ - + + =^ ^h h .

Solution: We consider cot cosec tan sec1 1i i i i+ - + +^ ^h h

= sincos

sin cossin

cos1 1 1 1

ii

i ii

i+ - + +c cm m

= (( ) )(( ) )sin cos

sin cos cos sin1 1i i

i i i i+ - + +

= ( )sin cos

sin cos 12

i ii i+ -

= sin cos

sin cos sin cos2 12 2

i ii i i i+ + -

= sin cossin cos1 2 1i ii i+ - ( 1)sin cos2 2a i i+ =

= sin cossin cos2 2i ii i =

(vii) sin cossin cos

sec tan11 1

i ii i

i i+ -- + =

-.

Solution: We consider sin cossin cos

11

i ii i+ -- +

= cossin

coscos

cos

cossin

coscos

cos1

1

+ -

- +

ii

ii

i

ii

ii

i

(Divide each term of bothnumerator and denominator by cos i )

= 1tan sec

tan sec1i ii i+ -- +

= tan sectan sec

11

i ii i+ -+ -

= ( )tan sec

tan sec sec tan1

2 2

i ii i i i

+ -+ - - ( 1 )sec tan2 2i i- =

= ( )( )tan sec

tan sec sec tan sec tan1i i

i i i i i i

+ -+ - + -

Aliter:

( ) ( 1)cosec cot cot coseci i i i+ - +

= cosec cot cosec cosec cot2 2

i i i i i- + +

cot cosec coti i i- +

= cot cosec cosec cosec 12 2

i i i i+ - + -

= cot cosec 1i i+ -

Aliter:Now ( )( )sec tan sin cos 1i i i i- - +

= tan seccossin sin tan1

2i i

ii i i- + - + -

= sincossin1 1 2

iii- + -c m

= sin cos1i i- +

Thus sin cossin cos

sec tan11 1

i ii i

i i- +- + =

-


= ( )( ( ))tan sec

tan sec sec tan1

1i i

i i i i

+ -+ - -

= ( )( )tan sec

tan sec sec tan11

i ii i i i

+ -+ - +

= tan seci i+

= ( )sec tansec tansec tan

#i ii ii i+--c m

= sec tansec tan2 2

i ii i--c m

= sec tan

1i i-

( 1)sec tan2 2a i i- =

(viii) tan

tan

sin

sin sin

1 2 90 1

900

0

2 2i

i

i

i i

-=

- -

-

^

^

h

h .

Solution: We consider tan

tan

12i

i

-= 1

cossin

cossin

2

2

-iiii

= cos

cos sincossin

2

2 2

ii iii

-

= cossin

cos sin

cos2 2

2

#ii

i i

i

-

= cos sin

sin cos2 2i i

i i

-

= ( )

cos cos

sin sin

1

902 2i i

i i

- -

-c

^ h ( )sin cos12 2a i i= -

= ( )

cos

sin sin

2 1

902i

i i

-

-c

= ( )

( )

sin

sin sin

2 90 1

902 i

i i

- -

-

c

c

(ix) cosec cot sin sin cosec cot

1 1 1 1i i i i i i-

- = -+

.

Solution: Now cosec cot 12 2i i- = ( )( ) 1cosec cot cosec cot& i i i i- + = ... (1)

We consider cosec cot sin

1 1i i i-

-

= ( )( )cosec cot

cosec cot cosec cotsin1

i ii i i i

i-- +

- by (1)

= ( )cosec cot coseci i i+ -

= coseccosec cot

1i

i i-

-^ h

= sin cosec cot1 1i i i-

+ by (1)


Aliter-I Aliter-II

cosec cot cosec cot

1 1i i i i-

++

cosec cot sin

1 1i i i-

-

= cosec cot

cosec cot cosec cot2 2i i

i i i i

-

+ + - = ( ) ( )cosec cot cosec cot

cosec cotsin1

i i i ii i

i- ++ -

= cosecsin sin sin

2 2 1 1ii i i

= = + = cosec cot

cosec cot cosec2 2i i

i i i-

+ -

Thus cosec cot sin

1 1i i i-

- =cosec cot coseci i i+ - =coti ...(1)

= sin cosec cot1 1i i i-

+ Now,

sin cosec cot1 1i i i-

+

= coseccosec cot

cosec cot2 2

ii i

i i-

-

-^ h

= cosec cosec coti i i- + = coti ...(2) From (1) and (2), the result follows.

(x) ( )tan cosec

cot sec sin cos tan cot2 2

2 2

i i

i i i i i i+

+ = +^ h.

Solution: We consider tan cosec

cot sec2 2

2 2

i i

i i

+

+

= ( )

( )

tan cot

cot tan

1

11

2 2

2 2

i i

i i

+ +

+ += ... (1)

( 1 , 1 )sec tan cosec cot2 2 2 2i i i i= + = +

Now, ( )sin cos tan coti i i i+

= sin coscossin

sincosi i

ii

ii+c m

= sin cos 12 2i i+ = ... (2)

Now (1) and (2) complete the proof.

4. If sec tanx a bi i= + and tan secy a bi i= + , then prove that .x y a b2 2 2 2- = -

Solution: We consider, x y2 2-

= ( ) ( )sec tan tan seca b a b2 2i i i i+ - +

= 2 ( 2 )sec tan sec tan tan sec tan seca b ab a b ab2 2 2 2 2 2 2 2i i i i i i i i+ + - + +

= sec tan tan seca a b b2 2 2 2 2 2 2 2i i i i- + -

= ( ) ( )sec tan tan seca b2 2 2 2 2 2i i i i- + -

= a b2 2- ( 1)sec tan2 2a i i- =

5. If tan tanni a= and ,sin sinmi a= then prove that cosn

m

1

12

2

2

i =-

- , 1n !! .Solution: Let us eliminate a from the given relations.

Given that tan tanni a= and sin sinmi a= .

Aliter:

[( ) ( )] ( )sin cos tan cot tan cosec2 2

i i i i i i+ +

= ( )( )sin cos tan cosec2 2 2 2i i i i+ +

= ( )( )sec cot1 1 12 2i i- + +

= sec cot2 2i i+


cottan

n& ai

= and cosecsin

mai

=

We know that

cosec cot2 2a a- = 1 1sin tan

m n2

2

2

2&

i i- =

& sin

cosm n2

2 2 2

ii- = 1

& cosm n2 2 2i- = sin2i

& cosm n2 2 2i- = 1 cos2i-

& m 12- = ( 1)cos n2 2i -

Thus, nm

11

2

2

-- = cos2i .

6. If ,sin cos tanandi i i are in G.P., then prove that 1.cot cot6 2i i- =

Solution: Given that ,sin cos tanandi i i are in G.P.

Thus, sincosii =

costanii

cos3& i = sin2i g (1)

We consider cot cot6 2i i- =

sincos

sincos

6

6

2

2

ii

ii- = ( )

sincos

sincos

6

3 2

2

2

i

i

ii-

= ( )sinsin

sincos

6

2 2

2

2

i

i

ii- using (1)

= sinsin

sincos

6

4

2

2

ii

ii- =

sin sincos1

2 2

2

i ii-

= sincos12

2

ii-

= sinsin

2

2

ii = 1.

Exercise 7.2

1. A ramp for unloading a moving truck, has an angle of elevation of 30°. If the top of the ramp is 0.9 m above the ground level, then find the length of the ramp.

Solution: Let C be the top of the ramp and AC be the length of the ramp.

Given that 0CAB 3+ = c and BC =0.9 m.

In the right 3CAB, sin 30° = ACBC

& sin

AC BC30

=c

= . . .m0 9 2 1 8# =

Thus, the length of the ramp is 1.8m. 2. A girl of height 150 cm stands in front of a lamp-post and casts a shadow of length

150 3 cm on the ground. Find the angle of elevation of the top of the lamp-post.

Solution: Let BC be the height of the girl and let AB be the length of her shadow on the ground. Let i be the angle of elevation of the top of the lamp-post.


Given that 150 , 150AB BC3 cm cm= =

In the right 3CAB, tani = ABBC

150 3

150

3

1= = = 30tan o

& i = 30°

Thus, the angle of elevation of the top of the lamp-post is 30o .

3. Suppose two insects A and B can hear each other up to a range of 2. The insect A is on the ground 1 m away from a wall and sees her friend B on the wall, about to be eaten by a spider. If A sounds a warning to B and if the angle of elevation of B from A is 30c, will the spider have a meal or not ?

(Assume that B escapes if she hears A calling)

Solution: Let A be the position of the insect A on the ground and B be the position of another insect B on the wall OB.

Given that 1 30OA m BAOand o+= =

In the right TBAO, cos 30° = ABAO

& AB = cosAO30c

& AB = 3

2

= 3

2

3

33

2 3# =

= . 2 0.5773

2 1 732# #= = 1.154m

That is, the distance between the insects is 1.154m which is less than 2m.

Thus, B can hear A’s warning and hence B escapes. So, the spider will not have a meal.

4. To find the cloud ceiling, one night an observer directed a spotlight vertically at the clouds. Using a theodolite placed 100 m from the spotlight and 1.5 m above the ground, he found the angle of elevation to be 60°. How high was the cloud ceiling? (Not for Examination)

Solution: Let B be the positon of the spotlight, AC be the height of the cloud ceiling and E be the position of the theodolite. Then

100 , 1.5 0BE m BC m AEB 6and o+= = = .

In the right ABET , 60tanBEABo

= 60tanAB BE& #= c

& AB = 3 100# = 1.732 100#

& AB = 173.2m

Now, AC = AB + BC = 173.2m + 1.5m = 174.7m

Thus, the cloud ceiling was 174.7m high.


Aliter: Height h of the cloud ceiling is given by tanh x y i= + where x is the distance from the observing point to the ground and y is the distance from the observing point to the theodolite and i is the angle of elevation.

h = 1.5 100 60tan#+ c

= 1.5 100 3#+ 1.5 100 1.732#= +

= 1.5 173.2+ = 174.7m

Thus, the cloud ceiling was at 174.7m high.

5. A simple pendulum of length 40 cm subtends 60° at the vertex in one full oscillation. What will be the shortest distance between the initial position and the final position of the bob?

Solution: Let O be the point about which the pendulum oscillates. Let A and B be the extrem ends of the oscillation where the angle of oscillation is 60°.

Given that OA = OB = 40cm and 0AOB 6+ = c. Draw OC perpendicular to AB. Since OA = OB, OC is the perpendicular bisector of AB and angle bisector of AOB+ 30AOC o`+ = .

In the right TCOA, sin 30° = OAAC

AC = 30 40 20sin cmOA21

#= =c

Since C is the mid-point of AB,

AB = 2AC = 2 20 40 cm# =

Thus, the shortest distance between the initial and the final position of the bob at that oscillation is 40cm.

6. Two crows A and B are sitting at a height of 15 m and 10 m in two different trees vertically opposite to each other . They view a vadai (an eatable) on the ground at an angle of depression 45° and 60° respectively. They start at the same time and fly at the same speed along the shortest path to pick up the vadai. Which bird will succeed in it?

Solution: Let A be the position of the crow A and B be the position of crow B. Let C and E be the foot of the trees. Let D be the position of the vadai.

Given that AC = 15m, BE = 10m, 45ADC o+ = and

BDE 60o+ = .In the right TADC, sin 45° =

ADAC &

2

1 = AD15

AD = 15 15 1.414 21.21m2 #= =


Thus, distance travelled by crow A to reach vadai is 21.21 m.

In the right TBDE, sin 60° = BDBE

& BD = sinBE60c

BD3

20( =

= 3

20

3

33

20 3# #=

= .3

20 1 732# = .20 0 574#

Thus, BD = 11.48m Thus, distance travelled by crow B to reach the vadai is 11.48m. Since crow B travels shortest distance to reach the vadai, Crow B will succeed in picking up the vadai.

7. A lamp-post stands at the centre of a circular park. Let P and Q be two points on the boundary such that PQ subtends an angle 90c at the foot of the lamp-post and the angle of elevation of the top of the lamp post from P is 30c. If PQ = 30 m, then find the height of the lamp post.Solution: Let O be the centre of the park and OR be the lamp post. P and Q are two points on the boundary of the circular park. Given that PQ = 30m, POQ 90o+ = .In the right TOPQ, POQ 90o+ = , OP = OQ = radius. So OPQ OQP 45+ += = c

OP = 45cosPQ o#

& OP = 2

30 = 2

30 2 15 2=

In the right TRPO, tan 30° = OPOR

OR = 30tanOP o#

& OR = 15 23

1# =

3

15 2

3

3#

= m3

15 6 5 6=

Thus, the height of the lamp post is 5 m6 .Aliter: Let O be the centre of the park and OR be the lamp post. P and Q are two points on the boundary of the circular park.Given that PQ = 30m, POQ 90o+ = , OP = OQ ( radius )In the right TOPQ, OP OQ PQ2 2 2

+ = & OP2 302 2

= & OP2

30 302 #=

& OP2

30=

In the right TRPO, 0tanOPOR3 o

= &OPOR

3

1=

& OR OP

3 2 3

30

#= = ( )OP

2

30a =

= 5 m6

30

6

6630 6 6# = =


8. A person in an helicopter flying at a height of 700 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30 45andc c. Find the width of the river. ( 1.7323 = )

Solution: Let C be the point of observation. Let A and B be two objects lying opposite to each other on either bank of a river. Draw CD AB= . CD is the distance between the observer and the river. Given that CD = 700m, 30 ,CAD CBD 45o o+ += = .

In the right , 45 700tanCDBDBCD DB CD&T = = =c m

In the right TCAD, tan 30° = ADCD &

30tanAD CD

o=

& AD = 700 m3

Now, the width of the river, AB = AD DB+

= 700 700 700( 1) 700( .732 )3 3 1 1+ = + = +

= 700(2.732) . . m1912 400 1912 4= =

Thus, the width of the river 1912.4m.

9. A person X standing on a horizontal plane, observes a bird flying at a distance of 100 m from him at an angle of elevation of 30c. Another person Y standing on the roof of a 20 m high building, observes the bird at the same time at an angle of elevation of 45c. If X and Y are on the opposite sides of the bird, then find the distance of the bird from Y.Solution: Let B be the position of the bird. Draw BD XA= and CY BD= .Then, CD = 20m. Given that BX = 100m, 30 , 45BXD BYCo o+ += = . Let the distance of the bird from the person Y is BY.In the right TBXD, sin 30° =

BXBD

& BD = 30sinBX o# & 50m

2100 =

Now, BC BD CD= - = 50 20 30m m m- =

In the right TYBC, sin45c = BYBC &

BY2

1 30=

& BY = 30 m2

Thus, the distance of the bird from the person Y is 30 m2 . 10. A student sitting in a classroom sees a picture on the black board at a height

of 1.5 m from the horizontal level of sight. The angle of elevation of the picture is 30c. As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation of 45c. Find the distance moved by the student.Solution: Let A and B be the position of the student when the angle of elevation of the picture to him is 30° and 45° respectively. AB is the distance moved by the student.


Given that 30 , 45DAC DBCo o+ += = and CD = 1.5m.

In the right TDBC, tan 45° = BCCD = .

BC1 5 & .BC m1 5=

In the right TDAC, tan 30° = ACCD

& 3

1 = .AC1 5 & 1.5AC m3=

So, AB = AC BC- 1.5 .3 1 5= - 1.5( 1)3= -

= 1.5(1.732 1) 1.5(0.732)- = = 1.098 mThus, the distance moved by the student is 1.098m.

11. A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. The angle of elevation from his eyes to the top of the building increases from 30 60toc cas he walks towards the building. Find the distance he walked towards the building.Solution: Suppose the boy initially stands at Al and moves to the position Bl towards the building .C Dl

In the figure, ABC is a horizontal at the level of the boy.

Given that . , ,AA BB CC m C D m1 5 30= = = =l l l l

30 60DAC DBCando o+ += =

Then, . .CD C D CC m30 1 5 28 5= - = - =l l .

The distance walked by the boy is AB.

In the right TDBC, tan 60° = BCCD &

tanBC CD

60o=

& BC = .

3

28 5 = .

3

28 5

3

3# = (28.5)

33 = 9.5 3^ h

In the right TDAC, tan 30° = ACCD &

tanAC CD

30=

c

& AC = 28.5 3

& AB BC+ = 28.5 3

& AB = . .28 5 3 9 5 3-^ ^h h 19 m3=

Thus, the distance walked by the boy is 19 m3 . 12. From the top of a lighthouse of height 200 feet, the lighthouse keeper observes a

Yacht and a Barge along the same line of sight . The angles of depression for the Yacht and the Barge are 45 and 30c c respectively. For safety purposes the two sea vessels should be atleast 300 feet apart. If they are less than 300 feet , the keeper has to sound the alarm. Does the keeper have to sound the alarm?Solution: Let A and B be the positions of Yatch and Barge respectively. Let D be the point of observation. Let CD be the light house.


Given that CD = 200feet, 30DAC DBC 45ando o+ += = . AB is the distance between the Yatch and Barge. In the right TDBC, DBC CDB 45+ += = c. So BC = CD = 200 feet

In the right TDAC, tan 30° = ACCD

& AC = tan30200

c

& AC = 200 3

Now, AB = AC – BC 200 2003= -

= 200( 1)3 - 200(1.732 1)= -

= 200(0.732) = 146.4 feetThe distance between the sea vessels is 146.4 feet which is less than 300 feet.So, the keeper has to sound the alarm.

13. A boy standing on the ground, spots a balloon moving with the wind in a horizontal line at a constant height . The angle of elevation of the balloon from the boy at an instant is 60c. After 2 minutes, from the same point of observation,the angle of elevation reduces to 30c. If the speed of wind is 29 3m/min. then, find the height of the balloon from the ground level.

Solution: Let A be the point of observation.

Let E and D be the positions of the balloon when its angles of elevation are 60o and 30o respectively.

Let B and C be the points on the ground such that

BC = ED and BE = CD.

Given that 60 30 .EAB DACando o+ += =

Speed of the wind = /m29 3 minute.

Distance covered by the balloon in 2 minutes,

BC = ED = 29 2 m3 58 3# = (Distance = Speed # Time)

In the right TEAB, tan 60° = ABBE

& BE = 60tanAB AB 3o= ...(1)

In the right , 30tanDACT c = ACCD

ACBE= CD BEa =^ h

& BE = 30tanAC CA AB BC

3 3= = +c = AB BC

3 3+

& BE = AB BE33 58

358+ = + ( From (1) and BC 58 3= )

& BE 131-` j = 58 58 87 .BE m

23& #= =

Thus, the distance of the ballon from the ground level is 87m.


14. A straight highway leads to the foot of a tower . A man standing on the top of the tower spots a van at an angle of depression of 30c. The van is approaching the tower with a uniform speed. After 6 minutes, the angle of depression of the van is found to be 60c. How many more minutes will it take for the van to reach the tower?

Solution: Let A be the point of observation. Let B and C be the positions of the van when the angle of depression from A are 30o and 60o respectively. Then 30 60 .ABD ACDando o+ += =

In the right TABD, 30tanBDADo

=

& BD = tanAD BD AD30

3& =c

... (1)

In the right TACD, 0tanCDAD6 o

= & 0tanAD CD6 o= CD3= ... (2)

Now, BC = BD – CD = AD CD3 -

= CD CD3 3 -^ ^h h = CD CD CD3 2- = ( Using (2) )

Now, the time taken to cover the distance BC is 6 minutes.

Thus, the time required to cover the distance CD = 26 = 3 minutes.

Hence, the van will take 3 more minutes to reach the tower.

15. The angles of elevation of an artificial earth satellite is measured from two earth stations, situated on the same side of the satellite, are found to be 30cand 60c. The two earth stations and the satellite are in the same vertical plane. If the distance between the earth stations is 4000 km, find the distance between the satellite and earth. ( ).3 1 732=

Solution: Let A and B be the positions of the two earth stations.

Let D be the position of the artificial earth satellite.

Let CD be the distance between the satellite and the earth.

Given that AB = 4000 km, 30 60DAC DBCando o+ += = .

In the right TDBC, tan 60° = BCDC

& DC = 60tanBC o

& DC = BC3 g (1)

In the right TDAC, tan 30° = ACDC

& DC = 30tanAC o

& DC = BC

3

4000 + g (2)

Now, (1) and (2) implies BC3 = BC

3

4000 +


& BC3 = 4000 BC+ & BC2 4000=

& BC = 2000 km

Using BC = 2000 in (1), we get

DC = 20003 #

= 1.732 2000# = 3464 km

Thus, the distance between the satellite and the earth is 3464 km.

16. From the top of a tower of height 60 m, the angles of depression of the top and the bottom of a building are observed to be 30 60andc crespectively. Find the height of the building.

Solution: Let AE be the building and BD be the tower.

Draw EC || AB such that AE = BC. Let AE = h metres.

Then, BC = h metres.

Given that BD = 60m, 30 60DEC DABando o+ += = .

Now, CD = BD BC- h60= -

In the right TDAB, tan 60° = ABBD

& AB = tanBD60o =

3

60 ...(1)

In the right TDEC, tan30c = tanEC

CD EC CD30

& =c

& AB = (60 )h 3- ( )EC AB= ...(2)

(1) and (2) & (60 )h 3- = 3

60

& 60 h- = 360

& 60 h- = 20 h& = 40 m

Thus, the height of the building is 40m.

17. From the top and foot of a 40 m high tower, the angles of elevation of the top of a lighthouse are found to be 30cand 60c respectively. Find the height of the lighthouse. Also find the distance of the top of the lighthouse from the foot of the tower.

Solution: Let AE be the tower and BD be the light house.

Draw EC || AB such that AE = BC. Then, AB = EC.

Given that AE = 40m, 0 0DAB DEC6 3ando o+ += = .


Let CD = x metre. Then, BD = BC + CD = 40 + x.

In the right TDAB, tan 60° = ABBD

& AB = tan

x AB x60

40

3

40&+ = +c

g (1)

In the right TDEC, tan 30° = ECCD =

ABCD AB EC=^ h

& AB = tan

x AB x30

3& =c

g (2)

(1) and (2) & x3 = x

3

40 + & x x3 40= +

& x2 = 40

Thus, the height of the light house, BD = 40 + x = 40 + 20 = 60 m.

In the right TDAB, sin 60° = ADBD

& AD = sinBD60c

40AD3

120 3& = =

Thus, the distance of the top of the light house from the foot of the tower is 40 3 m.

18. The angle of elevation of a hovering helicopter as seen from a point 45 m above a lake is 30c and the angle of depression of its reflection in the lake, as seen from the same point and at the same time, is 60c. Find the distance of the helicopter from the surface of the lake.

Solution: Let A be the point of observation which is 45m above the lake. Let BD be the surface of lake. Then AB = 45m. Let F be the position of the helicopter and C be its reflection in the lake. Let FD = h metre. DC = h metre. Draw AE || BD.

Then ED = 45m, 30 60 .FAE CAEando o+ += =

In the right TFAE, tan 30° = AEFE

& 3

1 = AE

h 45- ( FE = FD – ED )

& AE = 45h 3-^ h g (1)

In the right TACE, tan 60° = AEEC =

AEED DC+

& AE 3 = 45 + h

& h = 45 45AE h3 45 3- = - -^ ^h h ( From (1) )

& = 3h –180& 2h = 180 & h = 90Thus, the distance of the helicopter from the surface of the lake is 90 m.


Exercise 7.3


1. sin sec12 2i i-^ h =

(A) 0 (B) 1 (C) tan2i (D) cos2i

Solution: 1sin sec cos sec12 22 2

i i i i- = =^ h ( Ans (B) )

2. tan sin12 2i i+^ h =

(A) sin2i (B) cos2i (C) tan2i (D) cot2i

Solution: tan sin sec sincos

sin tan12 2

2

222 2

i i i ii

i i+ = = =^ h ( Ans (C) )

3. cos cot1 12 2i i- +^ ^h h =

(A) sin2i (B) 0 (C) 1 (D) tan2i

Solution: cos cot sin cosec1 1 12 2 2 2i i i i- + = =^ ^h h ( Ans (C) )

4. sin cos cos sin90 90i i i i- + -c c^ ^h h =

(A) 1 (B) 0 (C) 2 (D) –1

Solution: sin cos cos sin cos cos sin sin90 90 1i i i i i i i i- + - = + =c c^ ^h h ( Ans (A) )

5. 1cos

sin1

2

ii-

+ =

(A) cosi (B) tani (C) coti (D) coseci

Solution: 1 11

)(1 (1 )

cossin

cos

cos coscos cos

1

1 12

ii

i

i ii i-

+= -

+

+ -= - - =

^ h

( Ans (A) ) 6. cos sinx x

4 4- =

(A) 2 1sin x2

- (B) 2 1cos x2

- (C) 1 2sin x2

+ (D) 1 2 .cos x2

-

Solution: ( )( )cos sin cos sin cos sinx x x x x x2 2 2 24 4

- = + -

(1 ) 2cos sin cos cos cosx x x x x 12 2 2 2 2= - = - - = - ( Ans (B) )

7. If tani = xa , then the value of

a x

x2 2+

= (A) cosi (B) sini (C) coseci (D) seci

Solution: 1 tan sec

cosa x

x 1

1

1 1

x

a 22 2

2

2 i ii

+=

+=

+= = ( Ans (A) )

8. If secx a i= , tany b i= , then the value of ax

b

y2

2

2

2

- =

(A) 1 (B) –1 (C) tan2i (D) cosec2i

Solution: sec tan sec tanax

b

y

aa

bb 1

2

2 2

2

2 22 2

2

2

2

2i i i i- = - = - = . ( Ans (A) )


9. cot tan

seci ii

+ =

(A) coti (B) tani (C) sini (D) – coti

Solution: cot tan

sec sin

sin coscos sin

cos1

2 2i ii i

+= =

i ii ii

+ ( Ans (C) )

10. tan

sin sin

cot

cos cos90 90

i

i i

i

i i-+

-c c^ ^h h =

(A) tani (B) 1 (C) –1 (D) sini

Solution: tan

sin sin

cot

cos cos cos sin sin cos90 90

cossin

sincosi

i i

i

i i i i i i-+

-= +

ii

ii

c c^ ^h h

= cos sin 12 2i i+ = ( Ans (B) ) 11. In the adjoining figure, AC =

(A) 25 m (B) 25 3 m

(C) 3

25 m (D) 25 2 m

Solution: 60 25 60tan tanAC AC25

25 3o o&= = = m ( Ans (B) )

12. In the adjoining figure ABC+ =

(A) 45c (B) 30c

(C) 60c (D) 05 c

Solution: tan tanABC ABC ABC100

100 3 3 60o& &+ + += = = ( Ans (C) )

13. A man is 28.5 m away from a tower. His eye level above the ground is 1.5 m. The angle of elevation of the tower from his eyes is 45c. Then the height of the tower is

(A) 30 m (B) 27.5 m (C) 28.5 m (D) 27 m

Solution: Height of the tower = tanx y i+

= 1.5 28.5 45 1.5 28.5 30tan mo#+ = + = ( Ans (A) )

14. In the adjoining figure, sini = 1715 . Then BC =

(A) 85 m (B) 65 m

(C) 95 m (D) 75 m

Solution: sini = 1715 . From figure sini =

ACBC

ACBC BC

1715

1715 85 75&` #= = = m ( Ans (D) )


15. 1 tan sin sin1 12i i i+ - +^ ^ ^h h h =

(A) cos sin2 2i i- (B) sin cos2 2i i-

(C) sin cos2 2i i+ (D) 0

Solution: 1tan sin sin sec cos sin cos1 1 12 2 2 2 2i i i i i i i+ - + = = = +^ ^ ^h h h

( Ans (C) ) 16. cot cos cos1 1 12i i i+ - +^ ^ ^h h h =

(A) tan sec2 2i i- (B) sin cos2 2i i-

(C) sec tan2 2i i- (D) cos sin2 2i i-

Solution: 1cot cos cos cosec sin sec tan1 1 12 2 2 2 2i i i i i i i+ - + = = = -^ ^ ^h h h

( Ans (C) ) 17. 1 1 1cos cot2 2i i- + +^ ^h h =

(A) 1 (B) –1

(C) 2 (D) 0

Solution: 1cos cot sin cosec1 1 1 1 1 02 2 2 2#i i i i- + + =- + =- + =^ ^h h

( Ans (D) ) 18.

11

cottan

2

2

i

i

++ =

(A) cos2i (B) tan2i

(C) sin2i (D) cot2i

Solution: cottan

cosecsec

cossin tan

11

2

2

2

2

2

22

i

i

i

i

i

i i++ = = = ( Ans (B) )

19. 1

sintan12

2i

i+

+ =

(A) cosec cot2 2i i+ (B) cosec cot2 2i i-

(C) cot cosec2 2i i- (D) sin cos2 2i i-

Solution:

1sintan

sinsec

sin cos cosec cot1

1 122

22

2 2 2 2ii

ii

i i i i++

= + = + = = -

( Ans (B) ) 20. 9 9tan sec2 2i i- =

(A) 1 (B) 0

(C) 9 (D) –9

Solution: 9 9 9( ) 9tan sec sec tan2 2 2 2i i i i- =- - =- ( Ans (D) )

Solution - Mensuration 199

Exercise 8.1 1. A solid right circular cylinder has radius 14 cm and height 8 cm. Find its curved

surface area and total surface area.Solution: Given that radius r = 14 cm and height h = 8 cm Curved Surface Area, CSA = 2 rhr

= 2 × × ×722 14 8= 704 sq.cm

Total Surface Area, TSA = 2 ( )r h rr +

= 2 × 722 ×14(8+14)

Thus, the Total Surface Area = 88 × 22 = 1936 cm2. 2. The total surface area of a solid right circular cylinder is 660 sq.cm. If its diameter

of the base is 14 cm, find the height and curved surface area of the cylinder.Solution: Let r and h be the radius and height of a right circular cylinder respectively.Given that TSA = 660 cm2, 2r = 14 or r = 7 Total Surface Area, 2rr(h + r) = 660& 2×

722 ×7×(h + 7) = 660

& h = ×2 22660 – 7 = 8 cm

Thus, the Curved Surface Area 2 rhr = 2×722 ×7×8 = 352 cm2.

3. Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.Solution: Given that Curved SurfaceArea, CSA = 4400 cm2 Circumference of the base of the cylinder, 2 rr = 110 cm

& 2×722 ×r = 110

& diameter, 2r = ×22

110 7 = 35 cm

Now, Curved Surface Area, rh2r = 110×h = 4400

Thus, the height of the cylinder h = 1104400 = 40 cm.

4. A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost to paint the lateral surface of the pillars at ` 20 per sq.m.Solution: Let r and h be the radius and height of a right circular pillar respectively. Given that r = 50 cm = 0.5 m and h = 3.5 m

Curved surface area of the pillar = rh2r = 2×722 ×0.5×3.5 = 11 m2

Cost of painting per sq.m = ` 20 Hence, the cost of painting for 12 pillars = 12×20×11 = ` 2640.

Mensuration 8


5. The total surface area of a solid right circular cylinder is 231cm2 . Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.Solution: Given that Total Surface Area of a right circular cylinder = 231 cm2

Curved Surface Area = 32 × Total surface area

& rh2r = 32 ×231 = 154 cm2

Now, Total Surface Area ( )r h r2r + = 231

& rh2r +2 r2r = 231

& 2 r2r = 231–154 = 77

& r2 = 277r

= ××

××

2 2277 7

2 27 7=

Thus, the radius of the cylinder r = 27 cm

Now, rh2r = 154 & 2×722 ×

27 ×h = 154

& h = × ×× ×

2 22 7154 7 2 = 7

Thus, the height of the cylinder h = 7 cm.

6. The total surface area of a solid right circular cylinder is 1540 cm2. If the height is four times the radius of the base, then find the height of the cylinder.

Solution: Given that Total Surface Area, TSA = 1540 cm2 , height h = r4 or r h4

=

Total Surface Area, ( )r h r2r + = 1540

& 2×722 × h

4 h h

4+` j = 1540

& h45 2

= 2 22

1540 7 4## # & h2 =140×7×

54 = 28×7×4

Hence, the height of the cylinder, h = × ×28 7 4 = 28 cm.

7. The radii of two right circular cylinders are in the ratio of 3 : 2 and their heights are in the ratio 5 : 3. Find the ratio of their curved surface areas.

Solution: Let r1 , r2 be the radii of the cylinders and let h1 , h2 be their heights. Given that r1 : r2 = 3 : 2 and h1 : h2 = 5 : 3. The ratio of the curved surface areas = r h2 1 1r : 2 r h2 2r = 3 × 5 : 2 × 3 = 5 : 2.

8. The outer curved surface area of a hollow cylinder is 540r sq.cm. Its internal diameter is 16 cm and height is 15 cm. Find the total surface area. Solution: Let R and r be the outer and inner radii and h be the height of the hollow cylinder respectively. Given that height h = 15cm, internal diameter 2r = 16 cm i.e., r = 8cm


Outer curved surface area = 540rcm2 & Rh2r = 540r

Outer radius R = 2 15

540# # rr = 18cm

Thus, Total surface area = ( )( )R r R r h2r + - +

= 2 (18 8) 18 8 15r + - +^ h

= 2 26 25# # #r =1300r

Hence, the total surface area = 1300rcm2.

9. The external diameter of a cylindrical shaped iron pipe is 25 cm and its length is 20 cm. If the thickness of the pipe is 1cm, find the total surface area of the pipe.

Solution: Let R, r and h be the external, internal radii and length of the pipe respectively. Given that 2R = 25cm & R = 12.5cm and thickness w = 1 cm.

Internal radius, r = R – w = 12.5 – 1 = 11.5cm

Total surface area = ( )( )R r R r h2r + - +

= 2 ( . . ) ( . . )722 12 5 11 5 20 12 5 11 5# # + + -

Hence, the total surface area = 3168cm2.

10. The radius and height of a right circular solid cone are 7 cm and 24 cm respectively. Find its curved surface area and total surface area.

Solution: Given that, radius r = 7 cm and height h = 24 cm.

Slant height, l = h r2 2+ = 7 242 2

+ = 25cm

Curved surface area = rlr

= 7722 25# # = 550 cm2

Total surface area = ( )r l rr +

= 7 (2 )722 5 7# # + = 704 cm2.

11. If the vertical angle and the radius of a right circular cone are 60c and 15 cm respectively, then find its height and slant height.Solution: In the figure, OAB is the cone.Draw OC CB= . Given that the vertical angle AOB 60+ = c and AC = 15 cm. So

AOC+ = 30AOB2 2

60+ = =c c

Consider the right angled OACD . We have

tan 30° = OCAC

OC3

1 15& =

& OC = 15 3

Hence, the height of the cone is 15 3 cm.


Also, sin 30° = AOAC

AO21 15& =

& AO = 30Hence, the slant height of the cone is 30 cm.Aliter: TOAB is an equilateral triangle. Since AB = 2AC = 30 cm, we have the slant height AO = 30 cm. Height of the cone = 30

23 15 3# = cm.

( If a is a side of an equilateral triangle, then its height is a23 )

12. If the circumference of the base of a solid right circular cone is 236 cm and its slant height is 12 cm , find its curved surface area.

Solution: Given that slant height, l 12= cm and Circumference = 236 cm

r2r = 236 & r 118r = cm Curved surface area, rlr = 118 12 1416# = cm2.

13. A heap of paddy is in the form of a cone whose diameter is 4.2 m and height is 2.8 m. If the heap is to be covered exactly by a canvas to protect it from rain, then find the area of the canvas needed.

Solution: Let r and h be the radius and height of a heap of paddy.

Given that height h = 2.8m and diameter 2r = 4.2 m or r = 2.1m Slant height l = h r2 2

+ = . .2 8 2 12 2+ = 3.5 m

Area of the canvas, rlr = 2.1 3.5722

# # = 23.1 sq.m.

Hence, the area of the canvas needed to protect the heap of paddy from rain is 23.1m2

14. The central angle and radius of a sector of a circular disc are 180c and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.

Solution: Given that the central angle of the sector 180i = c and the radius of the sector r = 21 cm. By joining the edges of the sector, a hollow cone is formed.

Let R be the radius of the cone. Circumference of the base of the cone = Arc length of the sector

& R2r = r360

2#i r

Thus, radius of the cone R = 21360180

# = 10.5 cm.

15. Radius and slant height of a solid right circular cone are in the ratio 3 : 5. If the curved surface area is 60r sq.cm, then find its total surface area. Solution: Let r and l be the radius and slant height of a right circular cone respectively.


Ratio of radius to slant height = 3 : 5

& r : l = 3 : 5 lr

53& = & r l

53=

Curved surface area, rlr = 60r

& l l53

# #r = 60r & l2 = 6035

##

rr

= 100

Thus, the slant height of the cone, l = 10cm. So r = l53 = 6 cm

Hence, Total surface area ( )r l rr + = 6 (6 10)722

# # + = 7

2112 = 30175 cm2.

16. If the curved surface area of solid a sphere is 98.56 cm2 , then find the radius of the sphere.

Solution: Given that curved surface area of a solid sphere = 98.56cm2

Thus, 4 r2r = 98.56

& 4 r722 2

# # = 98.56

& r = 4 22

98.56 7#

# = . .7 84 2 8=

Hence, Radius r = 2.8cm.

17. If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.

Solution: Let r be the radius of the solid hemisphere. Given that curved surface area of the solid hemisphere 2 r2r = 2772cm2

& r2r = 2

2772 = 1386

Total surface area r3 2r = 3 × 1386 = 4158 cm2.

18. Radii of two solid hemispheres are in the ratio 3 : 5. Find the ratio of their curved surface areas and the ratio of their total surface areas.

Solution: Let r1 and r2 be the radii. Given that r1 : r2 = 3:5

Ratio of the curved surface area = 2 : 2r r1

2

2

2r r = 3 : 52 2 = 9 : 25

Ratio of the total surface area = 3 : 3r r1

2

2

2r r = 3 : 52 2= 9 : 25

19. Find the curved surface area and total surface area of a hollow hemisphere whose outer and inner radii are 4.2 cm and 2.1 cm respectively.

Solution: Let R and r be the outer and inner radii of the hollow hemisphere respectively. Given that R = 4.2cm and r = 2.1cm

Curved surface area, ( )R r2 2 2r + = 2 (4.2 2.1 )2 2r +

= 2 (17.64 4.41)r +

Hence, the curved surface area = 44.1r cm2.


Total surface area 2 ( ) ( )R r R r2 2 2 2r r+ + - = .44 1r + (4.2 2.1 )2 2r -

= .44 1r+ (17.64 4.41)r -

Thus, the total surface area = 44.1 13.23 57.33r r r+ = cm2.

20. The inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of `5 per sq. m.

Solution: Let r be the radius of the hemispherical dome.

Given that base circumference of the dome 2 rr = 17.6m

& r = . 2.82 2217 6 7

## =

Now, curved surface area, r2 2r = 2 2.8 2.8722

# # # = 49.28 m2

Cost of painting for 1m2 = ` 5

Hence, the total cost of painting for the dome = 49.28 × 5 = ` 246.40.

Exercise 8.2

1. Find the volume of a solid cylinder whose radius is 14 cm and height 30 cm.

Solution: Let r and h be the radius and height of the solid cylinder respectively.

Given that radius r = 14cm and height h = 30cm.

Volume of the cylinder = r h2r

= 14 14722 30 18480# # # = cm3.

2. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm.

If the bowl is filled with soup to a height of 4 cm, then find the quantity of soup to be prepared daily in the hospital to serve 250 patients?

Solution: Let r and h be the radius and height of a cylinderical bowl respectively.

Given that diameter 2r = 7cm, Radius r = 27 cm and Height h = 4cm.

The quantity of the soup = r h2r

= 722

27

27 4 154# # # = cm3

Quantity of the soup needed to

serve 250 patients daily3 = 154 × 250 = 38500 cm3

Thus, the quantity of the soup needed = 100038500 = 38.5 litres.

3. The sum of the base radius and the height of a solid right circular solid cylinder is 37 cm.If the total surface area of the cylinder is 1628 sq.cm, then find the volume of the cylinder.Solution: Let r and h be the radius and height of the right circular cylinder respectively. Given that the sum of the radius and height (r + h) = 37 cm.The total surface area of the cylinder = 1628 sq.cm.


Total surface area of the cylinder 2 r h rr +^ h = 628cm1 2

& 2 rr = 37

1628

& r = 37

162821

227

# #

Thus, the radius of the cylinder r = 7cm

Now, sum of the radius and height, r + h = 37 & h = 30 cm

Hence, the volume of the cylinder, r h2r = 7 30 4620cm722 2 3

# # = .

4. Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.Solution: Let r and h be the radius and height of the right circular cylinder respectively.Given that h = 4.5 cm. Volume of the solid cylinder, r h2r = 62.37 cm3

& r2 = .h

62 37r

= 62.37.

.227

4 51 4 41# # =

& r = . 2.14 41 =

Thus, the radius of the cylinder is 2.1cm.

5. The radii of two right circular cylinders are in the ratio 2 : 3. Find the ratio of their volumes if their heights are in the ratio 5 : 3.

Solution: Let r1 , r2 be the radii and h1 , h2 be the heights of the two right circular

cylinders respectively. Given that r1 : r2 = 2 : 3 and h1 : h2 = 5 : 3.

& r

r

2

1 = 32 & r

1 = r

32

2 and

h

h

2

1 = 35 & h

1= h

35

2

Now, the ratio of the volumes of the cylinders = 2 : 2r h r h1

2

1 2

2

2r r

= 2 : 2r h r h32

352

2

2

2 2

2

2#r r` j

Thus, the ratio of the volumes of the cylinders is : 12720 = 20 : 27

6. The radius and height of a cylinder are in the ratio 5 : 7. If its volume is 4400 cu.cm, find the radius of the cylinder.Solution: Let r and h be the radius and height of the right circular cylinder respectively.

Given that r : h = 5 : 7 & h = r57

Volume of the cylinder, r h2r = 4400

& r r722

572

# # = 4400

& r3 = 22 7

4400 7 5## # = 1000

Thus, the radius of the cylinder, r = 10 cm.


7. A rectangular sheet of metal foil with dimension 66 cm # 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.

Solution: Let r and h be the radius and height of the right circular cylinder respectively. Given that the dimension of the sheet is 66 12cm cm# . So, 66 , 12cm cml b= = (After the rectangular sheet rolled into a hollow cylinder , the length of the rectangular sheet is equal to the base circumference of the hollow cylinder.) Thus, base circumference, r2r = l

& 2 r722

# # = 66

& r2 2266 7

221

##= =

Height of the cylinder = length of the rectangular sheet & h = b = 12 cm.

Thus ,the volume of the cylinder = 221r h

722 122 2

# #r = ` j = 4158 cm3

8. A lead pencil is in the shape of right circular cylinder. The pencil is 28 cm long and its radius is 3 mm. If the lead is of radius 1 mm, then find the volume of the wood used in the pencil.Solution: Let R and h be the radius and height of the pencil which is in the form of a right circular cylinder. Let r be the radius of the lead. Given that R = 3 mm , h = 28 cm = 280 mm and r = 1 mm.

Now, the volume of wood = h R r2 2r -^ h = 280722 3 12 2

# # -^ h

= 22 40 8 7040# # =

Thus, the volume of the wood = 7040 mm3 = 7.04 cm3.

9. Radius and slant height of a right circular cone are 20 cm and 29 cm respectively. Find its volume.

Solution: Let r, h and l be the radius, height and slant height of the right circular cone respectively.

Given that r = 20 cm and l = 29 cm.

Now, h = l r2 2- = 29 202 2

- = 21 cm

Thus, the volume of the cone = r h31 2r = 20 21

31

722 2

# # # = 8800 cm3

10. The circumference of the base of a 12 cm high wooden solid cone is 44 cm. Find the volume of the wooden solid.Solution: Let r and h be the radius and height of the wodden solid cone respectively.Given that h = 12 cm. Base circumference of the wodden solid, 2 rr = 44

& r = 244r

= 2 2244 7## = 7 cm


Volume of the wooden solid = r h31 2r

= 7 131

722 22

# # # cm3.

Thus, the volume of the wooden solid = 166 cm3.

11. A vessel is in the form of a frustum of a cone. Its radius at one end and the height

are 8 cm and 14 cm respectively. If its volume is 3

5676 cm3, then find the radius at the other end. Solution: Let R, r be the radii and h be the height of the vessel which is in the form of

a frustum. Given that R = 8 cm, h = 14cm and volume = 3

5676 cm3

Now, Volume = ( )h R r Rr31 2 2r + + =

35676

& 14 (8 8 )r r31

722 2 2

# # # #+ + = 3

5676

& 64 8r r2+ + =

3 22 145676 3 7# #

# # = 129

& 8 64r r2+ + = 129 & 8 6 0r r 52

+ - =

& ( 13)( 5)r r+ - = 0 & r = –13 or r = 5

Since r cannot be negative, the radius at other end r = 5 cm.

12. The perimeter of the ends of a frustum of a cone are 44 cm and 8.4r cm. If the depth is 14 cm, then find its volume.Solution: Let R, r be the radii and h be the height of the frustum respectively. Given that 2 44cmRr = , 2 .r 8 4r r= cm and h = 14 cm. So,

R = 244

227 7

## = and . .r

28 4 4 2rr= = cm

Now, Volume of the frustum = ( )h R r Rr31 2 2r + +

= 14( . . )31

722 7 4 2 7 4 22 2

# # #+ +

= ( . . )344 49 29 4 17 64+ +

Thus, the volume of the frustum = 1408.57cm3. 13. A right angled ABCT with sides 5 cm, 12 cm and 13 cm. is revolved about the

fixed side of 12 cm. Find the volume of the solid generated.Solution: The sides of a right angled ABCT are 4cm, 12cm and 13cm.If the triangle is revolved about the side , then the volume generatedis a cone. The radius and height of the cone are respectively 5cm and 12 cm.

Volume of the cone = r h31 2r 5 5 12

31

722

# # # #= = 7

2200 = 31472 cm3.


14. The radius and height of a right circular cone are in the ratio 2 : 3. Find the slant height if its volume is 100.48 cu.cm. ( Take r = 3.14)Solution: Let r, h and l be the radius, height and slant height of the right circular cone

respectively. Given that r : h = 2:3 or hr r h

32

32&= =

Volume of the cone, r h31 2r = 100.48

& 3.14 h h31

32 2

# # #` j = 100.48

& h3 = .. 8 273 14 4

100 48 3 9## # #=

& h = 8 273# = 2 × 3 = 6cm

Now, r = h32 & r = 2

36 4# = cm.

Hence, slant height l r h2 2= + = 6 42 2

+ = 2 3 22 2+ = 2 cm13 .

15. The volume of a cone with circular base is 216r cu.cm. If the base radius is 9 cm, then find the height of the cone.Solution: Let r and h be the radius and height of the cone respectively. Given that r = 9 cm

Now, Volume of the cone, r h31 2r = 216r

& 9 h31 2# # #r = 216r

Thus, the height of the cone, h = 9 9

216 3# #

#rr = 8 cm

16. Find the mass of 200 steel spherical ball bearings, each of which has radius 0.7 cm, given that the density of steel is 7.95 g/cm3 . (Mass = Volume × Density) Solution: Let r be the radius of a spherical ball bearing. Given that r = 0.7cm

Volume of a ball bearing = r34 3r

= 0.7 0.7 0.734

722

# # # #

Thus, the volume of 200 ball bearings = . .3

88 0 049 200 287 46# # = cm3

Density of 1cm3 = 7.95 gThus, Mass of 200 ball bearings = 287.46 × 7.95 = 2285.316 g

= . .1000

2285 316 2 29= kg.

17. The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.Solution:Let R and r be the outer and inner radii of the hollow sphere.Given that R = 12 cm and r = 10 cm


Now, the volume of the hollow sphere = ( )R r34 3 3r -

= (12 10 )34

722 3 3

# - = (1728 1000)2188 -

= 7282188

# = 305032 cm3.

18. The volume of a solid hemisphere is 1152r cu.cm. Find its curved surface area.Solution: Let r be the radius of the hemisphere. Given that volume of the hemisphere = 1152rcm3

& r32 3r = 1152r

& r3 = 115223

# = 1728

& r = 17283 cm = 12 cm

Hence, the curved surface area = 2 r2r & 2 288r2# #r r= cm2.

19. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.Solution: Given that edge (side) of the cube = 14 cm.

If the largest circular cone is cut out from the cube,

then the radius of the cone = 2

side of the cube

& Radius r = 214 7= cm.

Now, Height of the cone h = 14cm

Hence, the volume of the cone = 7 7 14r h31

31

7222

# # # #r =

= 718.673

2156 = cm3.

20. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.Solution: Let r1 and r2 be the radii of the spherical balloon before and after air is being pumped into it. Given that r1 = 7 cm and r2 =14 cm.

So, : :r r 7 141 2

= .

Thus, the ratio of the volumes of the balloon in the two cases = 4 : 4r r

1

3

2

3r r

= 7 :143 3

= 7 7 7 :14 14 14# # # #

= 1 : 8


Exercise 8.3 1. A play-top is in the form of a hemisphere surmounted on a cone. The diameter

of the hemisphere is 3.6 cm. The total height of the play-top is 4.2 cm. Find its total surface area.Solution: Hemispherical portion: Diameter 2r = 3.6 cm & r = 1.8 cm

Conical portion : Radius r = 1.8cm, Height h = 4.2–1.8 = 2.4 cm

Slant height l = . .h r 2 4 1 82 2 2 2+ = +

= . ( . ) ( )0 6 4 3 0 6 52 2+ = = 3 cm.

Total surface area of the top = CSA of the hemispherical portion+ CSA of the conical portion = 2 r rl2r r+ = (2 )r r lr +

= 1.8(2 1.8 3)# #r +

= 1.8 6.6 11.88# #r r= cm2. 2. A solid is in the shape of a cylinder surmounted on a hemisphere. If the diameter

and the total height of the solid are 21 cm, 25.5 cm respectively, then find its volume.Solution: Hemispherical portion: Diameter 2r = 21cm & r =

221 cm

Conical portion: Radius r = 221 cm, height h = 15cm

The volume of the solid = Volume of the

hemisphere

Volume of the

cylinder+e eo o

= r r h32 3 2r r+ = ( )r r h

322r +

= 722

221

221

32

221 15# # # +` j

= 33 22 762221 3# # = cm3.

3. A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm, find its surface area.

Solution: Hemispherical portion: Radius r = 25 mm.

Cylindrical portion: Radius, r = 25 mm

Height, h = Total height – 2(Radius) = 14 – 5 = 9 mm

Curved surface area of the capsule = CSA of the

cylinder

CSA of the

hemisphere2 #+e eo o

= 2 4rh r2r r+ = 2 ( 2 )r h rr +

= 2722

25 9 2

25 220# # #+ =` j

Hence, the curved surface area of the capsule = 220 mm2.


4. A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.

Solution: Cylindrical portion: Height, h = 3 m, diameter 2r = 28m & r = 14mConical portion: Height, h

1 = Total height – Height of the cylinder

h1 = 13.5 – 3=10.5m

Radius, r = 14m

Slant height, l = .h r 10 5 141

2 2 2 2+ = +

= 0.7 ( . ) ( )15 20 0 7 252 2+ =

= 17.5 m

Total surface area of the tent = CSA of the

cylinder

CSA of the

cone+e co m

= 2 (2 )rh rl r h lr r r+ = + = 14(2 3 17.5)722

# # +

= 44 (6 17.5)# + = 44 23.5 1034# =

Hence, the total surface area of the tent = 1034 sq.m. 5. Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the

sphere.Solution: Let r1 and h be the radius and height of a right circular cone. Let r2 be the radius of the spherical shaped clay.Given that r1 = 12cm, h = 48cm.After the conical clay reshaped in to a spherical shaped clay, Volume of the sphere = Volume of the cone

r34

23r = r h

31

1

2r

r2

3 = 31 12 48

432

# # # #r

r = 123

Hence, the radius of the spherical clay = 12cm.

6. The radius of a solid sphere is 24 cm. It is melted and drawn into a long wire of uniform cross section. Find the length of the wire if its radius is 1.2 mm.

Solution: Let r1 and r2 be the radii of the solid sphere and wire respectively. Let h be length (height) of the wire.

Given that r1 = 24cm = 240 mm, r2 = 1.2mm = 1012 mm

The solid sphere is melted and recast into a long wire (cylindericalshape) Volume of the wire (cylinderical shape) = Volume of the solid sphere


& r h2

2r = r34

1

3r

& h1012

1012

# # #r = 240 240 24034# # # #r

& h = 34 240 240 240

1210

1210

# # # # # #r

r

= 12800000 mm

Thus, the length of the wire = 12.8100000012800000 = km.

7. A right circular conical vessel whose internal radius 5 cm and height 24 cm is full of water. The water is emptied into an empty cylindrical vessel with internal radius 10 cm. Find the height of the water level in the cylindrical vessel.Solution: Conical vessel : Radius r1 = 5 cm, height h1 = 24 cm Cylindrical vessel : Radius r2 = 10 cm. Volume of water in the cylinder = Volume of the conical vessel.

r h2

2

2r = r h

31

1

2

1r

& 10 h2

2# #r = 5 24

31 2# # #r

& h2 = 31

10 105 5 24

## #

# # #rr

Hence, the height of water level in the cylinder = 2 cm. 8. A solid sphere of diameter 6 cm is dropped into a right circular cylindrical vessel

with diameter 12 cm, which is partly filled with water. If the sphere is completely submerged in water, how much does the water level in the cylindrical vessel increase?Solution: Let h be the height of water raised in the cylinder.Given that diameter of the solid sphere 2r2 = 6 cm and Diameter of the cylinder 2r1 = 12 cm. The solid sphere is immersed into the cylinderical vessel. Then Volume of water raised = Volume of the solid sphere

& r h1

2r = r34

2

3r

& 6 6 h# # #r = 3 3 334# # # #r

& h = 34

6 63 3 3 1## #

# # #rr =

Thus, the water level in the cylindrical vessel is raised to 1 cm. 9. Through a cylindrical pipe of internal radius 7 cm, water flows out at the rate of

5 cm/sec. Calculate the volume of water (in litres) discharged through the pipe in half an hour.Solution: Given that radius of cylindrical pipe, r = 7cm Speed of water = 5 cm/sec. Time = 30 minutes = 30 × 60 = 1800 seconds.


Volume of water discharged in one second = r h2r

= 7 7 5722

# # #

= 770 cu. cm Volume of water discharged in half an hour = 770 × 1800 = 1386000 cu.cm

Hence, the volume of water discharged in half an hour = 1000

1386000

= 1386 litres.

Aliter: Volume of water discharged in half an hour = Area of the cross section × (time × speed)

= ( r2r ) × 5 × 1800 = 722 7 7# #` j × 5 × 1800

= 1386000 cu.cm

Hence, the volume of water discharged in half an hour = 1000

1386000

= 1386 litres.

10. Water in a cylindrical tank of diameter 4 m and height 10 m is released through a cylindrical pipe of diameter 10 cm at the rate of 2.5 Km/hr. How much time will it take to empty the half of the tank? Assume that the tank is full of water to begin with.Solution : Let r1 and h1 be the radius and height of a cylinderical water tank. Let r2 and h

2 be the radius and the height of the pipe.

Let T be the time taken to empty half of the cylinderical tank.

Cylindrical tank : Diameter 2r1 = 4m & r1 = 2m and height h1 = 10m

Cylindrical pipe : Diameter 2r2 = 10 cm & r2 = 1005 m

Speed = 2.5 km/hr = 2500 m/hr.

Thus, h2

= Time × Speed = T × 2500.

The volume of water discharged 3

through the cylindrical pipe = 21 Volume of water tank

& r h2

2

2r =

21 r h

1

2

1r` j

& T1005

1005 2500# # # #r = 2 2 10

21# # # #r

& T = 2

2 2 105 5

100 10025001# # # #

# ## #

rr

= .2580 3 2= hours

Thus, the time taken to empty half of the tank = 3 hr 12 min.


11. A spherical solid material of radius 18 cm is melted and recast into three small solid spherical spheres of different sizes. If the radii of two spheres are 2cm and 12 cm, find the radius of the third sphere.

Solution: Let R be the radius of the spherical solid material.

Let r1, r2 and r3 be the radii of the three solid spheres.

Given that, R = 18 cm. Let, r1 = 2 cm, r2 = 12cm. Volume of the three spheres = Volume of the solid sphere

& r r r34

34

34

1

3

2

3

3

3r r r+ + = R34 3r

& ( )r r r34

1

3

2

3

3

3r + + = R34 3# #r

& 2 12 r3 3

3

3+ + = 183

& r3

3 = 5832 1736- = 4096 = 163 & r3 = 16Thus, the radius of the third sphere = 16 cm.

12. A hollow cylindrical pipe is of length 40 cm. Its internal and external radii are 4 cm and 12 cm respectively. It is melted and cast into a solid cylinder of length 20 cm. Find the radius of the new solid.Solution : Let R, r and h be the external, internal radii and height of the hollow cylinderical pipe respectively. Let r1 and h1 be the radius and height of the solid cylinder respectively.Hollow Cylinder: R = 12cm, r = 4cm, h = 40 cm.Solid Cylinder: h1 = 20 cmVolume of the solid cylinder = volume of the hollow cylinder

& r h1

2

1r = ( )h R r2 2r -

& 20r1

2# #r = 40(12 4 )2 2

#r -

& r1

2 = ( )20

40 144 16#

#r

r - = 256 & r1 = 16.

Hence, the radius of the solid cylinder is 16 cm.

13. An iron right circular cone of diameter 8 cm and height 12 cm is melted and recast into spherical lead shots each of radius 4 mm. How many lead shots can be made?Solution: Let r and h be the radius and height of the iron right circular cone respectively. Let r1 be the radius of the spherical lead shot.

Now, r = 4m = 40mm, h = 12 cm = 120mm, r1 = 4mm


Let n be the number of lead shots.

Thus, n × (volume of the spherical shot) = Volume of the right circular cone

& n r34

1

3# # r = r h

31 2# # #r

& 4n34 3

# # #r = 40 12031 2# # #r

& n = 75031

4 4 440 40 120

43

## # #

# # # #r

r =

Hence, the number of spherical lead shots = 750.

14. A right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 6 cm, having a hemispherical shape on top. Find the number of such cones which can be filled with the ice cream available.Solution : Let r1 and h1 be the radius and height of a right circular cylinder respectively. Let r2 be the radius of the cone and hemispherical top and h2 be the height of the cone. Right circular cylinder: Diameter 2r1 = 12cm & r1 = 6cm and h1 =15cmCone: Diameter, 2r2 = 6cm & r2 = 3cm, h2 = 12cm No. of cones needed to fill with the ice cream

= Volume of the conical part Volume of the hemispherical partVolume of the cylinder

+

= r h r

r h

2

2

2 2

3

1

2

1

31

32r r

r

+ =

( )r h r2

6 6 15

2

2

2 231 # #

# # #

r

r

+

= ( )3 3 12 2 3

6 6 15

31 # # #

# #

+ =

36 6

1815 10

## # =

Thus, the number of the cones needed is 10.

15. A container with a rectangular base of length 4.4 m and breadth 2 m is used to collect rain water. The height of the water level in the container is 4 cm and the water is transferred into a cylindrical vessel with radius 40 cm. What will be the height of the water level in the cylinder?Solution:Cylindrical vessel: Let radius, r = 40 cm, and height of the water level, h = 4cmRectangular Let length, l = 4.4 m = 440 cm,breadth, b = 2 m = 200 cm and height h1= 4cm.

After the rain water transferred into the cylindrical vessel,


Volume of water in the cylinder = Volume of water in the rectangular container& r h2r = lbh1

& 40 40 h722

# # # = 440 200 4# #

& h = 40 40 22

440 200 4 7# #

# # # = 70

Hence, the height of water level in the cylinder is 70 cm.

16. A cylindrical bucket of height 32 cm and radius 18 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.Solution: Let r and h be the radius and height of the cylinderical bucket respectively.Let r1, h1 and l1 be the radius, height and slant height of a conical heap respectivelyCylinderical bucket: Radius r = 18cm, Height h = 32cmConical heap: Height h1 = 24cm

Volume of the conical heap = Volume of the sand in the cylindrical bucket

& r h31

1

2

1r = r h2r

& 24r31

1

2# # #r = 18 18 32# # #r

& r1 = 18 18 4# # = 18×2 = 36

Radius, r1 = 36cm

Slant height, l1 = r h 36 241

2

1

2 2 2+ = +

= 12 3 22 2+ = 12 13 cm.

17. A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out soil is evenly spread to form a cuboid-platform with base dimension .m m20 14# Find the height of the platform.

Solution: Let r and h be the radius and height of the well.

Let l1, b and h1 be the length, breadth and height of the platform respectively.

Cylinderical well : 2r = 14m & r = 7m, h = 20mPlatform: Length, l = 20 m, breadth, b = 14mVolume of the cuboid platform = Volume of the cylindrical well

& lbh1 = r h2r

& 20 × 14 × h1 = 7 7 20722

# # #

& h1 = 7 7722

20 1420 11#

## # =

Thus, the height of the platform is 11 m.


Exercise 8.4


1. The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to

(A) r cm2 (B) 2r cm2 (C) 3r cm3 (D) 2 cm2

Solution: CSA = 2 2 1rh 1# # #r r= =2r cm2 ( Ans. (B) ) 2. The total surface area of a solid right circular cylinder whose radius is half of its

height h is equal to

(A) h23 r sq. units (B) h

32 2r sq. units

(C) h23 2r sq.units (D) h

32 r sq.units

Solution: radius r = h2

TSA = 2 ( ) 22 2

3r h r h h h h h h2 2

3 2r r r r+ = + = =` `j j sq.units. ( Ans. (C) )

3. Base area of a right circular cylinder is 80 cm2 . If its height is 5 cm, then the volume is equal to

(A) 400 cm3 (B) 16 cm3 (C) 200 cm3 (D)3

400 cm3

Solution: Base area = r 802r = cm2 , Volume v = 80r h 5 4002#r = = cm3. ( Ans. (A) )

4. If the total surface area a solid right circular cylinder is 200 cm2r and its radius is 5 cm,

then the sum of its height and radius is

(A) 20 cm (B) 25 cm (C) 30 cm (D) 15 cm

Solution: TSA = 200 cm2r and radius = 5cm

2 ( )r h rr + 200r= cm 2 5( ) 200 ( )h r h r cm20& &#r r+ = + = ( Ans. (A) ) 5. The curved surface area of a right circular cylinder whose radius is a units and height

is b units, is equal to

(A) a b2r sq.cm. (B) 2rab sq.cm (C) 2r sq.cm (D) 2 sq.cm

Solution: Right circular cylinder: Radius, r = a, height, h = b,

CSA = rh ab2 2r r= sq.cm ( Ans. (B) ) 6. Radius and height of a right circular cone and that of a right circular cylinder are

respectively, equal. If the volume of the cylinder is 120 cm3, then the volume of the cone is equal to

(A) 1200 cm3 (B) 360 cm3 (C) 40 cm3 (D) 90 cm3

Solution: Volume of the cone = 31 Volume of the cylinder = (120) 40

31 = cm3

( Ans. (C) )


7. If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is

(A) 10 cm (B) 20 cm (C) 30 cm (D) 96 cm

Solution: Slant height, l = 6 8r h 100 102 2 2 2+ = + = = cm ( Ans. (A) )

8. If the circumference at the base of a right circular cone and the slant height are 120r cm and 10 cm respectively, then the curved surface area of the cone is equal to

(A) 1200r cm2 (B) 600r cm2 (C) 300r cm2 (D) 600 cm2

Solution: Circumference at the base r2 120r r= & r = 60

CSA = 60rl 10 600# #r r r= = cm2 ( Ans. (B) ) 9. If the volume and the base area of a right circular cone are 48r cm3 and 12r cm2

respectively, then the height of the cone is equal to

(A) 6 cm (B) 8 cm (C) 10 cm (D) 12 cm

Solution: Volume = r h31 482r r= m3

Base area = r 122r r= cm2.

48 12 48 12r h h h31

31

1248 32

& &# # #r r r rrr= = = = cm ( Ans. (D) )

10. If the height and the base area of a right circular cone are 5 cm and 48 sq. cm respectively, then the volume of the cone is equal to

(A) 240 cm3 (B) 120 cm3 (C) 80 cm3 (D) 480 cm3

Solution: Height h = 5cm, Base area = r 482r = cm2

Volume = r h31

31 48 5 802# #r = = cm3 ( Ans. (C) )

11. The ratios of the respective heights and the respective radii of two cylinders are 1:2 and 2:1 respectively. Then their respective volumes are in the ratio

(A) 4 : 1 (B) 1 : 4 (C) 2 : 1 (D) 1 : 2

Solution: h1 : h2 = 1 : 2, r1 : r2 = 2 : 1

Volumes : : 1 :1 2 4 : 2 2 :1V V r h r h31

31 2

1 2 12

1 22

22 2

& # #r r= = = ( Ans. (C) )

12. If the radius of a sphere is 2 cm , then the curved surface area of the sphere is equal to

(A) 8r cm2 (B) 16 cm2 (C) 12r cm2 (D) 16r cm2

Solution: CSA = 4 r 4 2 1622#r r r= = cm2 ( Ans. (D) )

13. The total surface area of a solid hemisphere of diameter 2 cm is equal to

(A) 12 cm2 (B) 12r cm2 (C) 4r cm2 (D) 3r cm2

Solution: TSA = r3 3 1 32 2#r r r= = cm2 ( Ans. (D) )


14. If the volume of a sphere is .cu cm169 r , then its radius is

(A) 34 cm (B)

43 cm

(C) 23 cm (D)

32 cm

Solution: Volume, V = .cu cm169 r

& r34 3r = r

169

109

43

64273

& #r r r= = r43& = ( Ans. (B) )

15. The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio

(A) 81 : 625 (B) 729 : 15625

(C) 27 : 75 (D) 27 : 125

Solution: S1 : S2 = 9 : 25

& :r r1

2

2

2 = 9 : 25

& r1 : r2 = 3 : 5

Now, :r r1

3

2

3 = 3 :53 3 = 27 :125

Thus, : :V V 27 1251 2

= . ( Ans. (D) ) 16. The total surface area of a solid hemisphere whose radius is a units, is equal to

(A) 2r a2 sq.units (B) 3ra2 sq.units

(C) 3ra sq.units (D) 3a2 sq.units

Solution: TSA = 3 3r a2 2r r= sq.units ( Ans. (B) ) 17. If the surface area of a sphere is 100r cm2, then its radius is equal to

(A) 25 cm (B) 100 cm

(C) 5 cm (D) 10 cm

Solution: Surface area = 100r

& 4 r2r = 100 25 5r r4

1002& &r

rr= = =

( Ans. (C) )

18. If the surface area of a sphere is 36r cm2, then the volume of the sphere is equal to

(A) 12r cm3 (B) 36r cm3

(C) 72r cm3 (D) 108r cm3.

Solution: Surface area = 4 36 9 3r r rcm2 2 2& &r r= = =

Volume = 3r34

34 363 3

#r r r= = cm3 ( Ans. (B) )


19. If the total surface area of a solid hemisphere is 12r cm2 then its curved surface area is equal to

(A) 6r cm2 (B) 24r cm2

(C) 36r cm2 (D) 8r cm2

Solution: TSA, r3 2r = 12r cm2 .

CSA, r2 2r = 8r

r

3

12 22

2#

r

r r r= cm2 ( Ans. (D) )

20. If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio

(A) 1 : 8 (B) 2: 1 (C) 1 : 2 (D) 8 : 1

Solution: r r2

12= .So, :V V

1 2=4 : 4 : : 1 : 8r r

rr

rr

2 813

23 2

3

23 2

3

23r r = = =c m . ( Ans. (A) )

21. Curved surface area of solid sphere is 24 cm2. If the sphere is divided into two hemispheres, then the total surface area of one of the hemispheres is

(A) 12 cm2 (B) 8 cm2

(C) 16 cm2 (D) 18 cm2

Solution: r4 2r = 24 cm2

TSA of the hemisphere, 3 r2r = 24 18r

r

4

3 cm22

2

#r

r = ( Ans. (D) )

22. Two right circular cones have equal radii. If their slant heights are in the ratio 4 : 3, then their respective curved surface areas are in the ratio

(A) 16 : 9 (B) 8 : 6 (C) 4 : 3 (D) 3 : 4

Solution: r1 = r

2 and :l l

1 2 = 4 : 3.

:r l r l1 1 2 2r r = :l l

1 2 = 4 : 3 ( Ans. (C) )

Do you know?Pythagorean triplets

We know that ( , , )a b c is a Pythagorean triplet if a b c2 2 2+ =

Thus, ( ,AB BC , AC ) is a Pythagorean triplet if AB BC AC2 2 2+ =

Some Pythagorean triplets are:, , . , , , , 1,2,3,Thus n n n n3 4 5 3 4 5 where g=^ ^h h

, , . , , , , 1,2,3,Thus n n n n5 12 13 5 12 13 where g=^ ^h h , , . , , , , 1,2,3,Thus n n n n7 24 25 7 24 25 where g=^ ^h h , , . , , , , 1,2,3,Thus n n n n8 15 17 8 15 17 where g=^ ^h h

, , . , , , , 1,2,3,Thus n n n n12 35 37 12 35 37 where g=^ ^h h , . , , , , 1,2,3,Thus n n n n20 21 29 20 21 29 where g=^ ^h h

In the same way, one can generate Pythagorean triplets.

Solution - Practical Geometry 221

Exercise 9.1

1. Draw a circle of radius 4.2 cm, and take any point on the circle. Draw the tangent at that point using the centre.Solution: Radius = 4.2 cm

Construction: (i) With O as the centre draw a circle of radius 4.2 cm.

(ii) Take a point P on the circle and join OP.

(iii) Draw an arc of a circle with centre at P cutting OP at L.

(iv) Mark M and N on the arc such that LM MN PL= =!!

(v) Draw the bisector PT of the angle MPN+ .

(vi) Produce TP to T l to get the required tangent T PTl .

Practical Geometry 9


2. Draw a circle of radius 4.8 cm. Take a point on the circle. Draw the tangent at that point using the tangent-chord theorem.Solution: The radius of the circle = 4.8 cm.

Construction: (i) With O as the centre, draw a circle of radius 4.8 cm.

(ii) Take a point P on the circle.

(iii) Through P, draw any chord PQ.

(iv) Mark a point R distinct from P and Q on the circle so that P, Q and R are in counter clockwise direction.

(v) Join PR and QR.

(vi) At R, draw an arc AB!

which intersects RQ and RP at A and B respectively.

(vii) With P as centre and RA (= RB) as radius draw an arc which intersect PQ at C.

(viii) With C as centre and AB as radius draw an arc which intersects the previous arc at D.

(ix) Produce PD to T l and T to get the required tangent line T lPT.


3. Draw a circle of diameter 10 cm. From a point P, 13 cm away from its centre, draw the two tangents PA and PB to the circle, and measure their lengths. Given: The radius of the circle = 5 cm, OP = 13 cm.

Construction: (i) With O as the centre draw a circle of radius 5 cm. (ii) Mark a point P at a distance of 13 cm from O and join OP.


(iii) Draw the perpendicular bisector of OP. Let it meet OP at M. (iv) With M as centre and MO as radius, draw another circle. (v) Let the two circles intersect at A and B. (vi) Join PA and PB . They are the required tangents. Length of the tangent, PA = 12 cm.Verification: In the right angled OPAT ,

PA = OP OA2 2- = 13 52 2

- = 169 25- = 144 ` PA = 12 cm .

4. Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 6 cm. Also, measure the lengths of the tangents.


Construction: (i) With O as the centre draw a circle of radius 6 cm. (ii) Mark a point P at a distance of 10 cm from O and join OP. (iii) Draw the perpendicular bisector of OP. Let it meet OP at M. (iv) With M as centre and MO as radius, draw another circle. (v) Let the two circles intersect at A and B. (vi) Join PA and PB . They are the required tangents. Length of the tangent, PA = 8cm.Verification: In the right angled OPAT ,

PA = OP OA2 2- = 10 62 2

-

= 100 36- = 64 ` PA = 8 cm.

5. Take a point which is 9 cm away from the centre of a circle of radius 3 cm, and draw the two tangents to the circle from that point.


Construction: (i) With O as the centre draw a circle of radius 3 cm.

(ii) Mark a point P at a distance of 9 cm from O and join OP.

(iii) Draw the perpendicular bisector of OP. Let it meet OP at M.

(iv) With M as centre and MO as radius, draw another circle.

(v) Let the two circles intersect at A and B.

(vi) Join PA and PB . They are the required tangents.

Length of the tangent, PA = 8.5 cm

Verification: In the right angled OPAT ,

PA = OP OA2 2- = 9 32 2

-

= 81 9- = 6 1.41472 36 2# #= = = = 8.484 cm

` Length of the tangents PA = PB = 8.5 cm.

Exercise 9.2

1. Construct a segment of a circle on a given line segment AB = 5.2 cm containing

an angle .48c

Given: AB = 5.2 cm angle of a sector = 48°.


Construction: (i) Draw a line segment AB = 5.2 cm. (ii) At A, make 48BAX+ = c. (iii) Draw AY AX= . (iv) Draw the perpendicular bisector of AB which meets AY at O. (v) With O as centre and OA as radius draw a circle. (vi) Take any point C on the circle. By the tangent chord theorem, the major arc

ACB is the required segment of the circle containing the angle 48°. (vii) The major arcs , ,AC B AC B AC B

1 2 3 are also having the same angle 48°.

2. Construct a PQRD such that PQ = 6 cm, 60R+ = c and the altitude from R to PQ is cm4 .

Given: PQ = 6 cm, R 60= c and the altitude from R to PQ is 4 cm.

Construction: (i) Draw a line segment PQ = 6 cm.

(ii) Draw PX such that QPX+ = 60°.

(iii) Draw PY PX= .

(iv) Draw the perpendicular bisector of PQ intersecting PY at O and PQ at M.


(v) With O as centre and OP as radius, draw the circle . (vi) The major arc PKQ contains the vertical angle 60°. (vii) On the perpendicular bisector MO, mark a point H such that MH = 4 cm. (viii) Draw RHRl parallel to PQ meeting the circle at Rand at Rl. (ix) Complete the PQRT , which is one of the required triangles.

3. Construct a PQRD such that PQ = 4 cm, R 25+ = c and the altitude from R to PQ is . cm4 5 .

Given: PQ = 4 cm, 5R 2= c and the altitude from R to PQ is 4.5 cm.

Construction: (i) Draw a line segment PQ = 4 cm. (ii) Draw PX such that QPX+ = 25c. (iii) Draw PY PX= . (iv) Draw the perpendicular bisector of PQ intersecting PY at O and PQ at M. (v) With O as centre and OP as radius, draw the circle . (vi) The major arc PKQ contains the vertical angle 25c. (vii) On the perpendicular bisector MO, mark a point H such that MH = 4.5 cm. (viii) Draw RHRl parallel to PQ meeting the circle at Rand at Rl. (ix) Complete the PQRT , which is one of the required triangles.


4. Construct a ABCD such that BC = 5 cm. 45A+ = c and the median from A to BC is cm4 .Given: BC = 5 cm, 45A+ = c, The median from A to BC = 4 cm.

Construction:

(i) Draw a line segment BC = 5 cm.

(ii) Through B draw BX such that CBX 45+ = c.

(iii) Draw BY=BX.

(iv) Draw the perpendicular bisector of BC intersecting BY at O and BC at M.

(v) With O as centre and OB as radius, draw the circle.

(vi) The major arc BKC of the circle, contains the vertical angle 45c.

(vii) With M as centre, draw an arc of radius 4 cm meeting the circle at A and Al.

(viii) Join AB and AC.

Now, ABC3 or A BCT l is the required triangle.


5. Construct a ABCD in which the base BC = 5 cm, 40BAC+ = c and the median from A to BC is 6 cm. Also, measure the length of the altitude from A.Given: BC = 5 cm, 40BAC+ = c, median from A to B = 6 cm.

Construction: (i) Draw a line segment BC = 5 cm.


(iii) Draw BY=BX.





(viii) ABC3 or A BCT l is the required triangle.

(ix) Produce CB to CZ.

(x) Draw AE CZ= .

(xi) Length of the altitude AE is 3.8 cm.


Exercise 9.3

1. Construct a cyclic quadrilateral PQRS, with PQ = 6.5 cm, QR = 5.5 cm, PR = 7 cm and PS = 4.5 cm.

Given: In the cyclic quadrilateral PQRS, PQ = 6.5 cm QR = 5.5 cm PR = 7 cm PS = 4.5 cm

Construction: (i) Draw a rough diagram and mark the measurements.

Draw a line segment PQ = 6.5 cm.

(ii) With P and Q as centres, draw arcs with radii 7 cm and 5.5 cm respectively, to intersect at R. Join PR and QR.

(iii) Draw the perpendicular bisectors of PQ and QR to intersect at O.

(iv) With O as the centre and OP (= OQ = OR) as radius draw the circumcircle of PQRD .

(v) With P as the centre and radius 4.5 cm. draw an arc intersecting the circumcircle at S.

(vi) Join PS and RS.

Now, PQRS is the required cyclic quadrilateral.


2. Construct a cyclic quadrilateral ABCD where AB = 6 cm, AD = 4.8 cm, BD = 8 cm and CD = 5.5 cm.

Given: In the cyclic quadrilateral ABCD, AB = 6 cm, AD= 4.8 cm, BD = 8 cm and CD = 5.5 cm.

Construction: (i) Draw a rough diagram and mark the measurements. Draw a line segment AB = 6 cm (ii) With A as centre and radius 4.8 cm, draw an arc. (iii) With B as centre and radius 8 cm, draw another arc meeting the previous

arc as in the figure at D. (iv) Join AD and BD. (v) Draw the perpendicular bisectors of AB and AD intersecting each other at O. (vi) With O as the centre OA(=OB=OC) as radius, draw the circumcircle of

ABDD . (vii) With D as centre and 5.5 cm radius , draw an arc intersecting the circle at C. (viii) Join CD and BC. (ix) Now,ABCD is the required cyclic quadrilateral.


3. Construct a cyclic quadrilateral PQRS such that PQ = 5.5 cm, QR = 4.5 cm, 45QPR+ = c and PS = 3 cm.

Given: In the cyclic quadrilateral PQRS, PQ = 5.5 cm, QR = 4.5 cm, 45QPR+ = c and PS = 3 cm.


Draw a line segment PQ = 5.5 cm.

(ii) Through P draw PX such that 45QPX+ = c.

(iii) With Q as centre and radius 4.5 cm, draw an arc intersecting PX at R and join QR.

(iv) Draw the perpendicular bisectors of PQ and QR intersecting each other at O.

(v) With O as centre and OP (= OQ= OR) as radius, draw the circumcircle of PQRD .

(vi) With P as centre and radius 3 cm, draw an arc intersecting the circle at S.

(vii) Join PS and RS.

(viii) Now, PQRS is the required cyclic quadrilateral.


4. Construct a cyclic quadrilateral ABCD with AB = 7 cm, 80A+ = c, AD = 4.5 cm and BC = 5 cm.

Given: In the cyclic quadrilateral ABCD, AB = 7 cm, AD = 4.5 cm, 80A+ = c and BC = 5 cm.


Draw a line segment AB = 7 cm.

(ii) Through A draw AX such that 80A+ = c.

(iii) With A as centre and radius 4.5 cm, draw an arc intersecting AX at D and join AD.

(iv) Draw the perpendicular bisectors of AD and BD intersecting each other at O.

(v) With O as centre and OA (= OB = OD) as radius, draw the circumcircle of ABDD .

(vi) With B as centre and radius 5 cm, draw an arc intersecting the circle at C.

(vii) Join CD and BC.

(viii) Now, ABCD is the required cyclic quadrilateral.


5. Construct a cyclic quadrilateral KLMN such that KL = 5.5 cm, KM = 5 cm, LM = 4.2 cm and LN = 5.3 cm.

Given: In the cyclic quadrilateral KLMN, KL = 5.5 cm, KM = 5 cm, LM = 4.2 cm and LN = 5.3 cm.

Construction:

(i) Draw a rough diagram and mark the measurements.

Draw a line segment KL = 5.5 cm

(ii) With K as centre and radius 5 cm, draw an arc.

(iii) With L as centre and radius 4.2 cm, draw another arc meeting the previous arc as in the figure at M.

(iv) Join KM and LM.

(v) Draw the perpendicular bisectors of KM and LM intersecting each other at O.

(vi) With O as the centre OK(=OL=OM) as radius, draw the circumcircle of KLMD .

(vii) With L as centre and 5.3 cm radius , draw an arc intersecting the circle at N.

(viii) Join KN and MN.

(ix) Now, KLMN is the required cyclic quadrilateral.


6. Construct a cyclic quadrilateral EFGH where EF = 7 cm, EH = 4.8 cm, FH = 6.5 cm and EG = 6.6 cm.

Given: In the cyclic quadrilateral EFGH, EF = 7 cm, EH = 4.8 cm, FH = 6.5 cm and EG = 6.6 cm.


Draw a line segment EF = 7 cm.

(ii) With E as centre and radius 4.8 cm, draw an arc.

(iii) With F as centre and radius 6.5 cm, draw another arc meeting the previous arc as in the figure at H.

(iv) Join EH and FH.

(v) Draw the perpendicular bisectors of EF and EH intersecting each other at O.

(vi) With O as the centre OE(=OF=OH) as radius, draw the circumcircle of EFHD .

(vii) With E as centre and 6.6 cm radius , draw an arc intersecting the circle at G.

(viii) Join HG and FG.

(ix) Now, EFGH is the required cyclic quadrilateral.


7. Construct a cyclic quadrilateral ABCD, given AB = 6 cm, ABC 70+ = c, BC = 5 cm and 30ACD+ = c

Given: In the cyclic quadrilateral ABCD, AB = 6 cm, BC = 5 cm, 70ABC+ = c and 30ACD+ = c.

Construction:


Draw a line segment AB = 6 cm.

(ii) From B, draw BX such that ABX 70+ = c.

(iii) With B as centre and radius 5 cm, draw an arc intersecting BX at C.

(iv) Join AC.

(v) Draw the perpendicular bisectors of AB and BC intersecting each other at O.

(vi) With O as centre and OA (= OB = OC) as radius, draw a circumcircle.

(vii) From C, draw CY such that 0ACD 3+ = c which intersects the circle at D.

(viii) Join AD.

Now, ABCD is the required cyclic quadrilateral.


8. Construct a cyclic quadrilateral PQRS given PQ = 5 cm, QR = 4 cm, 35QPR+ = c and 70PRS+ = c.

Given: In the cyclic quadrilateral PQRS, PQ = 5 cm, QR = 4 cm, QPR 35+ = c and 0PRS 7+ = c.


Draw a line segment PQ = 5 cm.

(ii) From P, draw PX such that 35QPX+ = c

(iii) With Q as centre and radius 4 cm, draw an arc intersecting PX at R.

(iv) Join QR.

(v) Draw the perpendicular bisectors of PQ and QR intersecting each other at O.

(vi) With O as centre and OP (= OQ = OR) as radius, draw a circumcircle.

(vii) From R, draw RY such that 70PRY+ = cwhich intersects the circle at S.

(viii) Join PS.

Now, PQRS is the required cyclic quadrilateral.


9. Construct a cyclic quadrilateral ABCD such that AB = 5.5 cm 50ABC+ = c, BAC 60+ = c and ACD 30+ = c.

Given: In the cyclic quadrilateral ABCD, AB = 5.5 cm, 50 , 60ABC BAC+ += =c c and 30ACD+ = c.

Construction:


Draw a line segment AB = 5.5 cm.

(ii) From B draw BX such that 0ABX 5+ = c.

(iii) From A draw AY such that 60BAY+ = c. Let AY meet BX at C.

(iv) Draw perpendicular bisectors of AB and BC intersecting each other at O.

(v) With O as centre and OA( = OB = OC ) as radius, draw a cicumcircle of ABC3 .

(vi) From C, draw CZ such that 30ACZ+ = c which intersects the circle at D.

(vii) Join AD.

Now, ABCD is the required cyclic quadrilateral.


10. Construct a cyclic quadrilateral ABCD, where AB = 6.5 cm, ABC 110+ = c, BC = 5.5 cm and AB || CD.

Given: In the cyclic quadrilateral ABCD, AB = 6.5 cm, BC = 5.5 cm and ABC 110+ = c.

and AB || CD.

Construction: (i) Draw a rough diagram and mark the measurements. Draw a line segment AB = 6.5 cm. (ii) From B, draw BX such that ABX+ = 110c. (iii) With B as centre and radius 5.5 cm, draw an arc intersecting BX at C. (iv) Draw perpendicular bisectors of AB and BC intersecting each other at O. (v) With O as centre, and OA (= OB = OC) as radius, draw a circumcircle of

ABCD . (vi) Draw CY such that CY AB< intersecting the circle at D. Join AD. (vii) Now, ABCD is the required cyclic quadrilateral.

Solution - Graph 241

Exercise 10.1

1. Draw the graph of the following functions.

(i) 3y x2

= .Solution:

x – 3 – 2 – 1 0 1 2 3

x2 9 4 1 0 1 4 9

y x3 2= 27 12 3 0 3 12 27

Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )3 27 2 12 1 3 0 0 1 3 2 12 3 27- - -

Graphs 10


(ii) 4y x2

=-

Solution:

x – 3 – 2 – 1 0 1 2 3

x2 9 4 1 0 1 4 9

y x4 2=- – 36 – 16 – 4 0 – 4 – 16 – 36

Points: ( 3, ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )36 2 16 1 4 0 0 1 4 2 16 3 36- - - - - - - - -

(iii) 6 8y x x x x2 4 2= + + = + +^ ^h h

Let y x x6 82= + +

Solution:

x – 5 – 4 – 3 – 2 – 1 0 1 2

x2 25 16 9 4 1 0 1 4

x6 – 30 – 24 – 18 – 12 – 6 0 6 128 8 8 8 8 8 8 8 8y 3 0 – 1 0 3 8 15 24

Points: ( 5,3), ( 4,0), ( 3, 1), ( 2,0), ( 1,3), (0,8), (1,15), (2,24)- - - - - -


(iv) 2 3y x x2

= - + .Solution:

x – 3 – 2 – 1 0 1 2 3

x2 2 18 8 2 0 2 8 18

x- 3 2 1 0 – 1 – 2 – 33 3 3 3 3 3 3 3y 24 13 6 3 4 9 18

Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )3 24 2 13 1 6 0 3 1 4 2 9 3 18- - -


2. Solve the following equations graphically

(i) 4 0x2- = .

Solution: Let y x 42= -

x – 3 – 2 – 1 0 1 2 3

x2 9 4 1 0 1 4 9– 4 – 4 – 4 – 4 – 4 – 4 – 4 – 4y 5 0 – 3 – 4 – 3 0 5

Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )3 5 2 0 1 3 0 4 1 3 2 0 3 5- - - - - -

Solve: y = x 42-

0 = x 42-

y = 0

The curve intersects the x-axis at ( , )2 0- and (2, 0).Thus, the x-coordinates of the points are – 2 and 2. Hence, the Solution set is {– 2, 2}.

(ii) 3 10 0x x2- - = .

Solution: Let y x x3 102= - -

x – 3 – 2 – 1 0 1 2 3 4 5 6

x2 9 4 1 0 1 4 9 16 25 36

x3- 9 6 3 0 – 3 – 6 – 9 – 12 – 15 – 18– 10 – 10 – 10 – 10 – 10 – 10 – 10 – 10 – 10 – 10 – 10

y 8 0 – 6 – 10 – 12 – 12 – 10 – 6 0 8


Points: ( 3,8), ( 2,0), ( 1, 6), (0, 10), (1, 12)- - - - - - (2, 12), (3, 10), (4, 6), (5,0), (6,8)- - -

Solve: y = x x3 102- -

0 = x x3 102- -

y = 0

The curve intersects the x-axis at ( , )2 0- and (5, 0).` The x-coordinates of the points are – 2 and 5. Thus, Solution set is {– 2, 5}.

(iii) 0x x5 1- - =^ ^h h .Solution: ( 5)( 1) 0x x &- - = x x6 5 02

- + = . Let y x x6 52= - +

x – 1 0 1 2 3 4 5 6

x2 1 0 1 4 9 16 25 36

x6- 6 0 – 6 – 12 – 18 – 24 – 30 – 365 5 5 5 5 5 5 5 5y 12 5 0 – 3 – 4 – 3 0 5

Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )1 12 0 5 1 0 2 3 3 4 4 3 5 0 6 5- - - - Solve: y = x x6 52

- +

0 = x x6 52- +

y = 0


(iv) 0x x2 1 3+ - =^ ^h h .Solution: (2 1)( 3) 0 2 5 3 0x x x x2

&+ - = - - = . Let y x x2 5 32= - -

x – 1 0 1 2 3 4

x2 2 2 0 2 8 18 32

x5- 5 0 – 5 – 10 – 15 – 20– 3 – 3 – 3 – 3 – 3 – 3 – 3y 4 – 3 – 6 – 5 0 9

Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , )1 4 0 3 1 6 2 5 3 0 4 9- - - - Solve: y = x x2 5 32

- -

0 = x x2 5 32- -

y = 0The curve intersects the x- axis at (– 0.5, 0) and (3, 0).So, the x-coordinates of the points are 1 and 5.Thus,Solution set is {–0.5, 3}.

The curve intersects the x-axis at (1, 0) and (5, 0).Thus, the x-coordinates of the points are 1 and 5. Hence, Solution set is {1, 5}.


3. Draw the graph of y x2

= and hence solve 4 5 0x x2- - = .

Solution: y x2=

x – 2 – 1 0 1 2 3 4 5 6

y x2= 4 1 0 1 4 9 16 25 36

Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )2 4 1 1 0 0 1 1 2 4 3 9 4 16 5 25 6 36- - Solve: y = x x0 02

+ +

0 = x x4 52- -

y = x4 5+

Let us draw the graph of the straight line y x4 5= + .Now, form the table for the line y x4 5= + .

x – 2 – 1 0 1 2y x4 5= + – 3 1 5 9 13

Points: ( , ), ( , ), ( , ), ( , ), ( , )2 3 1 1 0 5 1 9 2 13- - -

.The points of intersection of the line and the parabola are ( , )1 1- and ( , )5 25 .The x-coordinates of the points are – 1 and 5. Thus,Solution set is {– 1, 5}

4. Draw the graph of 2 3y x x2

= + - and hence find the roots of 6 0x x2- - = .

Solution: y x x2 32= + -

x – 3 – 2 – 1 0 1 2 3

x2 9 4 1 0 1 4 9

x2 – 6 – 4 – 2 0 2 4 6– 3 – 3 – 3 – 3 – 3 – 3 – 3 – 3y 0 – 3 – 4 – 3 0 5 12


Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )3 0 2 3 1 4 0 3 1 0 2 5 3 12- - - - - - Solve: y = x x2 32

+ -

0 = x x 62- -

y = x3 3+

Let us draw the graph of the straight line y x3 3= + .Now, form the table for the line y x3 3= + .

x – 2 – 1 0 1 2y x3 3= + – 3 0 3 6 9

Points: ( , ), ( , ), ( , ), ( , ), ( , )2 3 1 0 0 3 1 6 2 9- - -

The points of intersection of the line and the parabola are ( , )2 3- - and ( , 2)3 1 .The x-coordinates of the points are – 2 and 3. Thus, Solution set is {– 2, 3}.

5. Draw the graph of 2 6y x x2

= + - and hence solve 2 10 0x x2+ - = .

Solution: 6y x x2 2= + -

x – 3 – 2 – 1 0 1 2 3

x2 2 18 8 2 0 2 8 18

x – 3 – 2 – 1 0 1 2 3– 6 – 6 – 6 – 6 – 6 – 6 – 6 – 6y 9 0 – 5 – 6 – 3 4 15


Points: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )3 9 2 0 1 5 0 6 1 3 2 4 3 15- - - - - - Solve: y = x x2 62

+ -

0 = x x2 102+ -

y = 4

y 4= is a straight line parallel to x-axis.

The straight line and parabola intersect at points (–2.5, 4) and (2, 4).So, the x-coordinates of the points are – 2.5 and 2. Thus, Solution set is {– 2.5, 2}.

6. Draw the graph of 8y x x2

= - - and hence find the roots of 2 15 0x x2- - = .

Solution:

x – 4 – 3 – 2 – 1 0 1 2 3 4 5

x2 16 9 4 1 0 1 4 9 16 25

x- 4 3 2 1 0 – 1 – 2 – 3 – 4 – 5– 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8y 12 4 – 2 – 6 – 8 – 8 – 6 – 2 4 12Points: ( , ), ( , ), ( , ), ( , ), ( , )4 12 3 4 2 2 1 6 0 8- - - - - - - ( , ), ( , ), ( , ), ( , ), ( , )1 8 2 6 3 2 4 4 5 12- - -


Solve: y = x x 82- -

0 = x x2 152- -

y = x 7+

Let us draw the graph of the straight line y x 7= + .Now, form the table for the line y x 7= + .

x – 3 – 2 – 1 0 1 2 3 4 5y x 7= + 4 5 6 7 8 9 10 11 12Points: ( 3,4), ( 2,5), ( 1,6), (0,7), (1,8), (2,9), (3,10), (4,11), (5,12)- - -

The straight line and parabola intersects at points (– 3, 4) and (5, 12).The x-coordinates of the points are – 3 and 5.Thus, Solution set is {– 3, 5}.


7. Draw the graph of 12y x x2

= + - and hence solve 2 2 0x x2+ + = .

Solution:

x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4

x2 25 16 9 4 1 0 1 4 9 16

x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4– 12 – 12 – 12 – 12 – 12 – 12 – 12 – 12 – 12 – 12 – 12

y 8 0 – 6 – 10 – 12 – 12 – 10 – 6 0 8Points: ( , ), ( , ), ( , ), ( , ), ( , )5 8 4 0 3 6 2 10 1 12- - - - - - - - ( , ), ( , ), ( , ), ( , ), ( , )0 12 1 10 2 6 3 0 4 8- - -

Solve: y = x x 122+ -

0 = 2x x 22+ +

y = x 14- -

Let us draw the graph of the straight line y x 14=- - .


Now, form the table for the line y x 14=- - .

x – 3 – 2 – 1 0 1 2 3y x 14=- - – 11 – 12 – 13 – 14 – 15 – 16 – 17

Points ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )3 11 2 12 1 13 0 14 1 15 2 16 3 17- - - - - - - - - -

The straight line y x 14=- - does not intersect the curve y x x 122= + - .

Thus, x x2 2 02+ + = has no real roots.

Exercise 10.2

1. A bus travels at a speed of 40 km / hr. Write the distance-time formula and draw the graph of it. Hence, find the distance travelled in 3 hours.Solution:

x 1 2 3 4y 40 80 120 160

From the table, we found that as x increases, y also increases.

Thus, the variation is a direct variation.

y x\ & y kx= xy

k& =

where k is the constant of proportionality. From the given values, we have

k = 40140

280

3120

4160= = = =

Note: In this problem, the graph is a straight line passing through the origin. Thus, y x\ & y kx=

xy

k& = .


k` = 40; ,xy

y x40 40= =

The relation 4y x0= forms a straight line graph.From the graph, distance travelled in 3 hours = 120 km.

2. The following table gives the cost and number of notebooks bought.

No. of note books (x) 2 4 6 8 10 12

Cost ` y 30 60 90 120 150 180

Draw the graph and hence (i) Find the cost of seven note books. (ii) How many note books can be bought for ` 165.Solution:

x 2 4 6 8 10 12y 30 60 90 120 150 180



y x\ Let y kx= xy

k& =



k = 15230

460

690

8120

10150

12180= = = = = =

& 15y x= .

From the graph,(i) cost of seven note books = ` 105. (ii) Note books bought for ` 165 = 11.

3.

x 1 3 5 7 8y 2 6 10 14 16

Draw the graph for the above table and hence find

(i) the value of y if x = 4 , (ii) the value of x if y = 12.Solution:

x 1 3 5 7 8y 2 6 10 14 16



y x\ & y kx= xy

k& =



k = 212

36

510

714

816= = = = =

& y x2= .From the graph, (i) when ,x y4 8= = , (ii) when ,y x12 6= = .

4. The cost of the milk per litre is ` 15. Draw the graph for the relation between the quantity and cost . Hence find (i) the proportionality constant, (ii) the cost of 3 litres of milk.Solution:

No. of litres (x) 1 2 3 4 5Cost in ` 15 30 45 60 75



y x\ Let y kx= xy

k& =

where k is the constant of proportionality.

k = 15115

210

345

460

575= = = = =

k` = 15

& y = 15xIt forms a straight line graph. From the graph,(i) The proportionality constant = 15, (ii) cost of 3 litres of milk = ` 45.


5. Draw the Graph of xy 20= , ,x y > 0. Use the graph to find y when 5x = , and to find x when 10y = .

Solution: xy = 20, ,x y 0>

y = x20

x 1 2 4 5 10 20y 20 10 5 4 2 1

From the table, we observe that as x increases, y decreases.

This type of variation is called indirect variation.

From the graph, when ,x y5 4= = , when ,y x10 2= = .

6.

No. of workers (x) 3 4 6 8 9 16

No of days (y) 96 72 48 36 32 18

Draw graph for the data given in the table. Hence find the number of days taken by 12 workers to complete the work.

Solution:

No. of workers (x) 3 4 6 8 9 16No. of days (y) 96 72 48 36 32 18

From the table, we observe that as x increases y decreases.


Thus, the variation is a indirect variation.

yx

xy k1 &\ =

where k is the constant of proportionality.

k = 3 96 4 72# #=

= 6 48 8 36# #=

= 9 32 16 18 288# #= =

xy = 288

y = x

288

From the graph, number of days taken by 12 workers to complete the work is 24 days.


Exercise 11.1

1. Findtherangeandcoefficientofrangeofthefollowingdata. (i) 59, 46, 30, 23, 27, 40, 52,35, 29

Solution: (i) The largest value L = 59. The smallest value S = 23

Thus, Range = L - S = 59 -23 = 36

Coefficientofrange = L SL S+- =

2323

5959

+- =

8236

4118=

= 0.4390 = 0.44

(ii) 41.2, 33.7, 29.1, 34.5, 25.7, 24.8, 56.5, 12.5Solution: Fromthegivendata The largest value L = 56.5 The smallest value S = 12.5Now, Range = L - S = 56.5 -12.5 = 44

Coefficientofrange = L SL S+- =

12.512.5

56.556.5

+- =

6944

= 0.6376 = 0.64 2. The smallest value of a collection of data is 12 and the range is 59. Find the

largestvalueofthecollectionofdata.Solution: The smallest value, S = 12, Range, R = 59. We have R = L-S ` L = S + R = 12 + 59 = 71Thus, the largest value, L = 71

3. Thelargestof50measurementsis3.84kg.Iftherangeis0.46kg,findthesmallestmeasurement.Solution: The Largest value, L = 3.84 kg, Range R = 0.46 kg. So, Smallest value, S = L - R = 3.84 – 0.46 = 3.38Thus, the smallest value, S = 3.38 kg.

4. Thestandarddeviationof20observationsis 5 .Ifeachobservationismultipliedby2,findthestandarddeviationandvarianceoftheresultingobservations.

Solution: GiventhatSDof20observationsis 5 .

Ifeachobservationismultipliedby2,thenSDofnewdatais2 5 .

Variance = (SD)2 =(2 5 )2

= 4 5 20# = .

Statistics11

Solution - Statistics 259

5. Calculatethestandarddeviationofthefirst13naturalnumbers.

Solution:

TheSDofthefirstnnaturalnumbersv = n12

12-

Thus, SDofthefirst13naturalnumbersv = n12

12- =

1213 1

2-

= .12168 14 3 74= =

6. Calculatethestandarddeviationofthefollowingdata. (i) 10, 20, 15, 8, 3, 4 (ii) 38, 40, 34 ,31, 28, 26, 34.

Solution: (i) Firstwearrangethegivendatainascendingorder3,4,8,10,15,20

Arithmeticmean, xr = nxR = 10

63 4 8 10 15 20

660+ + + + + = =

d = 10x x x- = -r

x 10d x= - d2

348

101520

-7-6-2 0 5

10

493640

25100

d2

R = 214

(ii) 38, 40, 34 ,31, 28, 26, 34.

Firstwearrangethegivendatainascendingorder26,28,31,34,34,38,40

Arithmeticmean, xr = nxR

= 7

26 28 31 34 34 38 407

231 33+ + + + + + = =

x 33d x x x= - = -r d2

26283134343840

-7-5-2 1 1 5 7

4925411

2549

d2

R = 154

v = nd2R

= 6

214

b 5.97

v = nd2R

= 7

154

= 22

b 4.69


7. Calculatethestandarddeviationofthefollowingdata.

x 3 8 13 18 23

f 7 10 15 10 8

Solution: LetusfindtheStandardDeviationbyusingassumedmeanmethod

Let us take A =13astheassumedmeand x A x 13= - = -

x f d = x – 13 d2 fd fd2

38

131823

71015108

-10-5 0 5

10

100250

25100

-70-50 0 50 80

700250

0250800

fR =50 fdR =10 fd2

R =2000

Standarddeviation,v = f

fd

f

fd2 2

- e o//

//

= 50

20005010 2

- ` j

= 40251- =

25999 = .

531 61

Thus, v - 6.321

8. Thenumberofbooksboughtatabookfairby200studentsfromaschoolaregiveninthefollowingtable.

No.ofbooks 0 1 2 3 4No.ofstudents 35 64 68 18 15

Calculatethestandarddeviation.

Solution: LetuscalculatetheStandardDeviationbyusingassumedmeanmethod.

Let us take A=2astheassumedmeand x A x 2= - = -

x f d = x – 2 d2 fd fd2

01234

3564681815

-2-1 0 1 2

41014

-70-64 0 18 30

140640

1860

fR =200 fdR = – 86 fd2

R = 282

Standarddeviationv = ffd

ffd

2 2

R

R

R

R- c m


= 200282

20086 2

- -` j = ( )200

282

200

73962-

= ( )200

282 200 73962

# - = 20049004

2^ h = .

200221 4

Thus, v - 1.107

9. Calculatethevarianceofthefollowingdata

x 2 4 6 8 10 12 14 16f 4 4 5 15 8 5 4 5

Solution: Letusfindthevariancebyusingassumedmeanmethod.LetA = 10

x f d = x – 10 d2 fd fd2

2468

10121416

445

158545

-8-6-4-2 0246

643616404

1636

– 32– 24– 20– 30

0101630

25614480600

2064

180

fR =50 50fdR =- fd2

R = 804

2v =

ffd

ffd

2 2

R

R

R

R- c m

= 50804

5050 2

- -` j

= 15.0850804 1

50754- = =

Thus, Variance = 15.08

10. Thetime(inseconds)takenbyagroupofpeopletowalkacross apedestriancrossingisgiveninthetablebelow.

Time(insec.) 5-10 10-15 15-20 20-25 25-30No.ofpeople 4 8 15 12 11

Calculatethevarianceandstandarddeviationofthedata.

Solution: Let A=17.5,themid-valueoftheinterval15-20

Here, c = 5

Now, d = .c

x A x517 5- = -


Class interval

Midvalue

x f x–Ad =

.x517 5- d2 fd fd

2

5-1010-1515-2020-2525-30

7.512.517.522.527.5

48

151211

–10–505

10

-2-1 0 1 2

41014

-8-80

1222

1680

1244

f 50R = fd 18R = 80fd2

R =

Variance, 2v =

ffd

ffd

2 2

R

R

R

R- c m= G c2

#

= 5080

5018 5

2 2#- ` j; E =

5080

50

324 252 #-c m

= 2550

4000 3242 #- = 25

50 504000 324

##- =

1003676

2v = 36.76

Thus,StandardDeviation,v = 36.76 - 6.063

11. Agroupof45houseownerscontributedmoneytowardsgreenenvironmentoftheirstreet.Theamountofmoneycollectedisshowninthetablebelow.

Amount(`)

0-20 20-40 40-60 60-80 80-100

No. of house owners

2 7 12 19 5

Calculatethevarianceandstandarddeviation.Solution: LetustaketheassumedmeanA=50andc = 20.

d = c

x A x2050- = -

Class interval

midvalue

x f x–A

d= x2050- d2 fd fd

2

0-2020-4040-6060-80

80-100

1030507090

27

12195

– 40– 20

02040

-2-1012

41014

-4-70

1910

870

1920

f 45R = fd 18R = fd 542

R =

Variance, 2v =

ffd

ffd

c2 2

2#

R

R

R

R- c m= G


= 004554

4518 4

2

#- ` j; E = . ( . ) 4001 2 0 42#-^ h

= . . 4001 2 0 16 #-^ h = 1.04 400 416# =

Thus,StandardDeviation,v = 416 - 20.396

12. Findthevarianceofthefollowingdistribution

Classinterval 20-24 25-29 30-34 35-39 40-44 45-49Frequency 15 25 28 12 12 8

Solution: Let A=32,themidvalueoftheinterval30-34andc = 5.

d = c

x A x532- = -

Class interval

midvalue

x (f) x – A

d= x

532- d2 fd fd

2

20-2425-2930-3435-3940-4445-49

222732374247

15252812128

– 10– 505

1015

-2-1 0 1 2 3

410149

-30-25

0122424

60250

124872

100fR = fd/ =5 fd 2172

R =

Variance, 2v =

ffd

ffd

c2 2

2#

R

R

R

R- c m= G = 5

100217

1005 2 2

#- ` j; E

= . ( . 252 17 0 052#-^ h = . ( . 252 17 0 0025 #-^ h

= . 252 1675 #

Thus, 2v = 54.1875 - 54.19

13. Meanof100itemsis48andtheirstandarddeviationis10.Findthesumofalltheitemsandthesumofthesquaresofalltheitems.

Solution: TheMeanof100items,x = 48

Thesumof100items x/ = 48 × 100 = 4800 . xnxa R=r; E

GiventhatStandardDeviation,v = 10

Now,Variance, 2v =

nx

nx

2 2R R- c m = 100

& x100 100

48002 2R - ` j = 100 ( 2304x

100

2R - = 100

& x100

2R = 100 + 2304 = 2404

Thus, x2

R = 2404 × 100 = 2,40,400


14. Themeanandstandarddeviationof20itemsarefoundtobe10and2respectively.Atthetimeofcheckingitwasfoundthatanitem12waswronglyenteredas8.Calculatethecorrectmeanandstandarddeviation.

Solution: Letusfindthecorrectmean.

Meanof20items,xnx/

=r = 10

& x20/ = 10

& x/ = 10 × 20 = 200

Now, corrected xR = 200 + 4 = 204

Thus, thecorrectedMean = 20204 = 10.2

Variance, 2v =

nx

nx

2 2R R- c m = 4 2a v =6 @

& 10x20

22R - = 4

& x20

2R = 4 + 100 = 104

Thus, x2

R = 104 × 20 = 2080

So, corrected x2

R = 2080 + 122 – 82

= 2080 + 144 – 64 = 2160

Now, thecorrected 2v =

nxCorrected Corrected mean2

2R - ^ h

= (10.2)20

2160 2-

= 108 – 104.04 = 3.96

Correctedv = .3 96 - 1.99

Hence,thecorrectedMean =10.2andthecorrectedS.D.- 1.99

15. Ifn = 10, x = 12 and x2

R =1530,thencalculatethecoefficientofvariation.

Solution: Giventhatn = 10, x = nx/ =12and x

2R = 1530

Now,StandardDeviationv = nx

nx

2 2R R- c m

= 10

1530 122

- = 1153 44- = 3

Coefficientofvariation =x

100#vr

& C.V = 100123

#

= 10041# = 25.


16. Calculatethecoefficientofvariationofthefollowingdata:20,18,32,24,26.

Solution:

x d = 24x - d2

2018322426

-4-6 8 0 2

163664 0 4

120xR = d2

R = 120

Arithmeticmean,x = nx

5120 24R = =

Standarddeviation,v = nd2R C.V. = 100

x#v

v = 5

120 = . 100244 9

24490

# =

v = 24 = 4.9 = . .20 416 20 42b

17. Ifthecoefficientofvariationofacollectionofdatais57anditsS.Dis6.84,thenfindthemean.

Solution: Giventhatthecoefficientofvariation=57,v = 6.84

100x#v = 57 & . 100

x6 84

# = 57

x57684= = 12

18. Agroupof100candidateshavetheiraverageheight163.8cmwithcoefficientofvariation3.2.Whatisthestandarddeviationoftheirheights?

Solution: Theaverageheightof100candidates=163.8,i.e. x = 163.8

Coefficientofvariation = 3.2

& 10x

0#vr

= 3.2

& .

100163 8

#v = 3.2

v = . .100

3 2 163 8# = 5.2416 b 5.24

19. Given xR = 99 , n = 9 and ( 10)x 2R - = 79. Find ( )x x xand2 2R R - .

Solution: Giventhat xR =99andn = 9.

Thus, x = nxR =

999 11= .

Letusfind x2

R .

Now, ( )x 102

R - = 79

( )x x20 1002R - + = 79


x x20 100 12

R R R- + = 79 1 9/ =^ h

x 20 99 100 92

# #R - + = 79

x2

` R = 1159

Now, ( )x x 2R - = ( )x 112

R -

= x x22 1212

R - +^ h

= 1159 22 99 121 1# R- +

= 1159 2 121178 9#- +

= 1159 2178 1089- + = 70

Thus, x2

R = 1159and x x 2R -^ h = 70.

20. ThemarksscoredbytwostudentsA, Binaclassaregivenbelow.

A 58 51 60 65 66B 56 87 88 46 43

Whoismoreconsistent?

Solution:

Student-A Student-B

x = nx/ =

5300 60= x =

nx/ = 6

5320 4=

x d = 60x - d2 x d = 64x - d

2

5158606566

-9-2 0 5 6

8140

2536

4346568788

-21-18-8

23 24

441324 64529576

300xR = d2

R = 146 320xR = d2

R = 1934

v = nd2/ =

5146 = .29 2 = 5.4 v =

nd2/ = . 19.67

51934 386 8= =

C.V = x

100#v = .

605 4 100# = 9 g (1) C.V =

x100#

v = . 10064

19 67# = 30.73 g (2)

From(1)and(2),weseethatthecoefficientofvariationforAislessthanthecoefficientofvariationforB.

Thus,StudentAismoreconsistant.


Exercise 11.2

Choosethecorrectanswer.

1. Therangeofthefirst10primenumbers2,3,5,7,11,13,17,19,23,29is

(A)28 (B)26 (C)29 (D)27

Solution: R = 29 2 27L S- = - = (Ans.(D)) 2. Theleastvalueinacollectionofdatais14.1.Iftherangeofthecollectionis28.4,

thenthegreatestvalueofthecollectionis

(A)42.5 (B)43.5 (C)42.4 (D)42.1

Solution: S = 14.1, R = 28.4, R = L – S & L = R + S

L& = 28.4 14.1 42.5+ = (Ans.(A)) 3. Thegreatestvalueofacollectionofdata is72and the leastvalue is28.Then the

coefficientofrangeis

(A)44 (B)0.72 (C)0.44 (D)0.28

Solution: L = 72, S = 28

Co-efficientrange =72 2872 28 .

10044 0 44L S

L S+- =

+- = = (Ans.(C))

4. Foracollectionof11items, x 132R = ,thenthearithmeticmeanis

(A)11 (B)12 (C)14 (D)13

Solution: n = 11 xR = 132

xr = 12nx

11132R = = (Ans.(B))

5. Foranycollectionofnitems, ( )x xR - =

(A) xR (B)x (C)nx (D)0

Solution: Foracollectionofn items,wealwayshave ( )x x 0R - = . (Ans.(D)) 6. Foranycollectionofnitems,( )x xR - =

(A)nx (B)( 2)n x- (C)( 1)n x- (D)0

Solution: x xR - = ( 1)nx x n x- = -r r r (Ans.(C)) 7. If tisthestandarddeviationofx, y. z,thenthestandarddeviationofx + 5, y + 5, z +5is

(A) t3 (B)t+5 (C)t (D)x y z

Solution: TheSDofadistributionremainsunchangedwheneachvalueisadded(or)subtractedbythesamequantity. t` v= . (Ans.(C))

8. Ifthestandarddeviationofasetofdatais1.6,thenthevarianceis

(A)0.4 (B)2.56 (C)1.96 (D)0.04

Solution: Variance, . .1 6 2 562 2v = = (Ans.(B))


9. Ifthevarianceofadatais12.25,thentheS.Dis

(A)3.5 (B)3 (C)2.5 (D)3.25

Solution: SD = . 3.5Variance 12 25= = (Ans.(A))

10. Varianceofthefirst11naturalnumbersis

(A) 5 (B) 10 (C)5 2 (D)10

Solution: Varianceoffirst11naturalnumbers.

2v = n12

112

11 112120 10

2 2- = - = = (Ans.(D))

11. Thevarianceof10,10,10,10,10is

(A)10 (B) 10 (C)5 (D)0

Solution: x = 10, d x x 0= - =r .Thus,Variance 2v = 0nd2R = (Ans.(D))

(Notethatthevarianceofaconstantsequenceisalwayszero.) 12. Ifthevarianceof14,18,22,26,30is32,thenthevarianceof28,36,44,52,60is

(A)64 (B)128 (C)32 2 (D)32

Solution:Varianceof14,18,22,26,30is32.Thus,S.D.= 32 . Note that

28,36,44,52,60areobtainedbymultiplyingeachdataof14,18,22,26,30by2.

Thus,S.D.of28,36,44,52,60is2 32

Hence,varianceof28,36,44,52,60is 2v = 2 322^ h =128. (Ans.(B))

13. Standarddeviationofacollectionofdatais2 2 .Ifeachvalueismultipliedby3,thenthestandarddeviationofthenewdatais

(A) 12 (B)4 2 (C)6 2 (D)9 2

Solution:S.D.=2 2 .Ifeachdataismultipliedby3,then

newS.D. = 3 2 2 6 2# = (Ans.(C)) 14. Given ( ) ,x x x48 202

- = =/ andn=12.Thecoefficientofvariationis

(A)25 (B)20 (C)30 (D)10

Solution: v = nd

2R =

1248 4 2= =

Coefficientofvariation,C.V = 100x#vr

= 100202 10# = . (Ans.(D))

15. Meanandstandarddeviationofadataare48and12respectively.Thecoefficientofvariationis

(A)42 (B)25 (C)28 (D)48

Solution: Giventhatv = 12, x = 48


= 1004812 25# = . (Ans.(B))

Solution - Probability 269

Exercise 12.1 1. A ticket is drawn from a bag containing 100 tickets. The tickets are numbered

from one to hundred. What is the probability of getting a ticket with a number divisible by 10?Solution: 100 tickets are numbered from 1 to 100.

Sample space, S = {1,2,3, ,100} ; ( ) 100.n Sg =

Let A be the event of getting a ticket with a number divisible by 10.

So, {10,20, ,100} ; ( ) 10.A n Ag= =

Thus, ( )( )( )

P An Sn A

10010

101= = = .

2. A die is thrown twice. Find the probability of getting a total of 9.Solution: When a die is thrown twice, the sample space is , , , , , , , , , , , , , , , , , , , , ,S 1 1 1 2 1 6 2 1 2 2 2 6 6 1 6 2 6 6g g g g= ^ ^ ^ ^ ^ ^ ^ ^ ^h h h h h h h h h" ,

Thus, 36n S =^ h .

Let A be the event of getting a total of 9.

Then A = { (3,6), (4,5), (5,4), (6,3) } ; n(A)= 4.

Thus, P(A) = ( )( )

n Sn A

364

91= = .

3. Two dice are thrown together. Find the probability that the two digit number formed with the two numbers turning up is divisible by 3.Solution: When a die is thrown twice, the sample space is

, , , , , , , , , , , , , , , , , , , , ,S 1 1 1 2 1 6 2 1 2 2 2 6 6 1 6 2 6 6g g g g= ^ ^ ^ ^ ^ ^ ^ ^ ^h h h h h h h h h" ,; 36n S =^ h

Let A be the event of the two digit number formed with the two numbers turning up is divisible by 3.Thus, A = { 12, 15, 21, 24, 33, 36, 42, 45, 51, 54, 63, 66}. So, ( )n A = 12.Hence, ( )P A =

( )( )

n Sn A

3612

31= = .

4. Three rotten eggs are mixed with 12 good ones. One egg is chosen at random. What is the probability of choosing a rotten egg?

Solution: Number of good eggs =12; Number of rotten eggs 3=

So, total numbers of eggs 51= .

Let A be the event of choosing a rotten egg.

Thus, n(A) = 3 and ( )( )( )

.P An Sn A

153

51= = =

Probability 12


5. Two coins are tossed together. What is the probability of getting at most one head.

Solution: The sample space, { , , , } ; ( ) 4.S HH HT TH TT n S= =

Let A be the event of getting atmost one head.

Thus, A = , ,HT TH TT" , and 3n A =^ h .

Hence, ( )P A = n S

n A

43=

^^

hh .

6. Onecardisdrawnrandomlyfromawellshuffleddeckof52playingcards.Findthe probability that the drawn card is (i) a Diamond, (ii) not a Diamond, (iii) not an Ace.

Solution: Here, ( )n S 52=

(i) Let A be the event of drawing a diamond card.

Thus, ( ) 13 ; ( )( )( )

.n A P An Sn A

5213

41= = = =

(ii) Let B be the event of drawing a card which is not a diamond. Then B A= l

Thus, ( ) 1 ( ) 1P B P A41

43= - = - = .

(iii) Let C be the event of drawing an Ace card.

Thus, ( ) 4 ; ( )( )( )

.n C P Cn Sn C

524

131= = = =

Hence, ( ) 1 .P C131

1312= - =l

7. Three coins are tossed simultaneously. Find the probability of getting (i) atleast one head, (ii) exactly two tails, (iii) atleast two heads.

Solution: The sample space,

{ , , , , , , , } ; ( ) 8.S HHH HHT HTH THH HTT THT TTH TTT n S= =

(i) Let A be the event of getting atleast one head.

{ , , , , , , } ; ( ) 7A HHH HHT HTH THH HTT THT TTH n A= = and so, ( ) .P A87=

(ii) Let B be the event of getting exactly two tails.

Thus, { , , } ; ( ) 3B HTT THT TTH n B= = and so, ( )( )( )

.P Bn Sn B

83= =

(iii) Let C be the event of getting atleast two heads. Thus,

{ , , , } ; ( ) 4 ( )( )( )

.,C HHH HHT HTH THH n C P Cn Sn C

84

21and so= = = = =

8. A bag contains 6 white balls numbered from 1 to 6 and 4 red balls numbered from 7 to 10. A ball is drawn at random. Find the probability of getting (i) an even-numbered ball, (ii) a white ball.

Solution: The sample space, { , , , , , , , , , }S W W W W W W R R R R1 2 3 4 5 6 7 8 9 10

= ; ( )n S 10=

(i) Let A be the event of getting an even-numbered ball. Then, { , , , , }A W W W R R

2 4 6 8 10= ; ( )n A 5= .


Thus, ( )( )( )

.P An Sn A

105

21= = =

(ii) Let B be the event of getting a white ball { , , , , , }B W W W W W W

1 2 3 4 5 6= , ( )n B 6= .

Thus, ( )( )( )

P Bn Sn B

106

53= = = .

9. A number is selected at random from the integers 1 to 100. Find the probability that it is (i) a perfect square, (ii) not a perfect cube.

Solution: Given that ( )n S 100= .

(i) Let A be the event of getting a perfect square.

So, { , , , , , , , , , }A 1 4 9 16 25 36 49 64 81 100= ; ( )n A 10= .

Thus, ( )( )( )

P An Sn A

10010

101= = = .

(ii) Let B be the event of getting a perfect cube

So, { , , , }B 1 8 27 64= ; ( )n B 4= .

Thus, ( )( )( )

P Bn Sn B

1004

251= = = .

Hence, the probability that the selected number is not a cube is

( ) 1 ( )P B P B= -l 1251

2524= - = .

10. For a sightseeing trip, a tourist selects a country randomly from Argentina, Bangladesh, China, Angola, Russia and Algeria. What is the probability that the name of the selected country will begin with A ?

Solution: S = {Argentina, Bangladesh, China, Angola, Russia, Algeria}, ( )n S 6= .

Let A be the event of selecting a country whose name begins with the letter A.

So, A = {Argentina, Angola, Algeria} ; ( ) 3n A = .

Thus, ( )( )( )

P An Sn A

63

21= = = .

11. A box contains 4 Green, 5 Blue and 3 Red balls. A ball is drawn at random. Find the probability that the selected ball is (i) Red in colour (ii) not Green in colour.

Solution: Total number of balls, ( )n S 4 5 3= + + 12=

Let A be the event of drawing a Red coloured ball. Then, ( )n A 3= .

Thus, ( )( )( )

P An Sn A

123

41= = =

Let B be the event of drawing a Green Coloured ball. Then, ( )n B 4= .

( )( )( )

P Bn Sn B

124

31= = = .

Hence, ( ) 1 ( )P B P B= - 131

32= - = .


12. 20 cards are numbered from 1 to 20. One card is drawn at random. What is the probability that the number on the card is

(i) a multiple of 4, (ii) not a multiple of 6.


(i) Let A be the event of drawing a card such that a number on it, is a multiple of 4.

Thus, {4, 8,12,16, 20} ; ( ) 5.A n A= = So, ( )( )( )

.P An Sn A

205

41= = =

(ii) Let B be the event of drawing a card such that the number on the card is a multiple of 6.

Thus, {6,12,18}; ( ) 3.B n B= = So, ( )( )( )

.P Bn Sn B

203= =

Hence, ( ) 1 ( ) 1 .P B P B203

2017= - = - =

13. A two digit number is formed with the digits 3, 5 and 7. Find the probability that the number so formed is greater than 57 (repetition of digits is not allowed).

Solution: The sample space, { 35, 37, 53, , 73, 75 }S 57= ; ( ) 6.n S =

Let A be the event of getting the number greater than 57.

Then, A = { 73,75} ; ( ) 2n A = . Thus, ( )P A = ( )( )

.nn A5 6

231= =

14. Three dice are thrown simultaneously. Find the probability of getting the same number on all the three dice.Solution: The sample space S is the collection of all possible outcomes. ( ) 6n S 2163` = = .Let A be the event of getting the same number on all the three dice .Then, { ( , , ), ( , , ), ( , , ), ( , , ), ( , , ), ( , , )}A 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6= ; ( ) 6.n A = Thus, ( )

( )( )

P An Sn A

2166

361= = = .

15. Two dice are rolled and the product of the outcomes (numbers) are found. What is the probability that the product so found is a prime number?

Solution: When a die is thrown twice, the sample space,

, , , , , , , , , , , , , , , , , , , , ,S 1 1 1 2 1 6 2 1 2 2 2 6 6 1 6 2 6 6g g g g= ^ ^ ^ ^ ^ ^ ^ ^ ^h h h h h h h h h" , ; 36n S =^ h

Let A be the event of getting the product so found is a prime number.

Thus, {( , ), (2, ), ( ,3), ( , ), ( , ), (5, )}A 1 2 1 1 3 1 1 5 1= ; ( )n A 6= .

Hence, ( )( )( )

.P An Sn A

366

61= = =

16. A jar contains 54 marbles each of which is in one of the colours blue, green and white. The probability of drawing a blue marble is

31 and the probability of

drawing a green marble is 94 . How many white marbles does the jar contain?


Solution: Here, ( )n S 54= . Now, ( )P B31= .

Also, ( )P G94= . Now, ( )

( )P W

n W54

= .

We know that ( ) ( ) ( )P B P G P W 1+ + = ( )1

n W54 3

194& = - +` j.

( )1

n54 9

7W& = - & ( ) 54n

92W #= ( ) 12.n W& =

Thus, the jar contains 12 white marbles.

17. A bag contains of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. A trader A will accept only the shirt which are good, but the trader B will not accept the shirts which have major defects. One shirt is drawn at random. What is the probability that it is acceptable by (i) A (ii) B ?

Solution: Given that ( ) 100n S = .

Let G be the event of selecting a good shirt. Thus, ( )n G 88= .

Let M be the event of not selecting a major defective shirt. Thus, ( ) 88 8 .n M 96= + =

(i) Probability that the selected shirt is accepted by A is ( )( )

n Sn G

10088

2522= = .

(ii) Probability that the selected shirt is accepted by B is ( )( )n Sn M .

10096

2524= =

18. A bag contains 12 balls out of which x balls are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball. (ii) If 6 more white balls are put in the bag and if the probability of drawing a white ball will be twice that of in(i),thenfindx.

Solution: Let S1 and S2 be the sample spaces before and after adding the white balls.

Let W1 and W2 be the events of selecting white balls before and after adding 6 white balls.

Now ( ) , ( )n S n S12 12 6 181 2= = + = .

Let ( ) . , ( ) 6.Thenn W x n W x1 2= = +

Thus, ( ) ( ) .P W x P W x12 18

6and1 2= = +

Given that ( ) ( )P W P W22 1= x x18

6 212

& + = x x36& + = 3x& = .

Hence, ( ) 3.P W x123

41 and1 = = =

19. Piggybankcontains100fifty-paisecoins,50one-rupeecoins,20two-rupeescoinsand10five-rupeescoins.Onecoinisdrawnatrandom.Findtheprobabilitythatthedrawncoin(i)willbeafifty-paisecoin(ii)willnotbeafive-rupeescoin.

Solution: Let S be the sample space consisting of all coins.

Thus, ( ) 100 50 20 10 180.n S = + + + =


Let F1, O, T and F2 be the events of selecting a fifty paise, one-rupee, two-rupees and five-rupees coins respectively.

Thus, ( ) 100 ; ( ) 50;n F n O1 = = ( ) 20; ( ) 10n T n F2= = .

(i) Probability that the drawn coin is a fifty-paise coin,

( )( )( )

.P Fn Sn F

180100

95

11

= = =

(ii) The probability that the drawn coin is a five-rupees coin,

( )( )( )

.P Fn Sn F

18010

181

22

= = =

Probability that the drawn coin is not a five-rupees coin,

( ) 1 ( ) 1 .P F P F181

1817

2 2= - = - =

Exercise 12.2

(The problems in this exercise can also be solved directly using the definition of classical probability and without using the addition theorem.)

1. If A and B are mutually exclusive events such that ( ) ( )P A P B53

51and= = , then

find ( )P A B, .

Solution: If A and B are mutually exclusive events, then ( )P A B 0+ =

Now, ( ) ( ) ( ) ( )P A B P A P B P A B, += + - = 0 .53

51

54+ - =

2. If A and B are two events such that ( ) , ( )P A P B41

52= = and ( )P A B

21, = , then

find ( )P A B+ .

Solution: We know that ( ) ( ) ( ) ( )P A B P A P B P A B, += + - .

Thus, ( ) ( ) ( ) ( )P A B P A P B P A B+ ,= + -

= .41

52

21

203+ - =

3. If ( ) , ( ) , ( ) 1,P A P B P A B21

107 ,= = = thenfind(i) ( )P A B+ (ii) ( )P A B,l l .

Solution: (i) ( ) ( ) ( ) ( )P A B P A P B P A B+ ,= + -

= 121

107+ - .

105 7 10

102

51= + - = =

(ii) ( ) ( )P A B P A B, +=l l l ( ( )A B A B+ ,=l l l)

= 1 ( )P A B+- ( ( ) 1 ( )P A P A= -l )

= 151

54- = .

4. Ifadieisrolledtwice,findtheprobabilityofgettinganevennumberinthefirsttime or a total of 8.

Solution: Let S be the sample space consisting of all possible outcomes.

{(1,1), (1,2), . . , (1,6), (2,1), (2,2), . ., (2,6), (6,1), (6,2), . . , (6,6)}S = ; ( ) 36n S =


Let A be the event of getting an even number in the first time.

{ ( , ), ( , ), ( , ), ( , ), ( , ), ( , ),

( , ), ( , ), ( , ), ( , ), ( , ), ( , ),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) } . , ( ) 18.So

A

n A

2 1 2 2 2 3 2 4 2 5 2 6

4 1 4 2 4 3 4 4 4 5 4 6

=

=

Thus, ( )P A = ( )( )

n Sn A

3618

21= = .

Let B be the event of getting a total 8.

B = { (2,6), (3,5), (4,4), (5,3), (6,2) } . , ( ) 5.So n B =

Thus, ( )( )( )

P Bn Sn B

365= = .

Also, { ( , ), ( , ), ( , ) }A B 2 6 4 4 6 2+ = and ( )P A B363+ =

Hence, the required probability,

( ) ( ) ( ) ( )P A B P A P B P A B, += + - = 3618

365

363+ - .

3620

95= =

5. One number is chosen randomly from the integers 1 to 50. Find the probability that it is divisible by 4 or 6.

Solution: Sample space, { , , , , } .S 1 2 3 50g= So, ( )n S 50= .

Let A be the event of getting a number divisible by 4.

So, A = {4,8,12, ,48}g ; ( )n A 12= .

Thus, ( )( )( )

P An Sn A

5012= = .

Let B be the event of getting a number divisible by 6.

Then, { , , , , , , , }B 6 12 18 24 30 36 42 48= ; ( )n B 8= .

Thus, ( )( )( )

P Bn Sn B

508= = .

Also, { , , , }A B 12 24 36 48+ = . Then, ( )n A B 4+ = .

Thus, ( )P A B+ = ( )

( )n S

n A B504+

= .

Required probability, ( ) ( ) ( ) ( )P A B P A P B P A B, += + -

= 5012

508

504+ -

258= .

6. A bag contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted.Ifanitemischosenatrandom,findtheprobabilitythatitisrustedorthat it is a bolt.

Solution: Total number of items, ( )n S 200= . Number of bolts 50= .

Number of nuts 150= . Number of rusted bolts 25= . Number of rusted nuts 75= .

Let A be the event of getting rusted item.


Then, ( )n A = 75 25 100+ = .

Thus, ( )( )( )

.P An Sn A

200100= =

Let B be the event of getting a bolt. So, ( )n B = 50 .

Thus, ( )( )( )

.P Bn Sn B

20050= =

The number of bolts which are also rusted is, ( )n A B 25+ = .

Thus, ( )P A B+ = ( )

( ).

n Sn A B

20025+

=

Hence, the required probability, ( ) ( ) ( ) ( )P A B P A P B P A B, += + -

= 200100

20050

20025

85+ - = .

7. Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 3 nor by 4.

Solution: Sample space, {(1,1), (1,2), , (1,6), (2,1), (2,2), , (6,6)};S g g= ( ) 36.n S =

Let A be the event that the sum is divisible by 3.

Let B be the event that the sum is divisible by 4.

{(1,2), (2,1), (1,5), (5,1), (2,4), (4,2), (3,3), (3,6), (6,3), (4, ), (5,4), (6,6)}A 5=

{(1,3), (3,1), (2,2), (2,6), (6,2), (3,5), (5,3), (4,4), (6,6)}B = .

Thus, ( ) 12n A = ; ( )n B 9= ; ( , )A B 6 6+ = ; n A B 1+ =^ h .

Now, ( ) ( ) ( ) ( )P A B P A P B P A B, += + - .3612

369

361

3620= + - =

Hence, ( ' ') ( ) 'P A B P A B+ ,= 1 ( )P A B,= - 13620

3616

94= - = = .

8. A basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges arerotten.Ifapersontakesoutonefruitatrandom,findtheprobabilitythatthefruit is either an apple or a good fruit.

Solution: Total number of fruits in the basket, ( )n S 30= .

Total number of rotten fruits 8= ; Total number of good fruits 22= .

Let A be the event of getting an apple . So, n A 20=^ h

Thus, .P An S

n A

3020= =^

^^

hhh

Let B be the event of getting a good fruit. So, ( ) 2n B 2= .

Thus, ( )( )( )

P Bn Sn B

3022= = .

Number of good apples is, ( )n A B 15+ = .

Thus, ( )( )

( )P A B

n Sn A B

3015+

+= = .


Required probability, ( ) ( ) ( ) ( )P A B P A P B P A B, += + -

= .3020

3022

3015

109+ - =

9. In a class, 40% of the students participated in Mathematics-quiz, 30% in Science-quiz and 10% in both the quiz programmes. If a student is selected at random fromtheclass,findtheprobabilitythatthestudentparticipatedinMathematicsor Science or both quiz programmes.

Solution: Let S be the sample space.

Let M and S1 be the events that the selected student participated in mathematics and science quiz programmes respectively.

Given that ( )n S 100= , ( )n S 301 = , n M^ h = 40 and ( ) 10n M S1+ =

Thus, ( )P M10040= ; ( )P S

10030

1 = and ( )P M S10010

1+ = .

Now, the required probability, ( )P M S1

, = ( ) ( ) ( )P M P S P M S1 1

++ -

= 10040

10030

10010+ -

10060

53= = .

10. A card is drawn at random from a well-shuffled deck of 52 cards. Find theprobability that it will be a spade or a king.


Let A be the event of getting a spade. So, ( )n A 13= .

Thus, ( )( )( )

P An Sn A

5213= =

Let B be the event of getting a king. So, ( )n B 4= .

Thus, ( )( )( )

P Bn Sn B

524= =

Number of spade king, ( )n A B+ = 1. So, ( )P A B+ = ( )

( )n S

n A B521+

=

Required probability, ( )P A B, = ( ) ( ) ( )P A P B P A B++ -

= .5213

524

521

134+ - =

11. A box contains 10 white, 6 red and 10 black balls. A ball is drawn at random. Find the probability that the ball drawn is white or red.

Solution: Let S be the sample space,

Let W, R and B denote that the drawn ball is white, red and black respectively.Given that ( )n S 26= ; ( )n W 10= ; ( )n R 6= ; ( )n B 10= .

Thus, ( )( )( )

P Wn Sn W

2610= =

( )( )( )

P Rn Sn R

266= =


( )( )( )

P Bn Sn B

2610= = .

Here, W R+ is an impossible event. Thus, ( )P W R 0+ = .

Required probability, ( )P W R, = ( ) ( )P W P R+

= .2610

266

138+ =

12. A two digit number is formed with the digits 2, 5, 9 (repetition is allowed). Find the probability that the number is divisible by 2 or 5.


Let A and B be the events that the number is divisible by 2 and by 5 respectively.

Given that { , , , , , , , , }S 22 25 29 55 52 59 99 92 95= ; ( ) 9n S =

A = {22, 52, 92} and ( ) 3n A = . Thus, ( ) .P A93=

{25, 55, 95}B = and ( ) 3n B = . Thus, ( ) .P B93=

A B+ Q= and ( )n A B 0+ = . Thus, ( ) .P A B 0+ =

Hence, ( ) ( ) ( ) ( )P A B P A P B P A B, += + -

= 093

93+ - .

32=

13. Each individual letter of the word “ACCOMMODATION” is written in a piece of paper, and all 13 pieces of papers are placed in a jar. If one piece of paper is selected at random from the jar, find the probability that (i) the letter ‘A’ or ‘O’ is selected. (ii) the letter ‘M’ or ‘C’ is selected.


Thus, S contains all 13 pieces of papers. So, ( )n S 13= .

(i) Let H be the event of getting letter A . So, ( )n H 2=

Thus, ( )( )( )

P Hn Sn H

132= = .

Let B be the event of getting letter O. So, ( )n B 3= .

Hence, ( )( )( )

P Bn Sn B

133= = .

Required probability, ( ) ( ) ( )P H B P H P B, = + ( )H Ba + Q=

132

133

135= + =


(ii) Let C be the event of getting a letter M. So, ( )n C 2= . Hence, ( )

( )( )

P Cn Sn C

132= = .

Let D be the event of getting letter C. So, ( )n D 2= .

Hence, ( )P D = ( )( )n Sn D

132= .

Required probability, ( )P C D, = ( ) ( )P C P D+ ( )C Da + z=

= .132

132

134+ =

14. The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and theprobability that it will get both the awards is 0.15. Find the probability that (i) it will get atleast one of the two awards, (ii) it will get only one of the awards.

Solution: Let A be the event of getting award for design.

Let B be the event of getting award for efficient use of fuel.

Given ( ) .P A 0 25= , ( ) .P B 0 35= and ( ) .P A B 0 15+ =

(i) Probability of getting atleast one award,

( ) ( ) ( ) ( )P A B P A P B P A B, += + - . . .0 25 0 35 0 15= + - .0 45=

(ii) Probability of getting only one of the awards,

( ) ( )P A B P A B+ ++ [ ( ) ( )] [ ( ) ( )]P A P A B P B P A B+ += - + -

( . . ) ( . . )0 25 0 15 0 35 0 15= - + - . . .0 10 0 20 0 3= + =

15. The probability that A, B and C can solve a problem are ,54

32

73and respectively.

The probability of the problem being solved by A and B is 158 , B and C is

72 ,

A and C is 3512 . The probability of the problem being solved by all the three is

358 . Find the probability that the problem can be solved by atleast one of them.

Solution: Gi ven, ( ) , ( ) , ( )P A P B P C54

32

73= = = , ( ) ,P A B

158+ =

( ) ,P B C72+ = ( )P A C

3512+ = and ( )P A B C

358+ + = .

Now, ( )P A B C, , = ( ) ( ) ( ) ( )P A P B P C P A B++ + -

( ) ( ) ( ) .P B C P A C P A B C+ + + +- - +

= 54

32

73

158

72

3512

358+ + - - - + .

105101=


Exercise 12.3

Choose the correct answer

1. If z is an impossible event, then P z =^ h

(A) 1 (B) 41 (C) 0 (D)

21

Solution: Probability of an impossible event is 0. ( Ans. (C) ) 2. If S is the sample space of a random experiment, then P(S) =

(A) 0 (B) 81 (C)

21 (D) 1 ( Ans. (D) )

Solution: Every event is a subset of S. So, event S occurs always. Hence, P(S) = 1. 3. If p is the probability of an event A, then p satisfies

(A) p0 11 1 (B) p0 1# # (C) p0 11# (D) p0 11 #

( Ans. (B) ) 4. Let A and B be any two events and S be the corresponding sample space. Then

( )P A B+ =

(A) ( ) ( )P B P A B+- (B) ( ) ( )P A B P B+ -

(C) ( )P S (D) P A B, l^ h6 @ ( Ans. (A) )

5. The probability that a student will score centum in mathematics is 54 . The probability

that he will not score centum is

(A) 51 (B)

52 (C)

53 (D)

54

Solution: ( ) 1 ( )P A P A= - 154= -

51= ( Ans. (A) )

6. If A and B are two events such that ( ) 0.25, ( ) 0.05 ( ) 0.14P A P B P A Band += = =

( )P A Bthen , =

(A) 0.61 (B) 0.16 (C) 0.14 (D) 0.6 ( Ans. (B) )

Solution: ( ) ( ) ( ) ( ) 0.25 0.05 0.14P A B P A P B P A B, += + - = + - .0 16=

7. There are 6 defective items in a sample of 20 items. One item is drawn at random. The probability that it is a non-defective item is

(A) 107 (B) 0 (C)

103 (D)

32

Solution: Number of non-defective items 20 6= - 14= .

Probability of non-defective items2014=

107= . ( Ans. (A) )

8. If A and B are mutually exclusive events and S is the sample space such that ( ) ( )P A P B

31= and S A B,= , then ( )P A =

(A) 41 (B)

21 (C)

43 (D)

83

Solution: ( ) ( ) ( )P A B P A P B, = + (A , B are mutually exclusive events )

( ) ( ) ( )P S P A P A3= + 4 ( ) .P A 1& = Thus, ( )P A41= . ( Ans. (A) )


9. The probabilities of three mutually exclusive events A, B and C are given by

, ,31

41

125and . Then P A B C, ,^ h is

(A) 1219 (B)

1211

(C) 127 (D) 1

Solution: ( ) ( ) ( ) ( )P A B C P A P B P C, , = + + 31

41

125 1= + + = . ( Ans. (D) )

10. If ( ) 0.25, ( ) 0.50, ( ) 0.14, ( )thenP A P B P A B P A Bneither nor+= = = =

(A) 0.39 (B) 0.25 (C) 0.11 (D) 0.24

Solution: , ( ) 0.25 .50 .14Now P A B 0 0, = + - .0 61=

, ( ) 1 ( )Thus P A B P A B+ ,= - .1 0 61= - .0 39= . ( Ans. (A) )

11. A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected at random, the probability that it is not red is

(A) 125 (B)

124

(C) 123 (D)

43

Solution: ( ) 1 ( )P R P R= - 1123

43= - = . ( Ans. (D) )

12. Two dice are thrown simultaneously. The probability of getting a doublet is

(A) 361 (B)

31

(C) 61 (D)

32

Solution: ( )n S 36= . Let A be the event of getting a double. ( )n A 6=

{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} ; ( ) .A n A 6= = ( ) .P A366

61= = ( Ans. (C) )

13. A fair die is thrown once. The probability of getting a prime or composite number is

(A) 1 (B) 0

(C) 65 (D)

61

Solution: S = {1, 2, 3, 4, 5, 6}. Since 1 is neither a prime nor a composite number, required probability =

65 . ( Ans. (C) )

14. Probability of getting 3 heads or 3 tails in tossing a coin 3 times is

(A) 81 (B)

41 (C)

83 (D)

21

Solution: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}; n (S) = 8.

Required probability P{HHH, TTT} = 82

41= ( Ans. (B) )


15. A card is drawn from a pack of 52 cards at random. The probability of getting neither an ace nor a king card is

(A) 132 (B)

1311 (C)

134 (D)

138

Solution: n(Ace) = 4; n(King) = 4; n (non-ace and non-king cards) = 52 – 8 = 44

P (neither an ace nor a king) = 5244

1311= ( Ans. (B) )

16. The probability that a leap year will have 53 Fridays or 53 Saturdays is

(A) 72 (B)

71 (C)

74 (D)

73

Solution: Leap year contains 52 weeks and 2 more days.

Let S = { (Sun,Mon), (Mon, Tues), (Tues, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)}

n(S) = 7. Required probability = 72

72

71

73+ - = . ( Ans. (D) )

17. The probability that a non-leap year will have 53 Sundays and 53 Mondays is

(A) 71 (B)

72 (C)

73 (D) 0

Solution: Non-leap year contains 52 weeks and 1 more day. Thus, the event of 53 sundays and 53 mondays is impossible. So, the required probability = 0. ( Ans. (D) )

18. The probability of selecting a queen of hearts when a card is drawn from a pack of 52 playing cards is

(A) 521 (B)

5216 (C)

131 (D)

261 ( Ans. (A) )

Solution: 1 Queen of hearts in 52 cards. Thus, n(A) = 1, n(S) = 52, P(A) = 521

19. Probability of sure event is

(A) 1 (B) 0 (C) 100 (D) 0.1

Solution: Probability of sure event is 1. ( Ans. (A) )

20. The outcome of a random experiment results in either success or failure. If the probability of success is twice the probability of failure, then the probability of success is

(A) 31 (B)

32 (C) 1 (D) 0

Solution: Let p be the probability of success and q be the probability of failure.

Now, 1 .andp q p q p232&+ = = = ( Ans. (B) )

Classification of Questions - Sets and Functions 283

ClassifiCation of Questions1. sets anD funCtions

Two Mark Questions

example 1.1 For the given sets { 10,0,1, 9, 2, 4, 5}A = - and { 1, 2, 5, 6, 2,3,4}B = - - verify that (i) set union is commutative.

(ii) set intersettction is commutative. (each subdivision carries two mark)

example 1.2 Given, {1, 2, 3, 4, 5}, {3, 4, 5, 6} {5, 6, 7, 8}A B Cand= = = , show that (i) .A B C A B C, , , ,=^ ^h h

example 1.3 Let { , , , }, { , , } { , }A a b c d B a c e C a eand= = = . (i) Show that A B C+ +^ h = .A B C+ +^ h

exercise 1.1

1. If ,A B1 then show that A B B, = (use Venn diagram).

2. If ,A B1 then find A B+ and \A B (use Venn diagram).

3. Let { , , }, { , , , } { , , , }P a b c Q g h x y R a e f sand= = = . Find (iii) \R P Q+^ h.

4. If {4,6,7,8,9}, {2,4,6} {1,2,3,4,5,6}A B Cand= = = , then find (each subdivision carries two mark)

(i) A B C, +^ h (ii) A B C+ ,^ h (iii) \ \A C B^ h

5. Given { , , , , }, {1,3,5,7, 10}A a x y r s B= = - , verify the commutative property of set union.

6. Verify the commutative property of set intersection for

{ , , , , 2, 3, 4, 7} {2, 5, 3, 2, , , , }A l m n o B m n o pand= = - .

exercise 1.2 1. Represent the following using Venn diagrams (each subdivision carries two mark)

(i) {5,6,7,8, ......13}, {5,8,10,11}, {5,6,7,9,10}U A Band= = =

(ii) { , , , , , , , }, { , , , }, { , , , , }U a b c d e f g h M b d f g N a b d e gand= = =

3. Draw Venn diagram of three sets ,A B Cand illustrating the following: (each subdivisions carries two mark)

(i) A B C+ + (ii) A Band are disjoint but both are subsets of C

(iii) \A B C+ ^ h (iv) \B C A,^ h (v) A B C, +^ h

(vi) \C B A+ ^ h (vii) C B A+ ,^ h


5. Let U = { , , , , , , }4 8 12 16 20 24 28 , A= { , , }8 16 24

and B= { , , , }4 16 20 28 . Find 'A B A Band, + l^ ^h h . (each subdivision carries two mark)

exercise 1.3

1. If A and B are two sets and U is the universal set such that 700n U =^ h , 200, 300 100n A n B n A B n A Band , find+ += = = l l^ ^ ^ ^h h h h.

2. Given 285, 195, 500, 410,n A n B n U n A B n A Bfind, ,= = = = l l^ ^ ^ ^ ^h h h h h.

3. For any three sets A, B and C if n A 17=^ h , 17, 17, 7n B n C n A B+= = =^ ^ ^h h h

( ) , 5 2n B C n A C n A B C6 and+ + + += = =^ ^h h , find n A B C, ,^ h.

example 1.14 Let { , , , }A 1 2 3 4= and { , , , , , , , , , , }B 1 2 3 4 5 6 7 9 10 11 12= - .

Let R = {(1, 3), (2, 6), (3, 10), (4, 9)} A B#3 be a relation. Show that R is a function and find its domain, co-domain and the range of R .

example 1.16 (each subdivision carries two mark)

Let X = { 1, 2, 3, 4 }. Examine whether each of the relations given below is a function from X to X or not. Explain.

(i) f = { (2, 3), (1, 4), (2, 1), (3, 2), (4, 4) }

(ii) g = { (3, 1), (4, 2), (2, 1) }

(iii) h = { (2, 1), (3, 4), (1, 4), (4, 3) }


Which of the following relations are functions from A = { 1, 4, 9, 16 } to B = { –1, 2, –3, –4, 5, 6 }? In case of a function, write down its range.

(i) f1 = { (1, –1), (4, 2), (9, –3), (16, –4) }

(ii) f2 = { (1, –4), (1, –1), (9, –3), (16, 2) }

(iii) f3 = { (4, 2), (1, 2), (9, 2), (16, 2) }

(iv) f4 = { (1, 2), (4, 5), (9, –4), (16, 5) }

example 1.18: Let ifx

x x

x x

0

0if 1

$=

-' , where .x Rd Does the relation

{ ( ,x y ) | y = | x |, x R! } define a function? Find its range.



Use the vertical line test to determine which of the following graphs represent a function.

example 1.21 Let A = { 1, 2, 3, 4, 5 }, B = N and :f A B" be defined by ( )f x x2

= . Find the range of f . Identify the type of function.

exercise 1.4 1. State whether each of the following arrow diagrams define a function or not. Justify

your answer. (each subdivision carries two marks)

2. For the given function F = { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }, write the domain and range.

3. Let A = { 10, 11, 12, 13, 14 }; B = { 0, 1, 2, 3, 5 } and :f A Bi " , i = 1,2,3. State the type of function for the following (give reason): (each subdivision carries two marks)

(i) f1 = { (10, 1), (11, 2), (12, 3), (13, 5), (14, 3) }

(ii) f2 = { (10, 1), (11, 1), (12, 1), (13, 1), (14, 1) }

(iii) f3 = { (10, 0), (11, 1), (12, 2), (13, 3), (14, 5) }


4. If X = { 1, 2, 3, 4, 5 }, Y = { 1, 3, 5, 7, 9 } determine which of the following relations from X to Y are functions? Give reason for your answer. If it is a function, state its type. (each subdivision carries two marks)

(i) R1 = { ,x y^ h|y x 2= + , x X! , y Y! }

(ii) R2 = { (1, 1), (2, 1), (3, 3), (4, 3), (5, 5) }

(iii) R3 = { (1, 1), (1, 3), (3, 5), (3, 7), (5, 7) }

(iv) R4 = { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }

5. If R {( , 2), ( 5, ), (8, ), ( , 1)}a b c d= - - - represents the identity function, find the values of , ,a b c and d .

6. A = { –2, –1, 1, 2 } and , :f xx

x A1 != ` j$ .. Write down the range of f . Is f a function from A to A ?

8. Write the pre-images of 2 and 3 in the function

f = { (12, 2), (13, 3), (15, 3), (14, 2), (17, 17) }.

9. The following table represents a function from A= { 5, 6, 8, 10 } to

B = { 19, 15, 9, 11 } where f x^ h = x2 1- . Find the values of a and b .

x 5 6 8 10f(x) a 11 b 19

11. State whether the following graphs represent a function. Give reason for your answer. (each subdivision carries two marks)

12. Represent the function f = { (–1, 2), (– 3, 1), (–5, 6), (– 4, 3) } as (i) a table, (ii) an arrow diagram.


15. A function f : ,3 7- h6 " R is defined as follows (each subdivision carries two marks)

f x^ h = ;

;

;

x x

x x

x x

4 1 3 2

3 2 2 4

2 3 4 7

2 1

1 1

#

# #

- -

-

-

* .

Find (i) f f5 6+^ ^h h, (ii) f f1 3- -^ ^h h, (iii) f f2 4- -^ ^h h.

16. A function f : ,7 6- h6 " R is defined as follows (each subdivision carries two marks)

( )f x = ;

;

; .

x x x

x x

x x

2 1 7 5

5 5 2

1 2 6

2 1

1 1

#

# #

+ + - -

+ -

-

*

Find (i) ( ) ( )f f2 4 3 2- + , (ii) ( ) ( )f f7 3- - - .

Five Mark Questions

example 1.2 Given, {1, 2, 3, 4, 5}, {3, 4, 5, 6} {5, 6, 7, 8}A B Cand= = = , show that (ii) Verify .A B C A B C, , , ,=^ ^h h using Venn diagram.

example 1.3 Let { , , , }, { , , } { , }A a b c d B a c e C a eand= = = . (ii) Verify A B C+ +^ h = .A B C+ +^ h using Venn diagram.example 1.4 Given { , , , , }, { , , , , } { , , , }A a b c d e B a e i o u C c d e uand= = = . Show that (i) \ \ \ \A B C A B C!^ ^h h . (ii) Verify \ \ \ \A B C A B C!^ ^h h using Venn

diagram. (each subdivision carries five marks)example 1.5 Let {0,1,2,3,4}, {1, 2, 3,4,5,6} {2,4,6,7}A B Cand= = - = . Show that

(each subdivision carries five marks) (i) A B C, +^ h = .A B A C, + ,^ ^h h (ii) Verify A B C, +^ h = A B A C, + ,^ ^h h using Venn diagram.

example 1.6 For { 3 4, }, { 5, }A x x x B x x xR N1 1; # ! ; != - = and { 5, 3, 1,0,1,3}C = - - - , Show that A B C A B A C+ , + , +=^ ^ ^h h h.

exercise 1.1 7. For A = { 4 }x x 2is a prime factor of; , { 5 12, }B x x x N1; # != and

C = { , , , }1 4 5 6 , verify A B C A B C, , , ,=^ ^h h . 8. Given { , , , , }, { , , , , } { , , , }P a b c d e Q a e i o u R a c e gand= = = . Verify the associative

property of set intersection. 9. For {5,10,15, 20}; {6,10,12,18,24} {7,10,12,14,21,28},A B Cand= = = verify

whether \ \ \ \A B C A B C=^ ^h h . Justify your answer. 10. Let { 5, 3, 2, 1}, { 2, 1,0}, { 6, 4, 2}A B Cand= - - - - = - - = - - - . Find

\ \ ( \ ) \A B C A B Cand^ h . What can we conclude about set difference operation? 11. For { 3, 1, 0, 4,6,8,10}, { 1, 2, 3,4,5,6} { 1, 2,3,4,5,7},A B Cand= - - = - - = - show that (each subdivision carries five marks) (i) A B C, +^ h= A B A C, + ,^ ^h h (ii) A B C+ ,^ h= A B A C+ , +^ ^h h


(iii) Verify A B C, +^ h= A B A C, + ,^ ^h h using Venn diagram (iv) Verify A B C+ ,^ h= A B A C+ , +^ ^h h using Venn diagram.example 1.7 Use Venn diagrams to verify A B A B+ ,=l l l^ h .example 1.8 Use Venn diagrams to verify De Morgan’s law for set difference

\ \ \A B C A B A C+ ,=^ ^ ^h h h.example 1.9 Let { 2, 1, 0,1, 2, 3, ,10}, { 2, 2,3,4,5}U Ag= - - = - and

{1,3,5, ,9}B 8= . Verify De Morgan’s laws of complementation. (each law carries five marks)example 1.10 Let A = { , , , , , , , , , }a b c d e f g x y z , B = { , , , , }c d e1 2 and C = { , , , , , }d e f g y2 . Verify \ \ \A B C A B A C, +=^ ^ ^h h h.

exercise 1.2 4. Use Venn diagram to verify \A B A B A+ , =^ ^h h .

6. Given that U = { , , , , , , , }a b c d e f g h , { , , , }, { , , },A a b f g B a b cand= = verify De Morgan’s laws of complementation. (each law carries five marks)

7. Verify De Morgan’s laws for set difference using the sets given below: (each law carries five marks) {1, 3, 5, 7, 9,11,13,15}, {1, 2, 5, 7} {3,9, , ,13}A B C 10 12and= = = .

8. Let A = {10,15, 20, 25, 30, 35, 40, 45, }50 , B = { , ,10,15, 20, 30}1 5

and C = { , ,15,2 ,35,45, }7 8 0 48 . Verify \ \ \A B C A B A C+ ,=^ ^ ^h h h.

9. Using Venn diagram, verify whether the following are true: (each subdivision carries five marks) (i) A B C, +^ h = A B A C, + ,^ ^h h (ii) A B C+ ,^ h = A B A C+ , +^ ^h h(iii) A B, l^ h = A B+l l (iv) \A B C,^ h = \ \A B A C+^ ^h h.

example 1.11 In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. Find the number of students in the group. (Assume that each student in the group plays atleast one game.)

example 1.12 In a survey of university students, 64 had taken mathematics course, 94 had taken computer science course, 58 had taken physics course, 28 had taken mathematics and physics, 26 had taken mathematics and computer science, 22 had taken computer science and physics course, and 14 had taken all the three courses. Find the number of students who were surveyed. Find how many had taken one course only.

example 1.13 A radio station surveyed 190 students to determine the types of music they liked. The survey revealed that 114 liked rock music, 50 liked folk music, and 41 liked classical music, 14 liked rock music and folk music, 15 liked rock music and classical music, 11 liked classical music and folk music. 5 liked all the three types of music.

Find (i) how many did not like any of the 3 types? (ii) how many liked any two types only? (iii) how many liked folk music but not rock music?


exercise 1.3 4. Verify (each subdivision carries five marks)n A B C n A n B n C n A B, , += + + - -^ ^ ^ ^ ^h h h h h n B C n A C n A B C+ + + +- +^ ^ ^h h h

for the sets given below: (i) {4,5,6}, {5,6,7,8} {6,7,8,9}A B Cand= = =

(ii) { , , , , }, { , , } { , , }A a b c d e B x y z C a e xand= = =

5. In a college, 60 students enrolled in chemistry, 40 in physics, 30 in biology, 15 in chemistry and physics, 10 in physics and biology, 5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects.

6. In a town 85% of the people speak Tamil, 40% speak English and 20% speak Hindi. Also, 32% speak English and Tamil, 13% speak Tamil and Hindi and 10% speak English and Hindi, find the percentage of people who can speak all the three languages.

7. An advertising agency finds that, of its 170 clients, 115 use Television, 110 use Radio and 130 use Magazines. Also, 85 use Television and Magazines, 75 use Television and Radio, 95 use Radio and Magazines, 70 use all the three. Draw Venn diagram to represent these data. Find

(i) how many use only Radio? (ii) how many use only Television? (iii) how many use Television and magazine but not radio? 8. In a school of 4000 students, 2000 know French, 3000 know Tamil and 500 know

Hindi, 1500 know French and Tamil, 300 know French and Hindi, 200 know Tamil and Hindi and 50 know all the three languages.

(i) How many do not know any of the three languages? (ii) How many know at least one language? (iii) How many know only two languages? 9. In a village of 120 families, 93 families use firewood for cooking, 63 families use

kerosene, 45 families use cooking gas, 45 families use firewood and kerosene, 24 families use kerosene and cooking gas, 27 families use cooking gas and firewood Find how many use firewood, kerosene and cooking gas.

example 1.20 Let A= { 0, 1, 2, 3 } and B = { 1, 3, 5, 7, 9 } be two sets. Let :f A B" be a function given by ( )f x x2 1= + . Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph.

example 1.22 A function : [1, 6)f R$ is defined as follows ,

,

,

f x

x x

x x

x x

1 1 2

2 1 2 4

3 10 4 62

1

1

1

#

#

#

=

+

-

-

^ h * (Here, [1 , 6) = { x Re : 1# x 1 6} )

Find the value of (i) ( )f 5 , (ii) f 3^ h, (iii) f 1^ h, (iv) f f2 4-^ ^h h,

(v) 2 3f f5 1-^ ^h h


exercise 1.4 7. Let f = { (2, 7), (3, 4), (7, 9), (–1, 6), (0, 2), (5, 3) } be a function from

A = { –1, 0, 2, 3, 5, 7 } to B = { 2, 3, 4, 6, 7, 9 }. Is this (i) an one-one function (ii) an onto function (iii) both one-one and onto function?

10. Let A= { 5, 6, 7, 8 }; B = { –11, 4, 7, –10,–7, –9,–13 } and

f = {( ,x y ) : y = x3 2- , x A! , y B! }

(i) Write down the elements of f . (ii) What is the co-domain?

(iii) What is the range? (iv) Identify the type of function.

13. Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and :f A B" be defined by

f x^ h = x33- . Represent f by (i) an arrow diagram (ii) a set of ordered pairs

(iii) a table (iv) a graph .

14. Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If :f A B" is defined by f x x21 1= +^ h


15. A function f : ,3 7- h6 " R is defined as follows

f x^ h = ;

;

;

x x

x x

x x

4 1 3 2

3 2 2 4

2 3 4 7

2 1

1 1

#

# #

- -

-

-

* . Find (iv) ( ) ( )

( ) ( )f f

f f2 6 13 1

-+ - .


( )f x = ;

;

; .

x x x

x x

x x

2 1 7 5

5 5 2

1 2 6

2 1

1 1

#

# #

+ + - -

+ -

-

*

Find (iii) ( ) ( )( ) ( )

f ff f

6 3 14 3 2 4

- -- + .

Proof by pictureLet us illustrate the result : 1 2 3

( )n

n n2

1#g+ + + + =+ with the following diagram

Hence, 1 2 3 92

9 10#g+ + + + =

Classification of Questions - Sequences and Series of Real Numbers 291

2. SEQUENCES AND SERIES

Two Mark Questions

Example 2.1 Write the first three terms in a sequence whose th term is given by

cn n n

6

1 2 1n=

+ +^ ^h h , n N6 ! .

Example 2.2 (each subdivision carries two marks) Write the first five terms of each of the following sequences.

(i) 1,a1=- , 1a

n

an

2nn 1 2=+- and n N6 !

(ii) 1F F1 2= = and , 3,4, .F F F n

n n n1 2g= + =

- -

Exercise 2.1 1. Write the first three terms of the following sequences whose nth terms are given by

(each subdivision carries two marks)

(i) a n n

3

2n=

-^ h (ii) 3c 1n

n n 2= -

+^ h (iii) z n n

4

1 2n

n

=- +^ ^h h

2. Find the indicated terms in each of the sequences whose nth terms are given by (each subdivision carries two marks)

(i) ; ,ann a a2 3

2n 7 9=

++ (ii) 2 ; ,a n a a1 1

n

n n 3

5 8= - +

+^ ^h h

(iii) 2 3 1; ,a n n a a.n

2

5 7= - + (iv) ( 1) (1 ); ,a n n a an

n 2

5 8= - - +

3. Find the 18th and 25th terms of the sequence defined by

( ),

, .

if and even

if and odda

n n n n

n

n n n

3

1

2isis

N

Nn 2

!

!=

+

+*

4. Find the 13th and 16th terms of the sequence defined by

,

( ), .

if and even

if and oddb

n n n

n n n n2

isis

N

Nn

2!

!=

+)

5. Find the first five terms of the sequence given by

2, 3a a a1 2 1= = + and 2 5 2a a nfor

n n 12= +

-.

6. Find the first six terms of the sequence given by

1a a a1 2 3= = = and a a a

n n n1 2= +

- - for n 32 .

Example 2.3 Which of the following sequences are in an A.P.? (each subdivision carries two marks) (i) , , ,

32

54

76 g . (ii) , , , .m m m3 1 3 3 3 5 g- - -

Example 2.4 Find the first term and common difference of the A.P. (each subdivision carries two marks) (i) 5, 2, 1, 4,g- - . (ii) , , , , ,

21

65

67

23

617g


Example 2.5 Find the smallest positive integer n such that tn of the arithmetic sequence 20,19 ,18 ,

41

21 g is negative.?

Example 2.6 In a flower garden, there are 23 rose plants in the first row, 21 in the second row, 19 in the third row and so on. There are 5 rose plants in the last row. How many rows are there in the flower garden?

Example 2.7 If a person joins his work in 2010 with an annual salary of `30,000 and receives an annual increment of 600 every year, in which year, will his annual salary be `39,000?

Example 2.8 Three numbers are in the ratio 2 : 5 : 7. If the first number, the resulting number on the substraction of 7 from the second number and the third number form an arithmetic sequence, then find the numbers.

Exercise 2.2 1. The first term of an A.P. is 6 and the common difference is 5. Find the A.P. and its

general term.

2. Find the common difference and 15th term of the A.P. 125, 120, 115, 110, g .

3. Which term of the arithmetic sequence 24, 23 , 22 , 21 , .41

21

43 g is 3?

4. Find the 12th term of the A.P. , 3 , 5 , .2 2 2 g

5. Find the 17th term of the A.P. 4, 9, 14, g .

6. How many terms are there in the following Arithmetic Progressions? (each subdivision carries two marks) (i) 1, , , , .

65

32

310g- - - (ii) 7, 13, 19, g , 205.

10. How many two digit numbers are divisible by 13?

12. A man has saved `640 during the first month, `720 in the second month and `800 in the third month. If he continues his savings in this sequence, what will be his savings in the 25th month?

16. A person has deposited 25,000 in an investment which yields 14% simple interest annually. Do these amounts (principal + interest) form an A.P.? If so, determine the amount of investment after 20 years.

17. If , ,a b c are in A.P. then prove that ( ) 4( )a c b ac2 2

- = - .

18. If , ,a b c are in A.P. then prove that , ,bc ca ab1 1 1 are also in A.P.

Example 2.9 Which of the following sequences are geometric sequences? (each subdivision carries two marks)

(a) (i) 5, 10, 15, 20, g . (ii) 0.15, 0.015, 0.0015, g .

(b) (i) 5, 10, 15, 20, g . (iii) , , 3 , 3 , .7 21 7 21 g

(c) (ii) 0.15, 0.015, 0.0015, g . (iii) , , 3 , 3 , .7 21 7 21 g


Example 2.10 Find the common ratio and the general term of the following geometric sequences. (each subdivision carries two marks)

(i) , , ,52

256

12518 g . (ii) 0.02, 0.006, 0.0018, g .

Exercise 2.3

2. Find the 10th term and common ratio of the geometric sequence , ,1, 2,41

21 g- - .

3. If the 4th and 7th terms of a G.P. are 54 and 1458 respectively, find the G.P.

4. In a geometric sequence, the first term is 31 and the sixth term is

7291 , find the G.P.

5. Which term of the geometric sequence, (each subdivision carries two marks)

(i) 5, 2, , ,54

258 g , is

15625128 ? (ii) 1, 2, 4, 8,g , is 1024 ?

7. The fifth term of a G.P. is 1875. If the first term is 3, find the common ratio.

12. If 1000 is deposited in a bank which pays annual interest at the rate of 5% compounded annually, find the maturity amount at the end of 12 years .

13. A company purchases an office copier machine for `50,000. It is estimated that the copier depreciates in its value at a rate of 45% per year. What will be the value of the copier after 15 years?

Example 2.16 Find the sum of the arithmetic series 5 11 17 95g+ + + + .

Exercise 2.4 1. Find the sum of the first (each subdivision carries two marks) (i) 75 positive integers (ii) 125 natural numbers.

2. Find the sum of the first 30 terms of an A.P. whose nth term is n3 2+ .

3. Find the sum of each arithmetic series (each subdivision carries two marks) (i) 38 35 32 2g+ + + + . (ii) 6 5 4 25

41

21 g+ + + terms.

4. Find the Sn for the following arithmetic series described.

(each subdivision carries two marks) (i) ,a 5= ,n 30= l 121= (ii) ,a 50= ,n 25= d 4=-

7. In the arithmetic sequence 60, 56, 52, 48,g , starting from the first term, how many terms are needed so that their sum is 368?

13. A construction company will be penalised each day for delay in construction of a bridge. The penalty will be `4000 for the first day and will increase by `1000 for each following day. Based on its budget, the company can afford to pay a maximum of `1,65,000 towards penalty. Find the maximum number of days by which the completion of work can be delayed.

14. A sum of `1000 is deposited every year at 8% simple interest. Calculate the interest at the end of each year. Do these interest amounts form an A.P.? If so, find the total interest at the end of 30 years.


15. The sum of first n terms of a certain series is given as 3 2 .n n2- Show that the series

is an arithmetic series. 16. If a clock strikes once at 1 o’clock, twice at 2 o’clock and so on, how many times will

it strike in a day?

Example 2.22 Find the sum of the first 25 terms of the geometric series 16 48 144 432 g- + - + .

Example 2.23 Find Sn for each of the geometric series described below:

(each subdivision carries two marks) (i) ,a 2= t

6 = 486, n = 6 (ii) a = 2400, r = – 3, n = 5

Example 2.28 An organisation plans to plant saplings in 25 streets in a town in such a way that one sapling for the first street, two for the second, four for the third, eight for the fourth street and so on. How many saplings are needed to complete the work?

Exercise 2.5

1. Find the sum of the first 20 terms of the geometric series 25

65

185 g+ + + .

2. Find the sum of the first 27 terms of the geometric series 91

271

811 g+ + + .

3. Find Sn for each of the geometric series described below. (each subdivision carries two

marks) (i) ,a 3= 384,t8= n 8= . (ii) ,a 5= r 3= , n 12= .

5. How many consecutive terms starting from the first term of the series (each subdivision carries two marks)

(i) 3 9 27 g+ + + would sum to 1092 ? (ii) 2 6 18 g+ + + would sum to 728 ?

6. The second term of a geometric series is 3 and the common ratio is .54 Find the sum

of first 23 consecutive terms in the given geometric series.

Example 2.29 Find the sum of the following series (each subdivision carries two marks)(i) 26 27 28 60g+ + + + (iii) 31 33 53.g+ + +

Example 2.30 Find the sum of the following series (i) 1 2 3 252 2 2 2

g+ + + +

Example 2.31 Find the sum of the series. (i) 1 2 3 203 3 3 3

g+ + + +

Example 2.33 (ii) If 1 2 3 36100,n3 3 3 3

g+ + + + = then find 1 2 3 .ng+ + + +

Exercise 2.6 1. Find the sum of the following series. (each question carries two marks)

(i) 1 + 2 + 3 + g + 45 (iii) 2 + 4 + 6 + g + 100 (iv) 7 + 14 +21 g + 490 2. Find the value of k if (each question carries two marks) (i) 1 2 3 6084k

3 3 3 3g+ + + + = , (ii) 1 2 3 2025k

3 3 3 3g+ + + + =

3. If 1 2 3 171pg+ + + + = , then find 1 2 3 p3 3 3 3

g+ + + + .


Five Mark Questions

Exercise 2.2 7. If 9th term of an A.P. is zero, prove that its 29th term is double (twice) the 19th term.

8. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term.

9. Find n so that the nth terms of the following two A.P.’s are the same.

1, 7, 13, 19,g and 100, 95, 90, g .

11. A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year and in the 15th year.

13. The sum of three consecutive terms in an A.P. is 6 and their product is –120. Find the three numbers.

14. Find the three consecutive terms in an A. P. whose sum is 18 and the sum of their squares is 140.

15. If m times the mth term of an A.P. is equal to n times its nth term, then show that the ( )m n+ th term of the A.P. is zero.

19. If , ,a b c2 2 2 are in A.P. then show that , ,

b c c a a b1 1 1+ + +

are also in A.P.

20. If , , ,a b c x y z0 0 0x y z

! ! != = and b ac2= , then show that , ,

x y z1 1 1 are in A.P.

Example 2.11 The 4th term of a geometric sequence is 32 and the seventh term is

8116 . Find

the geometric sequence.

Example 2.12 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture initially, how many bacteria will be present at the end of 14th hour?

Example 2.13 An amount `500 is deposited in a bank which pays annual interest at the rate of 10% compounded annually. What will be the value of this deposit at the end of 10th year?

Example 2.14 The sum of first three terms of a geometric sequence is 1213 and their product

is -1. Find the common ratio and the terms.

Example 2.15 If , , ,a b c d are in geometric sequence, then prove that b c c a d b a d2 2 2 2- + - + - = -^ ^ ^ ^h h h h

Exercise 2.3

6. If the geometric sequences 162, 54, 18,g . and , , ,812

272

92 g have their nth term

equal, find the value of n.

8. The sum of three terms of a geometric sequence is 1039 and their product is 1. Find the

common ratio and the terms.


9. If the product of three consecutive terms in G.P. is 216 and sum of their products in pairs is 156, find them.

10. Find the first three consecutive terms in G.P. whose sum is 7 and the sum of their reciprocals is

47

11. The sum of the first three terms of a G.P. is 13 and sum of their squares is 91. Determine the G.P.

14. If , , ,a b c d are in a geometric sequence, then show that.a b c b c d ab bc cd- + + + = + +^ ^h h

15. If , , ,a b c d are in a G.P., then prove that , , ,a b b c c d+ + + are also in G.P.

Example 2.17 Find the sum of the first 2n terms of the following series. 1 2 3 4 ...2 2 2 2- + - + .

Example 2.18 In an arithmetic series, the sum of first 14 terms is 203- and the sum of the next 11 terms is –572. Find the arithmetic series.

Example 2.19 How many terms of the arithmetic series 24 21 18 15 g+ + + + , be taken continuously so that their sum is – 351.

Example 2.20 Find the sum of all 3 digit natural numbers, which are divisible by 8.

Example 2.21 The measures of the interior angles taken in order of a polygon form an arithmetic sequence. The least measurement in the sequence is 85c. The greatest measurement is 215c. Find the number of sides in the given polygon.

Exercise 2.4

5. Find the sum of the first 40 terms of the series 1 2 3 42 2 2 2

g- + - + .

6. In an arithmetic series, the sum of first 11 terms is 44 and that of the next 11 terms is 55. Find the arithmetic series.

8. Find the sum of all 3 digit natural numbers, which are divisible by 9.

9. Find the sum of first 20 terms of the arithmetic series in which 3rd term is 7 and 7th term is 2 more than three times its 3rd term.

10. Find the sum of all natural numbers between 300 and 500 which are divisible by 11.

11. Solve: 1 6 11 16 148xg+ + + + + = .

12. Find the sum of all numbers between 100 and 200 which are not divisible by 5.

17. Show that the sum of an arithmetic series whose first term is a , second term b and the

last term is c is equal to b a

a c b c a

2

2

-

+ + -

^^ ^

hh h .


18. If there are n2 1+^ h terms in an arithmetic series, then prove that the ratio of the sum of odd terms to the sum of even terms is :n n1+^ h .

19. The ratio of the sums of first m and first n terms of an arithmetic series is :m n2 2

show that the ratio of the mth and nth terms is :m n2 1 2 1- -^ ^h h

20. A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of 97 bricks. Each row must be decreased by 2 bricks on each end and the construction should stop at 25th row. How many bricks does he need to buy?

Example 2.24 In the geometric series 2 4 8 g+ + + , starting from the first term how many consecutive terms are needed to yield the sum 1022?

Example 2.25 The first term of a geometric series is 375 and the fourth term is 192. Find the common ratio and the sum of the first 14 terms.

Example 2.26 A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 8 and the sum of the last two terms is 72. Find the series.

Example 2.27 Find the sum to n terms of the series 6 + 66 + 666 +g

Exercise 2.5 4. Find the sum of the following finite series (each subdivision carries five marks)

(i) 1 0.1 0.01 0.001 .0 1 9g+ + + + + ^ h (ii) 1 11 111 g+ + + to 20 terms.

7. A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 9 and sum of the last two terms is 36. Find the series.

8. Find the sum of first n terms of the series (each subdivision carries five marks)

(i) 7 77 777 g+ + + . (ii) 0.4 0.94 0.994 g+ + + .

9. Suppose that five people are ill during the first week of an epidemic and each sick person spreads the contagious disease to four other people by the end of the second week and so on. By the end of 15th week, how many people will be affected by the epidemic?

10. A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have 1000 mangoes at once or he could get 1 mango on the first day, 2 on the second day, 4 on the third day, 8 mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes?

11. A geometric series consists of even number of terms. The sum of all terms is 3 times the sum of odd terms. Find the common ratio.

12. If ,S S Sand1 2 3

are the sum of first n, 2n and 3n terms of a geometric series respectively, then prove that S S S S S

1 3 2 2 12- = -^ ^h h .


Example 2.30 Find the sum of the following series (each subdivision carries five marks) (ii) 12 13 14 35

2 2 2 2g+ + + +

(iii) 1 3 5 512 2 2 2

g+ + + + .

Example 2.31 Find the sum of the series. (ii) 11 12 13 283 3 3 3

g+ + + +

Example 2.32 Find the value of k, if 1 2 3 k3 3 3 3

g+ + + + = 4356

Example 2.34 Find the total area of 14 squares whose sides are 11 cm, 12 cm, g , 24 cm, respectively.

Exercise 2.6 1. Find the sum of the following series. (each question carries five marks)

(ii) 16 17 18 252 2 2 2

g+ + + +

(v) 5 7 9 392 2 2 2

g+ + + +

(vi) 16 17 353 3 3

g+ + +

4. If 1 2 3 8281k3 3 3 3

g+ + + + = , then find 1 2 3 kg+ + + + .

5. Find the total area of 12 squares whose sides are 12 cm, 13cm, g , 23cm. respectively.

6. Find the total volume of 15 cubes whose edges are 16 cm, 17 cm, 18 cm, g , 30 cm respectively.

International Mathematical Olympaid (IMO) is the World Championship Mathematics Competition for high school students.

The first IMO was held in Romania in 1959. It has since been held annually. About 100 countries send teams of upto 6 students. Contestants have to solve 6 problems on branches of mathematics not conventionally covered at school. The participants are ranked based on their individual scores.

Mathematical Olympiad activity in our country is being undertaken by the National Board for Higher Mathematics (NBHM). Its main purpose is to support mathematical talent among high school students in the country.

NBHM trains the Indian team for participation in the IMO every year. All High School students upto class XII are eligible to appear for Olympiad.

For information and resources visit the website:http:\\www.imo-official.org

Classification of Questions - Algebra 299

3. ALGEBRA

Two Mark Questions

Example 3.1 Solve x y3 5- = –16 , x y2 5+ = 31

Example 3.2 The cost of 11 pencils and 3 erasers is ` 50 and the cost of 8 pencils and 3 erasers is ` 38. Find the cost of each pencil and each eraser.

Example 3.3 Solve by elimination method 3x y4+ = –25, x y2 3- = 6.

Exercise 3.1 Solve each of the following system of equations by elimination method.


1. x y2 7+ = , x y2 1- = 2. x y3 8+ = , x y5 10+ =

3. xy2

4+ = , x y3

2 5+ = 4. x y xy11 7- = , x y xy9 4 6- =

Example 3.6 Solve 2x + 7y – 5= 0; –3x + 8y = –11

Example 3.7 Using cross multiplication method, solve 3x + 5y = 25; 7x + 6y = 30.

Exercise 3.2 1. Solve the following systems of equations using cross multiplication method.


(i) x y3 4 24+ = , x y20 11 47- = (ii) . . .x y0 5 0 8 0 44+ = , . . .x y0 8 0 6 0 5+ =

(iii) ,x y x y23

35

23 2 6

13- =- + =

Example 3.11 Find the zeros of the quadratic polynomial 9 20x x2+ + , and verify the basic

relationships between the zeros and the coefficients.

Example 3.12 Find a quadratic polynomial if the sum and product of zeros of it are –4 and 3 respectively.

Example 3.13 Find a quadratic polynomial with zeros at x41= and x 1=- .

Exercise 3.3 1. Find the zeros of the following quadratic polynomials and verify the basic relationships

between the zeros and the coefficients. (each subdivision carries two marks)

(i) 2 8x x2- - (ii) 4 4 1x x

2- + (iii) 6 3 7x x

2- - (iv) 4 8x x

2+

(v) 15x2- (vi) 3 5 2x x

2- + (vii) 2 2 1x x2

2- + (viii) 2 143x x

2+ -


2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively. (each subdivision carries two marks)

(i) 3, 1 (ii) 2, 4 (iii) 0, 4 (iv) ,251

(v) ,31 1 (vi) ,

21 4- (vii) ,

31

31- (viii) ,3 2

Example 3.14 Find the quotient and remainder when 7 3x x x3 2+ - - is divided by x 3- .

Exercise 3.4 1. Find the quotient and remainder using synthetic division.


(i) ( 3 5x x x3 2+ - + ) ' ( x 1- ), (ii) (3 2 7 5x x x

3 2- + - ) ' ( x 3+ )

(iii) (3 4 10 6x x x3 2+ - + )' ( x3 2- ), (iv) (3 4 5x x

3 2- - ) ' ( 1x3 + )

(v) (8 2 6 5x x x4 2- + - )' ( 1x4 + ), (vi) (2 7 13 63 48x x x x

4 3 2- - + - )' ( 1x2 - )

Example 3.16 (each subdivision carries two marks)

(i) Prove that x 1- is a factor of 6 11 6x x x3 2- + - .

(ii) Prove that x 1+ is a factor of 6 11 6x x x3 2+ + + .

Example 3.19 Find the GCD of the following : (ii) 15x y z4 3 5 , 12x y z

2 7 2

Exercise 3.6 2. Find the GCD of the following (each subdivision carries two marks)

(i) c d2 2- , c c d-^ h (ii) 27x a x

4 3- , x a3 2-^ h

(iii) 3 18m m2- - , 5 6m m

2+ + (iv) 14 33x x

2+ + , 10 11x x x

3 2+ -

(v) 3 2x xy y2 2+ + , 5 6x xy y

2 2+ + (vi) 2 1x x

2- - , 4 8 3x x

2+ +

(x) a a1 35 2- +^ ^h h , a a a2 1 32 3 4- - +^ ^ ^h h h

Example 3.22 Find the LCM of the following. (each subdivision carries two marks)

(ii) 35a c b2 3 , 42a cb

3 2 , 30ac b2 3

(iii) a a1 35 2- +^ ^h h , a a a2 1 32 3 4- - +^ ^ ^h h h

Exercise 3.7 Find the LCM of the following. (each subdivision carries two marks)

4. 66a b c4 2 3 , 44a b c

3 4 2 , 24a b c2 3 4 6. x y xy

2 2+ , x xy

2+

7. a3 1-^ h, 2 a 1 2-^ h , a 12-^ h 9. x x4 32 3+ -^ ^h h , x x x1 4 3 2- + -^ ^ ^h h h


Exercise 3.8 1. Find the LCM of each pair of the following polynomials.

(i) 5 6x x2- + , 4 12x x

2+ - whose GCD is x 2- .

2. Find the other polynomial q x^ h of each of the following, given that LCM and GCD and one polynomial p x^ h respectively. (each subdivision carries two marks)

(i) x x1 22 2+ +^ ^h h , x x1 2+ +^ ^h h, x x1 22+ +^ ^h h.

(ii) x x4 5 3 73 3+ -^ ^h h , x x4 5 3 7 2+ -^ ^h h , x x4 5 3 73 2+ -^ ^h h .

Example 3.25 Simplify the rational expressions into lowest forms.

( each subdivision carries two marks )

(i) xx

7 285 20

++ , (ii)

x x

x x

3 2

53 4

3 2

+

- , (iii) x x

x x

9 12 5

6 5 12

2

+ -

- + , (iv) x x x

x x x

1 2 3

3 5 42

2

- - -

- - +

^ ^

^ ^

h h

h h

Exercise 3.9 Simplify the following into their lowest forms. (each subdivision carries two marks)

(i) x x

x x

3 12

6 92

2

-

+ (ii) x

x

1

14

2

-

+ (iii) x x

x

1

12

3

+ +

-

(iv) x

x

9

272

3

-

- (v) x x

x x

1

12

4 2

+ +

+ + (Hint: 1x x4 2+ + = x x1

2 2 2+ -^ h )

(vi) x x

x

4 16

84 2

3

+ +

+ (vii) x x

x x

2 5 3

2 32

2

+ +

+ - (viii) x x

x

9 2 6

2 1622

4

+ -

-^ ^h h

(ix) x x x

x x x

4 2 3

3 5 42

2

- - -

- - +

^ ^

^ ^

h h

h h (x)

x x x

x x x

10 13 40

8 5 502

2

+ - +

- + -

^ ^

^ ^

h h

h h (xi)

x x

x x

8 6 5

4 9 52

2

+ -

+ +

(xii) x x x

x x x x

7 3 2

1 2 9 142

2

- - +

- - - +

^ ^

^ ^ ^

h h

h h h

Example 3.26 Multiply (ii) a ab b

a b

22 2

3 3

+ +

+ by a ba b2 2

--

Example 3.27 Divide (each subdivision carries two marks)

(i) x

x

1

4 42-

- by xx

11

+- (ii)

xx

31

3

+- by

xx x3 9

12

++ +



(i) x

x xxx

22

23 6

2

#+-

-+ , (ii)

x

x

x x

x x

4

81

5 36

6 82

2

2

2

#-

-

- -

+ + , (iii) x x

x x

x

x x

20

3 10

8

2 42

2

3

2

#- -

- -

+

- +


2. Divide the following and write your answer in lowest terms.


(i) x

x

x

x1 1

2

2

'+ -

, (ii) x

xxx

49

3676

2

2

'-

-++

Example 3.28 Simplify (each subdivision carries two marks)

(i) xx

xx

32

21

++ +

-- (ii)

xx

11

2-+

^ h +

x 11+

Example 3.29 What rational expression should be added to x

x

2

12

3

+

- to get x

x x

2

2 32

3 2

+

- + ?

Exercise 3.11 1. Simplify the following as a quotient of two polynomials in the simplest form.


(i) x

xx2 2

83

-+

- (ii)

x x

x

x x

x

3 2

2

2 3

32 2+ +

+ +- -

-

(iii) x

x x

x x

x x

9

6

12

2 242

2

2

2

-

- - +- -

+ - (iv) x x

x

x x

x

7 10

2

2 15

32 2- +

- +- -

+

(v) x x

x x

x x

x x

3 2

2 5 3

2 3 2

2 7 42

2

2

2

- +

- + -- -

- -

2. Which rational expression should be added to x

x

2

12

3

+

- to get x

x x

2

3 2 42

3 2

+

+ + ?

3. Which rational expression should be subtracted from x

x x2 1

4 7 53 2

-- + to get

x x2 5 12- + ?

Example 3.31 Find the square root of (iii) (2 3 ) 24x y xy2

+ -

Example 3.32 Find the square root of (ii) 2xx

16

6+ -

Exercise 3.12 1. Find the square root of the following (each subdivision carries two marks)

(iii) 44x x11 2+ -^ h (iv) 4x y xy2- +^ h

(v) 121x y8 6 ' 81x y

4 8 (vi) x y a b b c

a b x y b c

25

644 6 10

4 8 6

+ - +

+ - -

^ ^ ^

^ ^ ^

h h h

h h h

2. Find the square root of the following: (each subdivisions carries two marks)

(i) 16 24 9x x2- + (iii) 4 9 25 12 30 20x y z xy yz zx

2 2 2+ + - + -

(iv) 2xx

14

4+ +


Example 3.36 Solve 6 5 25x x2- - = 0.

Exercise 3.14

Solve the following quadratic equations by factorization method.


(i) 81x2 3 2+ -^ h = 0, (ii) 3 5 12x x2- - = 0, (iii) 3x x5 2 5

2+ - = 0,

(v) xx

3 8- = 2, (vi) xx1+ =

526 , (viii) 1a b x a b x

2 2 2 2 2- + +^ h = 0,

(ix) 2 5x x1 12+ - +^ ^h h = 12, (x) 3 5x x4 42- - -^ ^h h = 12.

Exercise 3.15



(i) 7 12x x2- + = 0, (ii) 15 11 2x x

2- + = 0, (iii) x

x1+ = 2

21 ,

(iv) 3 2a x abx b2 2 2

- - = 0

Example 3.42 The sum of a number and its reciprocal is 551 . Find the number.

Example 3.45 Determine the nature of roots of the following quadratic equations


(i) 11 10 0x x2- - = , (ii) 4 28 49 0x x

2- + = , (iii) 2 5 5 0x x

2+ + =

Exercise 3.17 1. Determine the nature of the roots of the equation. (each subdivision carries two marks)

(i) 8 12 0x x2- + = , (ii), 2 3 4 0x x

2- + = , (iii) 9 12 4 0x x

2+ + = ,

(iv) 3 2 2 0x x62- + = , (v) 1 0x x

53

322

- + = , (vi) x a x b ab2 2 4- - =^ ^h h

2. Find the values of k for which the roots are real and equal in each of the following equations. (each subdivision carries two marks)

(i) 2 10 0x x k2- + = , (ii) 12 4 3 0x kx

2+ + = .

3. Show that the roots of the equation 2 2 0x a b x a b2 2 2+ + + + =^ ^h h are unreal.

4. Show that the roots of the equation 3 2 0p x pqx q2 2 2

- + = are not real.

Example 3.48 If one of the roots of the equation 3 10 0x x k2- + = is

31 , then find the

other root and also the value of k.

Example 3.49 If the sum and product of the roots of the quadratic equation 5 0ax x c2- + =

are both equal to 10, then find the values of a and c.


Example 3.50 If a and b are the roots of the equation 2 3 1 0x x2- - = , find the values of

(ii) ba

ab

+ , (iii) a b- if >a b , (iv) 2 2

ba

ab

+e o,

(v) 1 1ab a

b+ +c `m j (each subdivision carries two marks )

Example 3.51 Form the quadratic equation whose roots are 7 3+ and 7 3- .

Exercise 3.18

1. Find the sum and the product of the roots of the following equations.

(i) 6 5 0x x2- + = , (ii) 0kx rx pk

2+ + = ,

(iii) 3 5 0x x2- = , (iv) 8 25 0x

2- = . (each subdivision carries two marks)

2. Form a quadratic equation whose roots are

(i) 3 , 4 (ii) 3 7+ , 3 7- (iii) ,2

4 72

4 7+ -


3. If a and b are the roots of the equation 3 5 2x x2- + = 0 , then find the values of

(i) ba

ab

+ (ii) a b- (iii) 2 2

ba

ab

+


4. If a and b are the roots of the equation 3 6 4x x2- + = 0, find the value of 2 2

a b+ .

Five Mark Questions

Example 3.4 Using elimination method, solve x y101 99+ = 499, x y99 101+ = 501.

Example 3.5 Solve x y3 2 +^ h = xy7 ; x y3 3+^ h = xy11 using elimination method .

Exercise 3.1

Solve each of the following system of equations by elimination method.

(each question carries Five marks)

5. x y xy3 5 20+ = ,

x y xy2 5 15+ = , 0,x y 0! !

6. x y xy8 3 5- = , x y xy6 5 2- =-

7. x y13 11 70+ = , x y11 13 74+ =

8. x y65 33 97- = , x y33 65 1- =

9. x y15 2 17+ = , , ,

x yx y1 1

536 0 0! !+ =

10. x y2

32

61+ = , 0, 0, 0

x yx y3 2 ! !+ =


Example 3.8 In a two digit number, the digit in the unit place is twice of the digit in the tenth place. If the digits are reversed, the new number is 27 more than the given number. Find the number.

Example 3.9 A fraction is such that if the numerator is multiplied by 3 and the denominator

is reduced by 3, we get 1118 , but if the numerator is increased by 8 and the

denominator is doubled, we get 52 . Find the fraction.

Example 3.10 Eight men and twelve boys can finish a piece of work in 10 days while six men and eight boys can finish the same work in 14 days. Find the number of days taken by one man alone to complete the work and also one boy alone to complete the work.

Exercise 3.2 1. Solve the following systems of equations using cross multiplication method.

(iv) ,x y x y5 4 2 2 3 13- =- + =

2. Formulate the following problems as a pair of equations, and hence find their solutions: (each subdivision carries five marks)

(i) One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

(ii) The ratio of income of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3. If each of them manages to save ` 2000 per month, find their monthly income.

(iii) A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

(iv) Three chairs and two tables cost 700 and five chairs and three tables cost ` 1100. What is the total cost of 2 chairs and 3 tables?

(v) In a rectangle, if the length is increased and the breadth is reduced each by 2 cm then the area is reduced by 28 cm2 . If the length is reduced by 1 cm and the breadth increased by 2 cm , then the area increases by 33 cm2 . Find the area of the rectangle.

(vi) A train travelled a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken 4 hours less than the scheduled time. If the train were slower by 6 km/hr, then it would have taken 6 hours more than the scheduled time. Find the distance covered by the train.

Example 3.15 If the quotient on dividing 2 14 19 6x x x x4 3 2+ - - + by x2 1+ is

6x ax bx3 2+ - - . Find the values of a and b, also the remainder.

Exercise 3.4

2. If the quotient on dividing 10 35 50 29x x x x4 3 2+ + + + by x 4+ is 6x ax bx

3 2- + + ,

then find a, b and also the remainder.

3. If the quotient on dividing, 8 2 6 7x x x4 2- + - by x2 1+ is 4 3x px qx

3 2+ - + ,

then find p , q and also the remainder.


Example 3.17 Factorize 2 3 3 2x x x3 2- - + into linear factors.

Example 3.18 Factorize 3 10 24x x x3 2- - +

Exercise 3.5 1. Factorize each of the following polynomials. (each subdivision carries five marks)

(i) 2 5 6x x x3 2- - + (ii) 4 7 3x x3

- +

(iii) 23 142 120x x x3 2- + - (iv) 4 5 7 6x x x

3 2- + -

(v) 7 6x x3- + (vi) 13 32 20x x x

3 2+ + +

(vii) 2 9 7 6x x x3 2- + + (viii) 5 4x x

3- +

(ix) 10 10x x x3 2- - + (x) 2 11 7 6x x x3 2

+ - -

(xi) 14x x x3 2+ + - (xii) 5 2 24x x x

3 2- - +

Examples 3.19 Find the GCD of (iii) 6 x x2 3 22- -^ h, 8 x x4 4 1

2+ +^ h, 12 x x2 7 3

2+ +^ h

Example 3.20 Find the GCD of the polynomials 3 3x x x4 3+ - - and 5 3x x x

3 2+ - + .

Example 3.21 Find the GCD of the following polynomials

3 6 12 24x x x x4 3 2+ - - and 4 14 8 8x x x x

4 3 2+ + - .

Exercise 3.6 2. Find the GCD of the following (each subdivision carries five marks)

(vii) 2x x2- - , 6x x

2+ - , 3 13 14x x

2- +

(viii) 1x x x3 2- + - , 1x

4-

(ix) 24 x x x6 24 3 2- -^ h, 20 x x x2 3

6 5 4+ +^ h

3. Find the GCD of the following pairs of polynomials using division algorithm. (each subdivision carries five marks)

(i) 9 23 15x x x3 2- + - , 4 16 12x x

2- +

(ii) 3 18 33 18x x x3 2+ + + , 3 13 10x x

2+ +

(iii) 2 2 2 2x x x3 2+ + + , 6 12 6 12x x x

3 2+ + +

(iv) 3 4 12x x x3 2- + - , 4 4x x x x

4 3 2+ + +

Example 3.22 Find the LCM of the following. (iv) x y3 3+ , x y

3 3- , x x y y

4 2 2 4+ +

Exercise 3.7Find the LCM of the following. (each subdivision carries five marks)

8. 2 18x y2 2- , 5 15x y xy

2 2+ , 27x y

3 3+

10. 10 x xy y9 62 2+ +^ h, 12 x xy y3 5 2

2 2- -^ h, 14 x x6 2

4 3+^ h.


Example 3.23 The GCD of 3 5 26 56x x x x4 3 2+ + + + and 2 4 28x x x x

4 3 2+ - - + is

5 7x x2+ + . Find their LCM.

Example 3.24 The GCD and LCM of two polynomials are x 1+ and 1x6- respectively. If

one of the polynomials is 1x3+ , find the other.

Exercise 3.8 1. Find the LCM of each pair of the following polynomials. (each subdivision carries five marks)

(ii) 3 6 5 3x x x x4 3 2+ + + + , 2 2x x x

4 2+ + + whose GCD is 1x x

2+ + .

(iii) 2 15 2 35x x x3 2+ + - , 8 4 21x x x

3 2+ + - whose GCD is x 7+ .

(iv) 2 3 9 5x x x3 2- - + , 2 10 11 8x x x x

4 3 2- - - + whose GCD is x2 1- .

2. Find the other polynomial q x^ h of each of the following, given that LCM and GCD and one polynomial p x^ h respectively. (each subdivision carries five marks)

(iii) x y x x y y4 4 4 2 2 4- + +^ ^h h, x y

2 2- , x y

4 4- .

(iv) x x x4 5 13- +^ ^h h, x x5

2+^ h, x x x5 9 2

3 2- -^ h.

(v) x x x x1 2 3 32

- - - +^ ^ ^h h h, x 1-^ h, x x x4 6 33 2- + -^ h.

(vi) 2 x x1 42

+ -^ ^h h, x 1+^ h, x x1 2+ -^ ^h h.

Example 3.26 (iii) Multiply x

x

4

82

3

-

- by x x

x x

2 4

6 82

2

+ +

+ +

Example 3.27 (iii) Divide x

x

25

12

2

-

- by x x

x x

4 5

4 52

2

+ -

- -


(each subdivision carries five marks)

(iv) 4 16x x

xxx

x xx x

3 216

644

2 8

2

2

2

3

2

2# #- +-

+-

- -- +

(v) x x

x x

x x

x x

2

3 2 1

3 5 2

2 3 22

2

2

2

#- -

+ -

+ -

- -

(vi) x x

x

x x

x x

x x

x

2 4

2 1

2 5 3

8

2

32 2

4

2# #+ +

-

+ -

-

-

+

2. Divide the following and write your answer in lowest terms. (each subdivision carries five marks)

(iii) x

x x

x x

x x

25

4 5

7 10

3 102

2

2

2

'-

- -

+ +

- - (iv) x x

x x

x x

x x

4 77

11 28

2 15

7 122

2

2

2

'- -

+ +

- -

+ +


(v) x x

x x

x x

x x

3 10

2 13 15

4 4

2 62

2

2

2

'+ -

+ +

- +

- - (vi) x

x x

x x

x

9 16

3 4

3 2 1

4 42

2

2

2

'-

- -

- -

-

(vii) x x

x x

x x

x x

2 9 9

2 5 3

2 3

2 12

2

2

2

'+ +

+ -

+ -

+ -

Example 3.28 Simplify (iii) x

x x

x x

x x

9

6

12

2 242

2

2

2

-

- - +- -

+ -

Example 3.30 Simplify xx

xx

xx

12 1

2 11

12

-- -

++ +

++c m as a quotient of two polynomials in the

simplest form.

Exercise 3.11 1. Simplify the following as a quotient of two polynomials in the simplest form.


(vi) x x

x

x x

x x

6 8

4

20

11 302

2

2

2

+ +

- -- -

- +

(vii) xx

x

xxx

12 5

1

11

3 22

2

++ +

-

+ ---` j= G

(viii) x x x x x x3 2

15 61

4 32

2 2 2+ +

++ +

-+ +

.

4. If P = x yx+

, Q = x yy

+, then find

P Q P Q

Q1 22 2-

--

.

Example 3.32 Find the square root of (iii) x x x x x x6 2 3 5 2 2 12 2 2- - - + - -^ ^ ^h h h

Exercise 3.12

2. Find the square root of the following: (each subdivision carries five marks)

(ii) x x x x x25 8 15 2 152 2 2- + + - -^ ^ ^h h h

(v) x x x x x x6 5 6 6 2 4 8 32 2 2+ - - - + +^ ^ ^h h h

(vi) x x x x x x2 5 2 3 5 2 6 12 2 2- + - - - -^ ^ ^h h h

Example 3.33 Find the square root of 10 37 60 36x x x x4 3 2- + - + .

Example 3.34 Find the square root of x x x x6 19 30 254 3 2- + - +

Example 3.35 If 28 12 9m nx x x x2 3 4

- + + + is a perfect square,

then find the values of m and n.


Exercise 3.13

1. Find the square root of the following polynomials by division method.


(i) 4 10 12 9x x x x4 3 2- + - + (ii) 4 8 8 4 1x x x x

4 3 2+ + + +

(iii) 9 6 7 2 1x x x x4 3 2- + - + (iv) 4 25 12 24 16x x x x

2 3 4+ - - +

2. Find the values of a and b if the following polynomials are perfect squares.


(i) 4 12 37x x x ax b4 3 2- + + + (ii) 4 10x x x ax b

4 3 2- + - +

(iii) 109ax bx x x60 364 3 2+ + - + (iv) 40 24 36ax bx x x

4 3 2- + + +

Example 3.37 Solve x x x x7 216

6 9

1

9

12 2-

-- +

+-

= 0

Example 3.38 Solve x24 10- = x3 4- , x3 4 0>-

Exercise 3.14 Solve the following quadratic equations by factorization method.

(each subdivision carries fivc marks)

(iv) 3 x 62-^ h = x x 7 3+ -^ h (vii)

xx

xx

11

++ + =

1534

Example 3.39 Solve the quadratic equation 5 6 2x x2- - = 0 by completing the square.

Example 3.40 Solve the equation 3 2a x abx b2 2 2

- + = 0 by completing the square

Example 3.41 Solve the equation x x11

22

++

+ =

x 44+

, where x 1 0!+ , x 2 0!+

and x 4 0!+ using quadratic formula.

Exercise 3.15

1 Solve the following quadratic equations by completing the square .


(i) 6 7x x2+ - = 0 (ii) 3 1x x

2+ + = 0

(iii) 2 5 3x x2+ - = 0 (iv) 4 4x bx a b

2 2 2+ - -^ h = 0

(v) x x3 1 32- + +^ h = 0 (vi)

xx

15 7-+ = x3 2+




(v) a x 12+^ h = x a 1

2+^ h (vi) 36 12x ax a b

2 2 2- + -^ h = 0

(vii) xx

xx

11

43

+- +

-- =

310 (viii) a x a b x b

2 2 2 2 2+ - -^ h = 0

Example 3.43 The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq. cm, then find its base and altitude.

Example 3.44 A car left 30 minutes later than the scheduled time. In order to reach its destination 150 km away in time, it has to increase its speed by 25 km/hr from its usual speed. Find its usual speed.

Exercise 3.16

1. The sum of a number and its reciprocal is 865 . Find the number.

2. The difference of the squares of two positive numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.

3. A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences the sides of the rectangular garden letting his house compound wall act as the fourth side fence. Find the dimension of the garden.

4. A rectangular field is 20 m long and 14 m wide. There is a path of equal width all around it having an area of 111 sq. metres. Find the width of the path on the outside.

5. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

6. The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and return downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream.

7. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

8. A chess board contains 64 equal squares and the area of each square is 6.25 cm2 . A border around the board is 2 cm wide. Find the length of the side of the chess board.

9. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time that B would take to finish this work by himself.

10. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after two hours, they are 50 km apart, find the average speed of each train.

Example 3.46 Prove that the roots of the equation 2 0a b c x a b x a b c

2- + + - + - - =^ ^ ^h h h are rational numbers for all

real numbers a and b and for all rational c.


Example 3.47 Find the values of k so that the equation 2 7 0x x k k1 3 3 22- + + + =^ ^h h

has real and equal roots.

Exercise 3.17 2. Find the values of k for which the roots are real and equal in each of the following

equations. (each subdivision carries five marks)

(iii) 5 0x k x2 22+ - + =^ h (iv) 2 1 0k x k x1 1

2+ - - + =^ ^h h

5. If the roots of the equation 2 0a b x ac bd x c d2 2 2 2 2+ - + + + =^ ^h h ,

where , , are non zeroanda b c c , then prove that ba

dc= .

6. Show that the roots of the equation

x a x b x b x c x c x a 0- - + - - + - - =^ ^ ^ ^ ^ ^h h h h h h are always real and they cannot be equal unless a b c= = .

7. If the equation 2 0m x mcx c a12 2 2 2

+ + + - =^ h has equal roots, then prove that c a m12 2 2= +^ h.

Example 3.50 If a and b are the roots of the equation 2 3 1 0x x2- - = , find the values of

(all the three subdivisions carry five marks)

(i) 2 2a b+ (vi) 4 4

a b+ (vii) 3 3

ba

ab

+

Example 3.52 If anda b are the roots of the equation 3 4 1x x2- + = 0, form a quadratic

equation whose roots are 2

ba and

2

ab .

Exercise 3.18

5. If a , b are the roots of 2 3 5x x2- - = 0, form a equation whose roots are 2

a and 2b .

6. If a , b are the roots of 3 2x x2- + = 0, form a quadratic equation whose roots are a-

and b- .

7. If a and b are the roots of 3 1x x2- - = 0, then form a quadratic equation whose roots

are 1 1and2 2a b

.

8. If a and b are the roots of the equation 3 6 1x x2- + = 0, form an equation whose

roots are (each subdivision carries five marks)

(i) ,1 1a b

(ii) ,2 2a b b a (iii) 2 , 2a b b a+ +

9. Find a quadratic equation whose roots are the reciprocal of the roots of the equation 4 3 1x x

2- - = 0.

10. If one root of the equation 3 81x kx2+ - = 0 is the square of the other, find k.

11. If one root of the equation x ax2 64 02- + = is twice the other, then find the value of a

12. If a and b are the roots of 5 1x px2- + = 0 and a b- = 1, then find p.


4. MATRICES

Two Mark Questions

Example 4.1 The table shows a five-day forecastindicating high (H) and low (L)temperatures inFahrenheit.Organisethe temperatures in a matrix wherethe first and second rows representthe High and Low temperaturesrespectivelyandidentifywhichdaywillbethewarmest?

Example 4.2 Theamountoffat,carbohydrateandproteiningramspresentineachfooditemrespectivelyareasfollows:

Item 1 Item 2 Item 3 Item 4

Fat 5 0 1 10Carbohydrate 0 15 6 9

Protein 7 1 2 8

Usetheinformationtowrite3 4# and4 3# matrices.

Example 4.3 LetA a

1

6

3

9

4

2

7

2

8

5

0

1

ij= =

- -

J

L

KKKKK

N

P

OOOOO

6 @ .Find

(i)theorderofthematrix(ii)theelementsa13anda

42(iii)thepositionoftheelement2.

Example 4.4 Constructa2 3# matrixA aij

= 6 @whoseelementsaregivenbya i j2 3ij= -

Example 4.5 IfA 8

1

5

3

2

4=

-e o,thenfindAT and( )A

T T

Exercise 4.1 1. Theratesfortheentranceticketsatawaterthemeparkarelistedbelow:

WeekDaysrates(`)

WeekEndrates(`)

Adult 400 500Children 200 250SeniorCitizen 300 400

Writedownthematricesfortheratesofentranceticketsforadults,childrenandseniorcitizens.Alsofindthedimensionsofthematrices.

Classification of Questions - Matrices 313

2. Thereare6HigherSecondarySchools,8HighSchoolsand13PrimarySchoolsinatown.Representthesedataintheformof3 1# and1 3# matrices.

4. Amatrixhas8elements.Whatarethepossibleordersitcanhave?

5. Amatrixconsistsof30elements.Whatarethepossibleordersitcanhave?.

6. Constructa2 2# matrixA aij

= 6 @whoseelementsaregivenby (each subdivision carries two marks)

(i)a ijij

= (ii) 2a i jij= - (iii)a

i ji j

ij=

+-

7. Constructa3 2# matrixA aij

= 6 @whoseelementsaregivenby (each subdivisiion carries two marks)

(i)aji

ij= (ii) ( )

ai j

22

ij

2

=- (iii)a i j

2

2 3ij=

-

8. If A

1

5

6

1

4

0

3

7

9

2

4

8

=

-

-f p, (i) find the order of the matrix (ii) write down the

elements a24

and a32 (iii) inwhichrowandcolumndoestheelement7occur?

( combination of any two carries two marks [ex.(i,ii),(ii,iii),(i,iii)])

9. IfA2

4

5

3

1

0

= f p,thenfindthetransposeofA.

10. IfA1

2

3

2

4

5

3

5

6

=

-

-f p,thenverifythat( )A AT T

= .

Example 4.6 Findthevaluesofx,yandz if x

y

z

5

5

9

4

1

3

5

5

1=c cm m

Example4.7 Solve: y

x

x

y3

6 2

31 4=

-

+c em o

Example4.10 IfA 5

1

6

0

2

4

3

2=

-c mandB 3

2

1

8

4

2

7

3=

-c m,thenfindA + B.

Example 4.11 MatrixAshowstheweightoffourboysandfourgirlsinkgatthebeginningof a diet programme to loseweight. MatrixB shows the correspondingweightsafterthedietprogramme.

,A B35

42

40

38

28

41

45

30

32

40

35

30

27

34

41

27

BoysGirls

BoysGirls= =c cm m

FindtheweightlossoftheBoysandGirls.


Exercise 4.2 1. Findthevaluesofx,yandzfromthematrixequation

x y

z

5 2

0

4

4 6

12

0

8

2

+ -

+=

-e co m

2. Solveforxandyif x y

x y

2

3

5

13

+

-=e co m

3. IfA 2

9

3

5

1

7

5

1=

--

-e eo o,thenfindtheadditiveinverseofA.

4. LetA 3

5

2

1= c mandB 8

4

1

3=

-c m.FindthematrixCifC A B2= + .

5. IfA B4

5

2

9

8

1

2

3and=

-

-=

- -e eo ofind A B6 3- .

6. Findaandbifa b2

3

1

1

10

5+

-=c c cm m m.

9. If ,A B O3

5

2

1

1

2

2

3

0

0

0

0and= =

-=c c cm m m then


verify:(i)A B B A+ = + (ii) ( ) ( )A A O A A+ - = = - + .

Example 4.13 Solve x

y

3

4

2

5

8

13=c c cm m m

Example 4.17 If A 1

9

3

6=

-e o,thenverify AI IA A= = ,whereIistheunitmatrixof

order2.

Example 4.18 Prove that 3

1

5

2

2

1

5

3and

-

-c em o aremultiplicative inverses to each

other.

Exercise 4.3 2. Findtheproductofthematrices,ifexists,(each subdivision carries two marks)

(i) 2 15

4-^ ch m

(ii) 3

5

2

1

4

2

1

7

-c cm m

(iii) 2

4

9

1

3

0

4

6

2

2

7

1-

--

-

e fo p

(iv) 6

32 7

--e ^o h

Classification of Questions - Matrices 315

3. Afruitvendorsellsfruitsfromhisshop.SellingpricesofApple,MangoandOrangeare`20,`10and`5eachrespectively.Thesalesinthreedaysaregivenbelow

Day Apples Mangoes Oranges1 50 60 302 40 70 203 60 40 10

Writethematrixindicatingthetotalamountcollectedoneachdayandhencefindthetotalamountcollectedfromsellingofallthreefruitscombined.

4. Findthevaluesofxandyif x

y

x1

3

2

3 0

0

9

0

0=c c cm m m.

5. If ,A Xx

yC

5

7

3

5

5

11and= = =

-

-c c em m oandifAX C= ,thenfindthevaluesofx

andy.

10. ProvethatA B5

7

2

3

3

7

2

5and= =

-

-c em oareinversestoeachotherundermatrix

multiplication.

Five Mark Questions

Exercise 4.2

7. FindXandYif2 3X Y2

4

3

0+ = c mand3 2X Y

2

1

2

5+ =

-

-e o.

8. Solveforxandyif 3x

y

x

y

2 9

4

2

2 +-

=-

e e co o m.

10. If ,A B

4

1

0

1

2

3

2

3

2

2

6

2

0

2

4

4

8

6

= - =f fp pandC1

5

1

2

0

1

3

2

1

=

-

-

f p,then

verifythat ( ) ( )A B C A B C+ + = + + .

11. An electronic company records each type of entertainment device sold at three oftheirbranchstoressothattheycanmonitortheirpurchasesofsupplies.Thesalesintwoweeksareshowninthefollowingspreadsheets.

T.V. DVD Videogames CD Players

Week IStoreI 30 15 12 10StoreII 40 20 15 15StoreIII 25 18 10 12

Week IIStoreI 25 12 8 6StoreII 32 10 10 12StoreIII 22 15 8 10

Findthesumoftheitemssoldoutintwoweeksusingmatrixaddition.


12. Thefeesstructureforone-dayadmissiontoaswimmingpoolisasfollows:

Daily Admission Fees in `Member Children Adult

Before2.00p.m. 20 30After2.00p.m. 30 40

Non-MemberBefore2.00p.m. 25 35After2.00p.m. 40 50

Writethematrixthatrepresentstheadditionalcostfornon-membership.

Example 4.14 IfA a

c

b

dI

1

0

0

1and

2= =c cm m,thenshowthat ( ) ( )A a d A bc ad I

2

2- + = -

.

Example 4.15 IfA B

8

2

0

7

4

3

9

6

3

1

2

5and= -

-

=-

- -f ep o,thenfindABandBAiftheyexist.

Example 4.16 If ,A B C3

1

2

4

2

6

5

7

1

5

1

3and=

-=

-=

-e c eo m o

verifythat ( )A B C AB AC+ = +

Example 4.19 IfA B

2

4

5

1 3 6and=

-

= -f ^p h,thenverifythat( )AB B AT T T= .

Exercise 4.3

6. IfA 1

2

1

3=


2

2- + = .

7. IfA B3

4

2

0

3

3

0

2and= =c cm mthenfindABandBA.Aretheyequal?

8. If ,A B C1

1

2

2

1

3

0

1

2

2 1and=-

= =c f ^m p hverify( ) ( )AB C A BC= .

9. IfA B5

7

2

3

2

1

1

1and= =

-

-c em overifythat( )AB B A

T T T= .

11. Solve 0xx

11

2

0

3 5- -=^ e c ^h o m h.

12. IfA B1

2

4

3

1

3

6

2and=

-

-=

-

-e eo o,thenprovethat( ) 2A B A AB B

2 2 2!+ + + .

13. If ,A B C3

7

3

6

8

0

7

9

2

4

3

6and= = =


Is ( )A B C AC BC+ = + ?

Classification of Questions - Coordinate Geometry 317

5. COORDINATE GEOMETRY

Two Mark Questions

Example 5.2 Find the point which divides the line segment joining the points (3 , 5) and (8 , 10) internally in the ratio 2 : 3.

Example 5.5 Find the centroid of the triangle whose vertices are A(4, -6), B(3,-2) and C(5, 2).

Example 5.6 If , , , ,7 3 6 1^ ^h h ,8 2^ h and ,p 4^ h are the vertices of a parallelogram taken in order, then find the value of p.

Exercise 5.1 2. Find the centroid of the triangle whose vertices are (each subdivision carries two marks)

(i) , , , ,1 3 2 7 12 16and -^ ^ ^h h h (ii) , , , ,3 5 7 4 10 2and- - -^ ^ ^h h h

3. The centre of a circle is at (-6, 4). If one end of a diameter of the circle is at the origin, then find the other end.

4. If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then find the third vertex of the triangle .

6. Find the coordinates of the point which divides the line segment joining (3, 4) and (–6, 2) in the ratio 3 : 2 externally.

7. Find the coordinates of the point which divides the line segment joining (-3, 5) and (4, -9) in the ratio 1 : 6 internally.

Example 5.8 Find the area of the triangle whose vertices are (1, 2), (-3 , 4), and (-5 ,-6).

Example 5.9 If the area of the ABCT is 68 sq.units and the vertices are A(6 ,7), B(-4 , 1) and C(a , –9) taken in order, then find the value of a.

Example 5.10 Show that the points A(2 , 3), B(4 , 0) and C(6, -3) are collinear.

Example 5.11 If P ,x y^ h is any point on the line segment joining the points ,a 0^ h and , b0^ h then, prove that

ax

by

1+ = , where a, b 0! .

Exercise 5.2

1. Find the area of the triangle formed by the points (each subdivision carries two marks)

(i) (0, 0), (3, 0) and (0, 2) (ii) (5, 2), (3, -5) and (-5, -1)

(iii) (-4, -5), (4, 5) and (-1, -6)


2. Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a. (each subdivision carries two marks)

Vertices Area (in sq. units)

(i) ( , )0 0 , (4, a), (6, 4) 17

(ii) (a, a), (4, 5), (6,-1) 9

(iii) (a, -3), (3, a), (-1,5) 12

3. Determine if the following set of points are collinear or not. (each subdivision carries two marks)

(i) (4, 3), (1, 2) and (-2, 1) (ii) (-2, -2), (-6, -2) and (-2, 2)

(iii) 23 ,3-` j,(6, -2) and (-3, 4)

4. In each of the following, find the value of k for which the given points are collinear. (each subdivision carries two marks)

(i) (k, -1), (2, 1) and (4, 5) (ii) , , , ,and k2 5 3 4 9- -^ ^ ^h h h

(iii) , , , ,andk k 2 3 4 1-^ ^ ^h h h

6. If the three points , , ( , ) ,h a b k0 0and^ ^h h lie on a straight line, then using the area of the triangle formula, show that 1, , 0

ha

kb h kwhere !+ = .

Exercise 5.3

4. Find the angle of inclination of the line passing through the points (each subdivision carries two marks)

(i) ,1 2^ h and ,2 3^ h (ii) ,3 3^ h and ,0 0^ h (iii) (a , b) and (-a , -b)

5. Find the slope of the line which passes through the origin and the midpoint of the line segment joining the points ,0 4-^ h and (8 , 0).

7. The side BC of an equilateral 3ABC is parallel to x-axis. Find the slope of AB and the slope of BC.

Example 5.19 Find the equations of the straight lines parallel to the coordinate axes and passing through the point ,3 4-^ h.

Example 5.20 Find the equation of straight line whose angle of inclination is 45c and

y-intercept is 52 .

Example 5.21 Find the equation of the straight line passing through the point ,2 3-^ h with slope

31 .

Example 5.22 Find the equation of the straight line passing through the points ,1 1-^ h and ,2 4-^ h.

Example 5.24 If the x-intercept and y-intercept of a straight line are 32 and

43 respectively,

then find the equation of the straight line.


Exercise 5.4 2. Find the equations of the straight lines parallel to the coordinate axes and passing

through the point (-5,-2).

3. Find the equation of a straight line whose (each subdivision carries two marks)

(i) slope is -3 and y-intercept is 4.

(ii) angle of inclination is 600 and y-intercept is 3.

4. Find the equation of the line intersecting the y- axis at a distance of 3 units above the origin and tan

21i = , where i is the angle of inclination.

5. Find the slope and y-intercept of the line whose equation is (each subdivision carries two marks)

(i) y x 1= + (ii) x y5 3= (iii) x y4 2 1 0- + = (iv) x y10 15 6 0+ + =

6. Find the equation of the straight line whose (each subdivision carries two marks)

(i) slope is -4 and passing through (1, 2)

(ii) slope is 32 and passing through (5, -4)

7. Find the equation of the straight line which passes through the midpoint of the line segment joining (4, 2) and (3, 1) whose angle of inclination is 300 .

8. Find the equation of the straight line passing through the points (each subdivision carries two marks)

(i) (-2, 5) and (3, 6) (ii) (0, -6) and (-8, 2)

11. Find the equation of the straight line whose x and y-intercepts on the axes are given by (each subdivision carries two marks)

(i) 2 and 3 (ii) 31- and

23 (iii)

52 and

43-

12. Find the x and y intercepts of the straight line (each subdivision carries two marks)

(i) x y5 3 15 0+ - = (ii) 2 1 0x y 6- + = (iii) x y3 10 4 0+ + =

Example 5.26 Show that the straight lines x y3 2 12 0+ - = and x y6 4 8 0+ + = are parallel.

Example 5.27 Prove that the straight lines x y2 1 0+ + = and x y2 5 0- + = are perpendicular to each other.

Example 5.28 Find the equation of the straight line parallel to the line x y8 13 0- + = and passing through the point (2, 5).


Exercise 5.5 2. Show that the straight lines x y2 1 0+ + = and x y3 6 2 0+ + = are parallel.

3. Show that the straight lines x y3 5 7 0- + = and x y15 9 4 0+ + = are perpendicular.

4. If the straight lines 5 3y

x p ax y2

and= - + = are parallel, then find a .

5. Find the value of a if the straight lines x y5 2 9 0- - = and 11 0ay x2+ - = are perpendicular to each other.

8. Find the equation of the straight line parallel to the line x y3 7 0- + = and passing through the point (1, -2).

9. Find the equation of the straight line perpendicular to the straight line x y2 3 0- + = and passing through the point (1, -2).

Five Mark Questions

Example 5.3 In what ratio does the point P(-2 , 3) divide the line segment joining the points A(-3, 5) and B ( 4, -9) internally?

Example 5.4 Find the points of trisection of the line segment joining ,4 1-^ h and ,2 3- -^ h

Example 5.7 If C is the midpoint of the line segment joining A(4 , 0) and B(0 , 6) and if O is the origin, then show that C is equidistant from all the vertices of 3OAB.

Exercise 5.1 5. Using the section formula, show that the points A(1,0), B(5,3), C(2,7) and

D(-2, 4) are the vertices of a parallelogram taken in order.

8. Let A (-6,-5) and B (-6, 4) be two points such that a point P on the line AB satisfies

AP = 92 AB. Find the point P.

9. Find the points of trisection of the line segment joining the points A(2,- 2) and B(-7, 4).

10. Find the points which divide the line segment joining A(-4 ,0) and B (0,6) into four equal parts.

11. Find the ratio in which the x-axis divides the line segment joining the points (6, 4) and (1,-7).

12. In what ratio is the line joining the points (-5, 1) and (2 , 3) divided by the y-axis? Also, find the point of intersection .

13. Find the length of the medians of the triangle whose vertices are (1, -1) , (0, 4) and (-5, 3).

Example 5.12 Find the area of the quadrilateral formed by the points(-4, -2), (-3, -5), (3, -2) and (2 , 3).


Exercise 5.2 5. Find the area of the quadrilateral whose vertices are


(i) , , , , , ,and6 9 7 4 4 2 3 7^ ^ ^ ^h h h h (ii) , , , , , ,and3 4 5 6 4 1 1 2- - - -^ ^ ^ ^h h h h

(iii) , , , , , ,and4 5 0 7 5 5 4 2- - - -^ ^ ^ ^h h h h

7. Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are , , , ,and0 1 2 1 0 3-^ ^ ^h h h. Find the ratio of this area to the area of the given triangle.

Example 5.16 Using the concept of slope, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

Example 5.17 Using the concept of slope, show that the points (-2 , -1), (4 , 0), (3 , 3) and (-3 , 2) taken in order form a parallelogram.

Example 5.18 The vertices of a 3ABC are A(1 , 2), B(-4 , 5) and C(0 , 1). Find the slopes of the altitudes of the triangle.

Exercise 5.3

8. Using the concept of slope, show that each of the following set of points are collinear.

(i) (2 , 3), (3 , -1) and (4 , -5) (ii) (4 , 1), (-2 , -3) and (-5 , -5) (iii) (4 , 4), (-2 , 6) and (1 , 5)

9. If the points (a, 1), (1, 2) and (0, b+1) are collinear, then show that a b1 1+ = 1.

10. The line joining the points A(-2 , 3) and B(a , 5) is parallel to the line joining the points C(0 , 5) and D(-2 , 1). Find the value of a.

11. The line joining the points A(0, 5) and B(4, 2) is perpendicular to the line joining the points C(-1, -2) and D(5, b). Find the value of b.

12. The vertices of 3ABC are A(1, 8), B(-2, 4), C(8, -5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN is parallel to BC.

13. A triangle has vertices at (6 , 7), (2 , -9) and (-4 , 1). Find the slopes of its medians.

14. The vertices of a 3ABC are A(-5 , 7), B(-4 , -5) and C(4 , 5). Find the slopes of the altitudes of the triangle.

15. Using the concept of slope, show that the vertices (1 , 2), (-2 , 2), (-4 , -3) and (-1, -3) taken in order form a parallelogram.

16. Show that the opposite sides of a quadrilateral with vertices A(-2 ,-4), B(5 , -1), C(6 , 4) and D(-1, 1) taken in order are parallel.


Example 5.23 The vertices of a 3ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.

Example 5.25 Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5.

Exercise 5.4

9. Find the equation of the median from the vertex R in a 3PQR with vertices at P(1, -3), Q(-2, 5) and R(-3, 4).

10. By using the concept of the equation of the straight line, prove that the given three points are collinear. (each subdivision carries five marks)

(i) (4, 2), (7, 5) and (9, 7) (ii) (1, 4), (3, -2) and (-3, 16)

13. Find the equation of the straight line passing through the point (3, 4) and has intercepts which are in the ratio 3 : 2.

14. Find the equation of the straight lines passing through the point (2, 2) and the sum of the intercepts is 9.

15. Find the equation of the straight line passing through the point (5, -3) and whose intercepts on the axes are equal in magnitude but opposite in sign.

16. Find the equation of the line passing through the point (9, -1) and having its x-intercept thrice as its y-intercept.

17. A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2), then find the equation of AB.

18. Find the equation of the line passing through (22, -6) and having intercept on x-axis exceeds the intercept on y-axis by 5.

19. If A(3, 6) and C(-1, 2) are two vertices of a rhombus ABCD, then find the equation of straight line that lies along the diagonal BD.

20. Find the equation of the line whose gradient is 23 and which passes through P, where P

divides the line segment joining A(-2, 6) and B (3, -4) in the ratio 2 : 3 internally.

Example 5.29 The vertices of ABC3 are A(2, 1), B(6, –1) and C(4, 11). Find the equation of the straight line along the altitude from the vertex A.

Exercise 5.5

6. Find the values of p for which the straight lines px p y8 2 3 1 0+ - + =^ h and px y8 7 0+ - = are perpendicular to each other.

7. If the straight line passing through the points ,h 3^ h and (4, 1) intersects the line x y7 9 19 0- - = at right angle, then find the value of h .



11. Find the equation of the straight line passing through the point of intersection of the lines x y2 3 0+ - = and x y5 6 0+ - = and parallel to the line joining the points (1, 2) and (2, 1).

12. Find the equation of the straight line which passes through the point of intersection of the straight lines x y5 6 1- = and x y3 2 5 0+ + = and is perpendicular to the straight line x y3 5 11 0- + = .

13. Find the equation of the straight line joining the point of intersection of the lines 3 0x y 9- + = and x y2 4+ = and the point of intersection of the lines 2 0x y 4+ - = and x y2 3 0- + = .

14. If the vertices of a 3ABC are A(2, -4), B(3, 3) and C(-1, 5). Find the equation of the straight line along the altitude from the vertex B.

15. If the vertices of a 3ABC are A(-4,4 ), B(8 ,4) and C(8,10). Find the equation of the straight line along the median from the vertex A.

16. Find the coordinates of the foot of the perpendicular from the origin on the straight line x y3 2 13+ = .

17. If 2 7x y+ = and x y2 8+ = are the equations of the lines of two diameters of a circle, find the radius of the circle if the point (0, -2) lies on the circle.

18. Find the equation of the straight line segment whose end points are the point of intersection of the straight lines x y2 3 4 0- + = , x y2 3 0- + = and the midpoint of the line joining the points (3, -2) and (-5, 8).

19. In an isosceles 3PQR, PQ = PR. The base QR lies on the x-axis, P lies on the y- axis and x y2 3 9 0- + = is the equation of PQ. Find the equation of the straight line along PR.

Proof by picture

Let us illustrate the result :31

3

1

3

121

2 3g+ + + = with the following diagram


6. GEOMETRY

Two Mark Questions

Example 6.1 In ABC3 , DE BC< and DBAD

32= . If AE = 3.7 cm, find EC.

Example 6.2 In PQR3 , given that S is a point on PQ such that

ST QR< and SQPS

53= . If PR = 5.6 cm, then find PT.

Example 6.5 In ABC3 , the internal bisector AD of A+ meets the side BC at D. If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, then find DC.

Example 6.6 In ABC3 , AE is the external bisector of A+ , meeting BC produced at E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then find CE.

Exercise 6.1 1. In a ABCD , D and E are points on the sides AB and AC respectively such that DE BC< .


(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, then find AC.

(ii) If AD = 8 cm, AB = 12 cm and AE = 12 cm, then find CE.

3. E and F are points on the sides PQ and PR respectively, of a PQR3 . For each of the following cases, verify EF QR< . (each subdivision carries two marks)

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.

9. In a ABC3 , AD is the internal bisector of A+ , meeting BC at D. (each subdivision carries two marks)

(i) If BD = 2 cm, AB = 5 cm, DC = 3 cm find AC.

(ii) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm find BC.

(iii) If AB = x, AC = x–2, BD = x+2 and DC = x–1 find the value of x.

10. Check whether AD is the bisector of A+ of ABC3 in each of the following. (each subdivision carries two marks)

(i) AB = 4 cm, AC = 6 cm, BD = 1.6 cm, and CD = 2.4 cm.

(ii) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 3 cm.

11. In a MNO3 , MP is the external bisector of M+ meeting NO produced at P. If MN = 10 cm, MO = 6 cm, NO = 12 cm, then find OP.

Example 6.8 In PQRT , AB ;;QR. If AB is 3 cm, PB is 2cm and PR is 6 cm, then find the length of QR.

Theorem 6.5 State Pythagoras theorem (Baudhayan theorem)

Classification of Questions - Geometry 325

Theorem 6.6 State Converse of Pythagoras theorem.

Theorem 6.7 State Tangent-Chord theorem.

Theorem 6.8 State the Converse of Tangent-Chord theorem.

Example 6.12 Let PQ be a tangent to a circle at A and AB be

a chord. Let C be a point on the circle such that

54 62 . .BAC BAQ ABCand Find+ + += =c c .

Example 6.13 Find the value of x in each of the following diagrams. (each subdivision carries two marks)

(i) (ii)

Exercise 6.3 1. In the figure TP is a tangent to a circle. A and B are two points on the

circle. If +BTP = 72c and +ATB = 43c find +ABT.

2. AB and CD are two chords of a circle which intersect each other internally at P. (each subdivision carries two marks)

(i) If CP = 4 cm, AP = 8 cm, PB = 2 cm, then find PD.

(ii) If AP = 12 cm, AB = 15 cm, CP = PD, then find CD

3. AB and CD are two chords of a circle which intersect each other externally at P (each subdivision carries two marks)

(i) If AB = 4 cm BP = 5 cm and PD = 3 cm, then find CD.

(ii) If BP = 3 cm, CP = 6 cm and CD = 2 cm, then find AB

Five Mark Questions

Theorem 6.1 State and prove the Basic Proportionality (Thales) Theorem (or)

If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio - prove.

Theorem 6.2 State and prove the converse of Basic Proportionality (Thales) Theorem (or) If a straight line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side - prove.

A

BC

P Q62c

62c

54c

A B

T P27 c

43c


Theorem 6.3 State and prove the Angle Bisector Theorem (or)

The internal (external) bisector of an angle of a triangle divides the opposite side internally (externally) in the ratio of the corresponding sides containing the angle - prove.

Theorem 6.4 State and Prove the Converse of Angle Bisector Theorem (or)

If a straight line through one vertex of a triangle divides the opposite side internally (externally) in the ratio of the other two sides, then the line bisects the angle internally (externally) at the vertex.

Example 6.3 In a ABC3 , D and E are points on AB and AC respectively such that

DBAD

ECAE= and ADE DEA+ += . Prove that ABC3 is isosceles.

Example 6.4 The points D, E and F are taken on the sides AB, BC and CA of a ABC3

respectively, such that DE AC< and FE AB< .Prove that ADAB = FC

AC

Example 6.7 D is the midpoint of the side BC of ABC3 . If P and Q are points on AB and on AC such that DP bisects BDA+ and DQ bisects ADC+ , then prove that PQ BC< .

Exercise 6.1 1. In a ABCD , D and E are points on the sides AB and AC respectively such that DE BC< .

(iii) If AD = 4x–3, BD = 3x–1 , AE = 8x–7 and EC = 5x–3, then find the value of x.

2. In the figure, AP = 3 cm, AR = 4.5cm, AQ = 6cm, AB = 5 cm and AC = 10 cm. Find the length of AD.

4. In the figure, AC BD< and CE DF< . If OA =12cm, AB = 9 cm, OC = 8 cm and EF = 4.5 cm , then find FO.

5. ABCD is a quadrilateral with AB parallel to CD. A line drawn parallel to AB meets AD

at P and BC at Q. Prove that PDAP

QCBQ

= .

6. In the figure, PC QK< and BC HK< . If AQ = 6 cm, QH = 4 cm, HP = 5 cm, KC = 18cm, then find AK and PB.

A

P Q

BCD

R

AE

B

F

D

O

C

A

K

CB

PH

Q


7. In the figure, DE AQ< and DF AR< Prove that EF QR< .

8. In the figure DE AB< andDF AC< . Prove that EF BC< .

12. In a quadrilateral ABCD, the bisectors of B+ and D+ intersect on AC at E. Prove that BC

ABDCAD= .

13. The internal bisector of A+ of ABCT meets BC at D and the external bisector of A+

meets BC produced at E. Prove that BEBD

CECD= .

14. ABCD is a quadrilateral with AB =AD. If AE and AF are internal bisectors of BAC+ and DAC+ respectively, then prove that EF BD< .

Example 6.9 A man of height 1.8 m is standing near a Pyramid. If the shadow of the man is of length 2.7 m and the shadow of the Pyramid is 210 m long at that instant, find the height of the Pyramid.

Example 6.10 A man sees the top of a tower in a mirror which is at a distance of 87.6 m from the tower. The mirror is on the ground, facing upward. The man is 0.4 m away from the mirror, and the distance of his eye level from the ground is 1.5 m. How tall is the tower? (The foot of man, the mirror and the foot of the tower lie along a straight line).

Example 6.11 The image of a tree on the film of a camera is of length 35 mm, the distance from the lens to the film is 42 mm and the distance from the lens to the tree is 6 m. How tall is the portion of the tree being photographed?

FPE

D

A

CB

P

E F

R

A

Q

D


Exercise 6.2 1. Find the unknown values in each of the following figures. All lengths are given in

centimetres. (All measures are not in scale) (each subdivision carries five marks)

(i) (ii) (iii)

2. The image of a man of height 1.8 m, is of length 1.5 cm on the film of a camera. If the film is 3 cm from the lens of the camera, how far is the man from the camera?

3. A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 0.6 m/sec. If the lamp is 3.6 m above the ground level, then find the length of her shadow after 4 seconds.

5. P and Q are points on sides AB and AC respectively, of ABCD . If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.

6. In ABCD , AB = AC and BC = 6 cm. D is a point on the side AC such that AD = 5 cm and CD = 4 cm. Show that BCD ACB+D D and hence find BD.

7. The points D and E are on the sides AB and AC of ABCD respectively, such thatDE || BC. If AB = 3 AD and the area of ABCD is 72 cm2, then find the area of the quadrilateral DBCE.

8. The lengths of three sides of a triangle ABC are 6 cm, 4 cm and 9 cm. PQR ABC3 3+. One of the lengths of sides of 3PQR is 35cm. What is the greatest perimeter possible for 3PQR?

9. In the figure, DE || BC and BDAD

53= , calculate the value of

(i) ABCADE

area ofarea of

DD , (ii)

ABC

BCED

area ofarea of trapezium

D

10. The government plans to develop a new industrial zone in an unused portion of land in a city.

The shaded portion of the map shown on the right, indicates the area of the new industrial zone.

Find the area of the new industrial zone.

GA

B FC

D E24

8 6

8

x

y

A

D G

HFE

B C

6

9

53

4

xy

z

A

E

B D C

F

x

y

5 7

6G


11. A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?

12. A student wants to determine the height of a flagpole. He placed a small mirror on the ground so that he can see the reflection of the top of the flagpole. The distance of the mirror from him is 0.5 m and the distance of the flagpole from the mirror is 3 m. If his eyes are 1.5 m above the ground level, then find the height of the flagpole.(The foot of student, mirror and the foot of flagpole lie along a straight line).

13. A roof has a cross section as shown in the diagram,

(i) Identify the similar triangles

(ii) Find the height h of the roof.

Example 6.14 In the figure, tangents PA and PB are drawn to a circle with centre O from an external point P. If CD is a tangent to the circle at E and AP = 15 cm, find the perimeter of TPCD

Example 6.15 ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6 cm, BC = 6.5 cm and CD = 7 cm , then find the length of AD.

Exercise 6.3 4. A circle touches the side BC of TABC at P, AB and AC produced at Q and R respectively,

prove that AQ = AR = 21 ( perimeter of TABC)


6. A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water 40 cm away from the original position of the stem. How much of the stem was below the water surface originally?

7. A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that OA OC OB OD

2 2 2 2+ = + .

D

CA

PE

B

O


7. TRIGONOMETRY

Two Mark Questions

Example 7.1 Prove the identity 1cosecsin

seccos

ii

ii+ =

Example 7.2 Prove the identity coscos cosec cot

11

ii i i

+- = -

Example 7.7 Prove the identity 1 3 .sin cos sin cos6 6 2 2i i i i+ = -^ h

Example 7.8 Prove the identity 2

.cos cos

sin sin tan23

3

i i

i i i-

- =

Example 7.10 Prove that secsec

cossin1

1

2

ii

ii+ =

-.

Exercise 7.1 1. Determine whether each of the following is an identity or not.


(i) 2cos sec2 2i i+ = +sini (ii) cot cos sin

2 2i i i+ =

2. Prove the following identities (each subdivision carries two marks)

(i) sec cosec sec cosec2 2 22

i i i i+ = (ii) cos

sin cosec cot1 i

i i i-

= +

(iii) sinsin sec tan

11

ii i i

+- = - (iv) 1

sec tancos sini ii i

-= +

(v) sec cosec tan cot2 2i i i i+ = + (vi) 1

sin coscos sin cot

1

2

i ii i i+

+ - =^ h

(vii) 1sec sin sec tan1i i i i- + =^ ^h h (viii) cosec cot

sini ii+

= cos1 i-

Example 7.14 Akiteisflyingwithastringoflength200m.Ifthethreadmakesanangle30c withtheground,findthedistanceofthekitefromthegroundlevel.(Here,assume that the string is along a straight line)

Example 7.15 A ladder leaning against a vertical wall, makes an angle of 60c with the ground. The foot of the ladder is 3.5 m away from the wall. Find the length of the ladder.

Example 7.16 Find the angular elevation (angle of elevation from the ground level) of the Sunwhenthelengthoftheshadowofa 30mlongpoleis10 3 m.

Example 7.17 Theangleofelevationofthetopofatowerasseenbyanobserveris30c. The observerisatadistanceof30 3 mfromthetower.If the eye level of the observeris1.5mabovethegroundlevel,thenfindtheheightofthetower.

Classification of Questions - Trigonometry 331

Exercise 7.2

1. Arampforunloadingamovingtruck,hasanangleofelevationof30c.Ifthetopoftherampis0.9mabovethegroundlevel,thenfindthelengthoftheramp.

2. Agirlofheight150cmstandsinfrontofalamp-postandcastsashadowoflength150 3 cmontheground.Findtheangleofelevationofthetopofthelamp-post.

3. SupposetwoinsectsAandBcanheareachotheruptoarangeof2..TheinsectAisontheground1mawayfromawallandseesherfriendBonthewall,abouttobeeatenbyaspider.IfAsoundsawarningtoBandiftheangleofelevationofBfromAis30c,willthespiderhaveamealornot?(AssumethatBescapesifshehearsAcalling)

5. Apendulumoflength40cmsubtends60c at the vertex in one full oscillation. What willbetheshortestdistancebetweentheinitialpositionandthefinalpositionofthebob?

Five Mark Questions

Example 7.3 Prove the identity [ ] [ ]cosec sin cosec sin tan cot90 90 1i i i i i i- - - - + =c c^ ^h h6 @

Example 7.4 Prove that 1tan sectan sec

cossin

11

i ii i

ii

- ++ - = +

Example 7.5 Prove the identity 1cot

tantan

cot tan cot1 1i

iii i i

-+

-= + + .

Example 7.6 Prove the identity 7sin cosec cos sec tan cot2 2 2 2i i i i i i+ + + = + +^ ^h h

Example 7.9 Prove the identity 1 2 2 .sec tansec tan sec tan tan

2

i ii i i i i+- = - +

Example 7.11 Prove the identity cosec sin sec costan cot

1i i i ii i

- - =+

^ ^h h .

Example 7.12 If tan sin mi i+ = , tan sin ni i- = and m n! , then show that

4 .m n mn2 2- =

Example 7.13If ,tan cos sin2 2 2a b b= - thenprovethatcos sin tan2 2 2a a b- = .

Exercise 7.1 3. Prove the following identities. (each subdivision carries five marks)

(i) sin

sin

coscos

1

90

1 90i

i

ii

+

-+

- -

c

c^

^h

h = 2seci

(ii) cot

tantan

cot sec cosec1 1

1ii

ii i i

-+

-= +


(iii) tan

sincot

coscos sin

190

190

0 0

ii

ii

i i-

-+

--

= +^ ^h h

(iv) 902 .

cosectan

cotcosec sec

11

0

ii

ii i

+-

+ + =^ h

(v) .cot coseccot cosec cosec cot

11

i ii i i i- ++ - = +

(vi) 2cot cosec tan sec1 1i i i i+ - + + =^ ^h h

(vii) sin cossin cos

sec tan11 1

i ii i

i i+ -- + =

-

(viii) 1 2 1tan

tan

sin

sin sin

90

900

0

2 2i

i

i

i i

-=

- -

-

^

^

h

h

(ix) cosec cot sin sin cosec cot

1 1 1 1i i i i i i-

- = -+

.

(x) ( )tan cosec

cot sec sin cos tan cot2 2

2 2

i i

i i i i i i+

+ = +^ h.

4. If sec tanx a bi i= + and tan secy a bi i= + ,thenprovethatx y a b2 2 2 2- = - .

5. Iftan tanni a= and ,sin sinmi a= thenprovethat1

1cosn

m2

2

2

i =-

- .

6. If ,sin cos tanandi i i areinG.P.,thenprovethat 1.cot cot6 2i i- =

Example 7.18 Averticaltreeisbrokenbythewind.Thetopofthetreetouchesthegroundand makes an angle 30cwithit.Ifthetopofthetreetouchestheground30mawayfromitsfoot,thenfindtheactualheightofthetree.

Example 7.19 A jet fighter at a height of 3000 m from the ground, passesdirectly over another jet fighter at an instance when theirangles of elevation from the same observation point are 60c and 45crespectively.Findthedistanceofthefirstjetfighterfromthesecondjet at that instant. ( .3 1 732= )

Example 7.20 The angle of elevationofthetopofahillfromthefootofatoweris60c andtheangleofelevationofthetopofthetowerfromthefootofthehillis30c . Ifthetoweris50mhigh,thenfindtheheightofthehill.

Example 7.21 Averticalwallandatowerareontheground.Asseenfromthetopofthetower,theanglesofdepressionofthetopandbottomofthewallare45c and 60crespectively.Findtheheightofthewalliftheheightofthetoweris90m.( 3 .1 732= )

Classification of Questions - Trigonometry 333

Example 7.22 Agirlstandingonalighthousebuiltonacliffneartheseashore,observestwoboatsdueEastofthelighthouse.Theanglesofdepressionofthetwoboatsare30c and 60c.Thedistancebetweentheboatsis300m.Findthedistanceofthetopofthelighthousefromthesealevel.

Example 7.23 Aboyspotsaballoonmovingwiththewindinahorizontallineataheightof88.2 m from the ground level. The distance of his eye level from the ground is 1.2m.Theangleofelevationoftheballoonfromhiseyesataninstantis60 .cAftersometime,fromthesamepointofobservation,theangleofelevationoftheballoonreducesto30c.Findthedistancecoveredbytheballoonduringthe interval.

Example 7.24 Aflagpoststandsonthetopofabuilding.Fromapointontheground,theanglesofelevationof the topandbottomof theflagpostare60c and 45c respectively. If theheightof theflagpost is10 m ,find the height of the building.( 3 .1 732= )

Example 7.25 A man on the deck ofaship,14mabovethewaterlevel,observesthattheangleofelevationofthetopofacliffis60candtheangleofdepressionofthebaseofthecliffis30c Find the height of the cliff.

Example 7.26 TheangleofelevationofanaeroplanefromapointA on the ground is 60 .c After a flight of 15 seconds horizontally, the angle of elevation changes to 30c.Iftheaeroplaneisflyingataspeedof200m/s,thenfindtheconstantheightatwhichtheaeroplaneisflying.

Exercise 7.2

6. Two crows A and B are sitting at a height of 15mand10m in twodifferent treesverticallyoppositetoeachother.Theyviewavadai(aneatable)onthegroundatanangleofdepression45c and 60crespectively.Theystartatthesametimeandflyatthesamespeedalongtheshortestpathtopickupthevadai.Whichbirdwillsucceedinit?

7. Alamp-poststandsatthecentreofacircularpark.LetP and QbetwopointsontheboundarysuchthatPQ subtendsanangle90catthefootofthelamp-postandtheangleofelevationofthetopofthelamppostfromP is 30c.IfPQ =30m,thenfindtheheightofthelamppost.

8. A person in an helicopter flying at a height of 700m, observes two objects lyingopposite to each other on either bank of a river. The angles of depression of theobjectsare30 45andc c. Find the width of the river. ( .3 1 732= )

9. ApersonX standingonahorizontalplane,observesabirdflyingatadistanceof100mfromhimatanangleofelevationof30c. Anotherperson Y standing on theroofofa20mhighbuilding,observes thebirdat thesametimeatanangleof elevation of 45c.IfX and Yareontheoppositesidesofthebird,thenfindthedistanceofthebirdfromY.


10. Astudentsittinginaclassroomseesapictureontheblackboardataheightof1.5mfromthehorizontallevelofsight.Theangleofelevationofthepictureis30 .c Asthepictureisnotcleartohim,hemovesstraighttowardstheblackboardandseesthepictureatanangleofelevationof45c.Findthedistancemovedbythestudent.

11. Aboyisstandingatsomedistancefroma30mtallbuildingandhiseyelevelfromthegroundis1.5m.Theangleofelevationfromhiseyestothetopofthebuildingincreases from 30 60toc cas he walks towards the building. Find the distance hewalkedtowardsthebuilding.

12. Fromthetopofalighthouseofheight200feet,thelighthousekeeperobservesaYachtandaBargealongthesamelineofsight.TheanglesofdepressionfortheYachtandtheBargeare45 and 30c crespectively.Forsafetypurposesthetwoseavesselsshouldbeatleast300feetapart.Iftheyarelessthan300feet,thekeeperhastosoundthealarm.Doesthekeeperhavetosoundthealarm?

13. Aboystandingontheground,spotsaballoonmovingwiththewindinahorizontallineataconstantheight .Theangleofelevationoftheballoonfromtheboyataninstant is 60c. After 2 minutes, from the same point of observation,the angle ofelevation reduces to 30c.Ifthespeedofwindis29 3m/min. then,findtheheightoftheballoonfromthegroundlevel.

14. Astraighthighwayleadstothefootofatower.Amanstandingonthetopofthetowerspotsavanatanangleofdepressionof30c.Thevanisapproachingthetowerwithauniformspeed.After6minutes,theangleofdepressionofthevanisfoundtobe60c.Howmanymoreminuteswillittakeforthevantoreachthetower?

15. Theanglesofelevationofanartificialearthsatelliteismeasuredfromtwoearthstations,situatedonthesamesideofthesatellite,arefoundtobe30cand 60c. The two earth stations andthesatelliteareinthesameverticalplane.Ifthedistancebetweentheearthstationsis4000km,findthedistancebetweenthesatelliteandearth.( .3 1 732= )

16. Fromthetopofatowerofheight60m,theanglesofdepressionofthetopandthebottomofabuildingareobservedtobe30 60andc crespectively.Findtheheightofthebuilding.

17. Fromthetopandfootofa40mhightower,theanglesofelevationofthetopofalighthousearefoundtobe30cand 60crespectively.Findtheheightofthelighthouse.Alsofindthedistanceofthetopofthelighthousefromthefootofthetower.

18. Theangleofelevationofahoveringhelicopterasseenfromapoint45mabovealakeis 30candtheangleofdepressionofitsreflectioninthelake,asseenfromthesamepointandatthesametime,is60c.Findthedistanceofthehelicopterfromthesurfaceof the lake.

Classification of Questions - Mensuration 335

8. MENSURATION

Two Mark Questions

Example 8.1 A solid right circular cylinder has radius 7 cm and height 20 cm. Find its (i) curved surface area and (ii) total surface area. ( Take

722r = )

(each subdivision carrries two marks)

Example 8.6 Radius and slant height of a solid right circular cone are 35 cm and 37 cm respectively. Find the curved surface area and total surface area of the cone. ( Take

722r = )

Example 8.9 A hollow sphere in which a circus motorcyclist performs his stunts, has an inner diameter of 7 m. Find the area available to the motorcyclist for riding. (Take

722r = )

Example 8.10 Total surface area of a solid hemisphere is 675r sq.cm. Find the curved surface area of the solid hemisphere.

Example 8.11 The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.( Take

722r = )

Exercise 8.1

1. A solid right circular cylinder has radius of 14 cm and height of 8 cm . Find its curved surface area and total surface area.


12. If the circumference of the base of a solid right circular cone is 236 cm and its slant height is 12 cm, find its curved surface area.

16. If the curved surface area of solid a sphere is 98.56 cm2, then find the radius of the sphere..

17. If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.

18. Radii of two solid hemispheres are in the ratio 3 : 5. Find the ratio of their curved surface areas and the ratio of their total surface areas.

Example 8.15 The volume of a solid right circular cone is 4928 cu. cm. If its height is 24 cm, then find the radius of the cone. ( Take

722r = )

Example 8.16 The radii of two circular ends of a frustum shaped bucket are 15 cm and 8 cm. If its depth is 63 cm, find the capacity of the bucket in litres. ( Take

722r = )

Example 8.17 Find the volume of a sphere-shaped metallic shot-put having diameter of 8.4 cm. ( Take

722r = )


Example 8.18 A cone, a hemisphere and cylinder have equal bases. If the heights of the cone and a cylinder are equal and are same as the common radius, then find the ratio of their respective volumes.

Example 8.19 If the volume of a solid sphere is 7241 71 cu.cm, then find its radius.

(Take 722r = )

Example 8.20 Volume of a hollow sphere is 7

11352 cm3 . If the outer radius is 8 cm, find

the inner radius of the sphere. ( Take 722r = )

Exercise 8.2 1. Find the volume of a solid cylinder whose radius is 14 cm and height 30 cm.

2. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, then find the quantity of soup to be prepared daily in the hospital to serve 250 patients?

4. Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.


8. A lead pencil is in the shape of right circular cylinder. The pencil is 28 cm long and its radius is 3 mm. If the lead is of radius 1 mm, then find the volume of the wood used in the pencil.

10. The circumference of the base of a 12 cm high wooden solid cone is 44 cm. Find the volume.

13. A right angled 3ABC with sides 5 cm, 12 cm and 13 cm is revolved about the fixed side of 12 cm. Find the volume of the solid generated.

15. The volume of a cone with circular base is 216r cu.cm. If the base radius is 9 cm, then find the height of the cone.

17. The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.

19. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.

20. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.

Five Mark Questions

Example 8.2 If the total surface area of a solid right circular cylinder is 880 sq.cm and its radius is 10 cm, find its curved surface area. ( Take

722r = )

Example 8.3 The ratio between the base radius and the height of a solid right circular cylinder is 2 : 5. If its curved surface area is

73960 sq.cm, find the height and radius.

( use 722r = )


Example 8.4 The diameter of a road roller of length 120 cm is 84 cm. If it takes 500 complete revolutions to level a playground, then find the cost of levelling it at the cost of 75 paise per square metre. (Take

722r = )

Example 8.5 The internal and external radii of a hollow cylinder are 12 cm and 18 cm respectively. If its height is 14 cm, then find its curved surface area and total surface area. (Take

722r = )

Example 8.7 Let O and C be the centre of the base and the vertex of a right circular cone. Let B be any point on the circumference of the base. If the radius of the cone is 6 cm and if OBC 60o+ = , then find the height and curved surface area of the cone.

Example 8.8 A sector containing an angle of 120c is cut off from a circle of radius 21 cm and folded into a cone. Find the curved surface area of the cone. ( Take

722r = )

Exercise 8.1 2. The total surface area of a solid right circular cylinder is 660 sq.cm. If its diameter of

the base is 14 cm, find the height and curved surface area of the cylinder.

3. Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.

4. A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost to paint the lateral surface of the pillars at ` 20 per square metre.

5. The total surface area of a solid right circular cylinder is 231 cm2. Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.

6. The total surface area of a solid right circular cylinder is 1540 cm2. If the height is four times the radius of the base, then find the height of the cylinder.`

8. The outer curved surface area of a hollow cylinder is 540r sq.cm. Its internal diameter is 16 cm and height is 15 cm. Find the total surface area.

9. The external diameter of a cylindrical shaped iron pipe is 25 cm and its length is 20 cm. If the thickness of the pipe is 1cm, find the total surface area of the pipe.

10. The radius and height of a right circular solid cone are 7 cm and 24 cm respectively. Find its curved surface area and total surface area.

11. If the vertical angle and the radius of a right circular cone are 60c and 15 cm respectively, then find its height and slant height.

13. A heap of paddy is in the form of a cone whose diameter is 4.2 m and height is 2.8 m. If the heap is to be covered exactly by a canvas to protect it from rain, then find the area of the canvas needed.

14. The central angle and radius of a sector of a circular disc are 180c and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.

15. Radius and slant height of a solid right circular cone are in the ratio 3 : 5. If the curved surface area is 60r sq.cm, then find its total surface area.


19. Find the curved surface area and total surface area of a hollow hemisphere whose outer and inner radii are 4.2 cm and 2.1 cm respectively.

20. The inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of `5 per sq. m.

Example 8.12 If the curved surface area of a right circular cylinder is 704 sq.cm, and height is 8 cm, find the volume of the cylinder in litres. ( Take

722r = )

Example 8.13 A hollow cylindrical iron pipe is of length 28 cm. Its outer and inner diameters are 8 cm and 6 cm respectively. Find the volume of the pipe and weight of the pipe if 1 cu.cm of iron weighs 7 gm.( Take

722r = )

Example 8.14 Base area and volume of a solid right circular cylinder are 13.86 sq.cm, and 69.3 cu.cm respectively. Find its height and curved surface area.( Take

722r = )

Exercise 8.2 3. The sum of the base radius and the height of a solid right circular solid cylinder is 37 cm.

If the total surface area of the cylinder is 1628 sq.cm, then find the volume of the cylinder.

6. The radius and height of a cylinder are in the ratio 5 : 7. If its volume is 4400 cu.cm, find the radius of the cylinder.

7. A rectangular sheet of metal foil with dimension 66 cm # 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.

9. Radius and slant height of a cone are 20 cm and 29 cm respectively. Find its volume.

11. A vessel is in the form of a frustum of a cone. Its radius at one end and the height are 8 cm and 14 cm respectively. If its volume is

35676 cm3, then find the radius at the other end.

12. The perimeter of the ends of a frustum of a cone are 44 cm and 8.4r cm. If the depth is 14 cm., then find its volume.

14. The radius and height of a right circular cone are in the ratio 2 : 3. Find the slant height if its volume is 100.48 cu.cm. ( Take r = 3.14)

16. Find the mass of 200 steel spherical ball bearings, each of which has radius 0.7 cm, given that the density of steel is 7.95 g/cm3 . (Mass = Volume # Density)

18. The volume of a solid hemisphere is 1152r cu.cm. Find its curved surface area.

Example 8.21 A solid wooden toy is in the form of a cone surmounted on a hemisphere. If the radii of the hemisphere and the base of the cone are 3.5 cm each and the total height of the toy is 17.5 cm, then find the volume of wood used in the toy. ( Take r =

722 )


Example 8.22 A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8 cm and the total height of the cup is 11.5 cm. Find the total surface area of the cup. ( Take r =

722 )

Example 8.23 A circus tent is to be erected in the form of a cone surmounted on a cylinder. The total height of the tent is 49 m. Diameter of the base is 42 m and height of the cylinder is 21 m. Find the cost of canvas needed to make the tent, if the cost of canvas is `12.50/m2 . ( Take r =

722 )

Example 8.24 A hollow sphere of external and internal diameters of 8 cm and 4 cm respectively is melted and made into another solid in the shape of a right circular cone of base diameter of 8 cm. Find the height of the cone.

Example 8.25 Spherical shaped marbles of diameter 1.4 cm each, are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Example 8.26 Water is flowing at the rate of 15 km / hr through a cylindrical pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. In how many hours will the water level in the tank raise by 21 cm? ( Take r =

722 )

Example 8.27 A cuboid shaped slab of iron whose dimensions are 55 cm#40 cm#15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe.(Take r =

722 )

Exercise 8.3 1. A play-top is in the form of a hemisphere surmounted on a cone. The diameter of the

hemisphere is 3.6 cm. The total height of the play-top is 4 . 2 c m. Find its total surface area.

2. A solid is in the shape of a cylinder surmounted on a hemisphere. If the diameter and the total height of the solid are 21 cm, 25.5 cm respectively, then find its volume.

3. A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm, find its surface area.


5. Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere.

6. The radius of a solid sphere is 24 cm. It is melted and drawn into a long wire of uniform cross section. Find the length of the wire if its radius is 1.2 mm.


7. A right circular conical vessel whose internal radius is 5 cm and height is 24 cm is full

of water. The water is emptied into an empty cylindrical vessel with internal radius 10 cm. Find the height of the water level in the cylindrical vessel.

8. A solid sphere of diameter 6 cm is dropped into a right circular cylindrical vessel with diameter 12 cm, which is partly filled with water. If the sphere is completely submerged in water, how much does the water level in the cylindrical vessel increase?.

9. Through a cylindrical pipe of internal radius 7 cm, water flows out at the rate of 5 cm/sec. Calculate the volume of water (in litres) discharged through the pipe in half an hour.

10. Water in a cylindrical tank of diameter 4 m and height 10 m is released through a cylindrical pipe of diameter 10 cm at the rate of 2.5 Km/hr. How much time will it take to empty the half of the tank? Assume that the tank is full of water to begin with.


12. A hollow cylindrical pipe is of length 40 cm. Its internal and external radii are 4 cm and 12 cm respectively. It is melted and cast into a solid cylinder of length 20 cm. Find the radius of the new solid.

13. An iron right circular cone of diameter 8 cm and height 12 cm is melted and recast into spherical lead shots each of radius 4 mm. How many lead shots can be made?.

14. A right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 6 cm, having a hemispherical shape on top. Find the number of such cones which can be filled with the ice cream available.

15. A container with a rectangular base of length 4.4 m and breadth 2 m is used to collect rain water. The height of the water level in the container is 4 cm and the water is transferred into a cylindrical vessel with radius 40 cm. What will be the height of the water level in the cylinder?

16. A cylindrical bucket of height 32 cm and radius 18 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

17. A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out soil is evenly spread to form a cuboid-platform with base dimension 20 m # 14 m. Find the height of the platform.

Classification of Questions - Statistics 341

11. STATISTICS

Two Mark Questions

Example 11.1 Find the range and the coefficient of range of 43, 24, 38, 56, 22, 39, 45.

Example 11.2 The weight (in kg) of 13 students in a class are 42.5, 47.5, 48.6, 50.5, 49, 46.2, 49.8, 45.8, 43.2, 48, 44.7, 46.9, 42.4. Find the range and coefficient of range.

Example 11.3 The largest value in a collection of data is 7.44. If the range is 2.26, then find the smallest value in the collection.

Example 11.11 Find the standard deviation of the first 10 natural numbers.

Example 11.19 The mean of 30 items is 18 and their standard deviation is 3. Find the sum of all the items and also the sum of the squares of all the items.

Exercise 11.1 1. Find the range and coefficient of range of the following data.


(i) 59, 46, 30, 23, 27, 40, 52,35, 29

(ii) 41.2, 33.7, 29.1, 34.5, 25.7, 24.8, 56.5, 12.5

2. The smallest value of a collection of data is 12 and the range is 59. Find the largest value of the collection of data.

3. The largest of 50 measurements is 3.84 kg. If the range is 0.46 kg, find the smallest measurement.

4. The standard deviation of 20 observations is 5 . If each observation is multiplied by 2, find the standard deviation and variance of the resulting observations.

5. Calculate the standard deviation of the first 13 natural numbers.

13. Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the items and the sum of the squares of all the items.

15. If n = 10, x = 12 and x2/ = 1530, then calculate the coefficient of variation .

17. If the coefficient of variation of a collection of data is 57 and its S.D is 6.84, then find the mean.

18. A group of 100 candidates have their average height 163.8 cm with coefficient of variation 3.2. What is the standard deviation of their heights?

Five Mark Questions

Example 11.4 The number of books read by 8 students during a month are 2, 5, 8, 11, 14, 6, 12, 10. Calculate the standard deviation of the data.


Example 11.5 A test in General Knowledge was conducted for a class. The marks out of 40, obtained by 6 students were 20, 14, 16, 30, 21 and 25. Find the standard deviation of the data.

Example 11.6 Find the standard deviation of the numbers 62, 58, 53, 50, 63, 52, 55.

Example 11.7 The marks obtained by 10 students in a test in Mathematics are : 80, 70, 40, 50, 90, 60, 100, 60, 30, 80. Find the standard deviation.

Example 11.8 Find the standard deviation of the data 3, 5, 6, 7. Then add 4 to each item and find the standard deviation of the new data.

Example 11.9 Find the standard deviation of 40, 42 and 48. If each value is multiplied by 3, find the standard deviation of the new data.

xample 11.10 Prove that the standard deviation of the first n natural numbers is

v = n12

12- .

Example 11.12 The following table shows the marks obtained by 48 students in a Quiz competition in Mathematics. Calculate the standard deviation.

Data x 6 7 8 9 10 11 12Frequency f 3 6 9 13 8 5 4

Example 11.13 Find the standard deviation of the following distribution.

x 70 74 78 82 86 90f 1 3 5 7 8 12

Example 11.14 Find the variance of the following distribution.Class interval 3.5-4.5 4.5-5.5 5.5-6.5 6.5-7.5 7.5-8.5Frequency 9 14 22 11 17

Example 11.15 The following table gives the number of goals scored by 71 leading players in International Football matches. Find the standard deviation of the data.

Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70Frequency 8 12 17 14 9 7 4

Example 11.16 Length of 40 bits of wire, correct to the nearest centimetre are given below. Calculate the variance.

Length cm 1-10 11-20 21-30 31-40 41-50 51-60 61-70No. of bits 2 3 8 12 9 5 1

Example 11.17 Find the coefficient of variation of the following data. 18, 20, 15, 12, 25.

Classification of Questions - Statistics 343

Example 11.18 Following are the runs scored by two batsmen in 5 cricket matches.Who is more consistent in scoring runs.

Batsman A 38 47 34 18 33Batsman B 37 35 41 27 35

Example 11.20 The mean and the standard deviation of a group of 20 items was found to be 40 and 15 respectively. While checking it was found that an item 43 was wrongly written as 53. Calculate the correct mean and standard deviation.

Example 11.21 For a collection of data, if x/ = 35, n = 5, x 9 822- =^ h/ , then find x x xand2 2-^ h/ / .

Example 11.22 The coefficient of variations of two series are 58 and 69. Their standard deviations are 21.2 and 15.6. What are their arithmetic means?

Exercise 11.1 6. Calculate the standard deviation of the following data. (each subdivision carries five

marks)

(i) 10, 20, 15, 8, 3, 4 (ii) 38, 40, 34 ,31, 28, 26, 34

7. Calculate the standard deviation of the following data.

x 3 8 13 18 23f 7 10 15 10 8

8. The number of books bought at a book fair by 200 students from a school are given in the following table.

No. of books 0 1 2 3 4No of students 35 64 68 18 15

Calculate the standard deviation.

9. Calculate the variance of the following data

x 2 4 6 8 10 12 14 16f 4 4 5 15 8 5 4 5

10. The time (in seconds) taken by a group of people to walk across a pedestrian crossing is given in the table below.

Time (in sec.) 5-10 10-15 15-20 20-25 25-30No. of people 4 8 15 12 11

Calculate the variance and standard deviation of the data.


11. A group of 45 house owners contributed money towards green environment of their street. The amount of money collected is shown in the table below.

Amount (`) 0-20 20-40 40-60 60-80 80-100No. of house owners 2 7 12 19 5

Calculate the variance and standard deviation.

12. Find the variance of the following distribution

Class interval 20-24 25-29 30-34 35-39 40-44 45-49Frequency 15 25 28 12 12 8

14. The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking it was found that an item 12 was wrongly entered as 8. Calculate the correct mean and standard deviation.

16. Calculate the coefficient of variation of the following data: 20, 18, 32, 24, 26.

19. Given xR = 99 , n = 9 and ( )x 102

R - = 79. Find ( )x x xand2 2R R - r .

20. The marks scored by two students A, B in a class are given below.

A 58 51 60 65 66B 56 87 88 46 43

Who is more consistent?

Proof by pictureLet us illustrate the result : 1 ( )n n3 5 7 2 1

2g+ + + + + - =

with the following diagram

Thus, , , ,1 3 2 1 3 5 3 1 3 5 7 42 2 2

g+ = + + = + + + =

Classification of Questions - Probability 345

12. PROBABILITY

Two Mark Questions

Example 12.3 An integer is chosen from the first twenty natural numbers. What is the probability that it is a prime number?

Example 12.4 There are 7 defective items in a sample of 35 items. Find the probability that an item chosen at random is non-defective.

Example 12.7 There are 20 boys and 15 girls in a class of 35 students . A student is chosen at random. Find the probability that the chosen student is a (i) boy (ii) girl.

Example 12.8 The probability that it will rain on a particular day is 0.76. What is the probability that it will not rain on that day?

Example 12.9 A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of drawing a red ball, then find the number of blue balls in the bag.

Example 12.10 Find the probability that (each subdivision carries two marks)

(i) a leap year selected at random will have 53 Fridays

(ii) a leap year selected at random will have only 52 Fridays

(iii) a non-leap year selected at random will have 53 Fridays.

Example 12.11 If A is an event of a random experiment such that ( ) : ( ) 7 :12P A P A = ,then find P(A).

Exercise 12. 1 1. A ticket is drawn from a bag containing 100 tickets. The tickets are numbered from

one to hundred. What is the probability of getting a ticket with a number divisible by 10?

2. A die is thrown twice. Find the probability of getting a total of 9.

3. Two dice are thrown together. Find the probability that the two digit number formed with the two numbers turning up is divisible by 3.

4. Three rotten eggs are mixed with 12 good ones. One egg is chosen at random. What is the probability of choosing a rotten egg?

5. Two coins are tossed together. What is the probability of getting at most one head.

6. One card is drawn randomly from a well shuffled deck of 52 playing cards. Find the probability that the drawn card is (each subdivision carries two marks)

(i) a Diamond (ii) not a Diamond (iii) not an Ace.


8. A bag contains 6 white balls numbered from 1 to 6 and 4 red balls numbered from

7 to 10. A ball is drawn at random. Find the probability of getting (each subdivision carries two marks)

(i) an even-numbered ball (ii) a white ball.

9. A number is selected at random from integers 1 to 100. Find the probability that it is (each subdivision carries two marks)

(i) a perfect square (ii) not a perfect cube.

11. A box contains 4 Green, 5 Blue and 3 Red balls. A ball is drawn at random. Find the probability that the selected ball is (i) Red in colour (ii) not Green in colour.

12. 20 cards are numbered from 1 to 20. One card is drawn at random. What is the probability that the number on the card is

(i) a multiple of 4 (ii) not a multiple of 6.

13. A two digit number is formed with the digits 3, 5 and 7. Find the probability that the number so formed is greater than 57 (repetition of digits is not allowed).

14. Three dice are thrown simultaneously. Find the probability of getting the same number on all the three dice.

17. A bag consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. A trader A will accept only the shirt which are good, but the trader B will not accept the shirts which have major defects. One shirt is drawn at random. What is the probability that it is acceptable by (i) A (ii) B ?

19. Piggy bank contains 100 fifty-paise coins, 50 one-rupee coins, 20 two-rupees coins and 10 five- rupees coins. One coin is drawn at random. Find the probability that the drawn coin (i) will be a fifty-paise coin (ii) will not be a five-rupees coin.

Example 12.15 A letter is chosen at random from the letters of the word “ENTERTAINMENT”. Find the probability that the chosen letter is a vowel or T. (repetition of letters is allowed)

Exercise 12.2

1. If A and B are mutually exclusive events such that ( ) ( )P A P B53

51and= = , then find

( )P A B, .


52= = and ( )P A B

21, = ,then find

( )P A B+ .

11. A box contains 10 white, 6 red and 10 black balls. A ball is drawn at random. Find the probability that the ball drawn is white or red.


12. A two digit number is formed with the digits 2, 5, 9 (repetition is allowed). Find the probability that the number is divisible by 2 or 5.

13. Each individual letter of the word “ACCOMMODATION” is written in a piece of paper, and all 13 pieces of papers are placed in a jar. If one piece of paper is selected at random from the jar, find the probability that (each subdivision carries two marks) (i) the letter ‘A’ or ‘O’ is selected. (ii) the letter ‘M’ or ‘C’ is selected.

Five Mark Questions

Example 12.1 A fair die is rolled. Find the probability of getting (i) the number 4 (ii) an even number

(iii) a prime factor of 6 (iv) a number greater than 4.

Example 12.2 In tossing a fair coin twice, find the probability of getting

(i) two heads (ii) atleast one head (iii) exactly one tail

Example 12.5 Two unbiased dice are rolled once. Find the probability of getting (i) a sum 8 (ii) a doublet (iii) a sum greater than 8.

Example 12.6 From a well shuffled pack of 52 playing cards, one card is drawn at random. Find the probability of getting

(i) a king (ii) a black king (iii) a spade card (iv) a diamond 10.

Exercise 12. 1 7. Three coins are tossed simultaneously. Find the probability of getting

(i) at least one head (ii) exactly two tails (iii) at least two heads.


16. A jar contains 54 marbles each of which is in one of the colours blue, green and white. The probability of drawing a blue marble is

31 and the probability of drawing a green

marble is 94 . How many white marbles does the jar contain?

18. A bag contains 12 balls out of which x balls are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball. (ii) If 6 more white balls are put in the bag and if the probability of drawing a white ball will be twice that of in (i), then find x.

Example 12.12 Three coins are tossed simultaneously. Using addition theorem on probability, find the probability that either exactly two tails or at least one head turn up.


Example 12.13 A die is thrown twice. Find the probability that at least one of the two throws comes up with the number 5 (use addition theorem).

Example 12.14 The probability that a girl will be selected for admission in a medical college is 0.16. The probability that she will be selected for admission in an engineering college is 0.24 and the probability that she will be selected in both, is 0.11

(i) Find the probability that she will be selected in at least one of the two colleges.

(ii) Find the probability that she will be selected either in a medical college only or in an engineering college only.

Example 12.16 Let A, B, C be any three mutually exclusive and exhaustive events such that ( )P B = ( )P A

23 and ( ) ( )P C P B

21= . Find P(A).

Example 12.17 A card is drawn from a deck of 52 cards. Find the probability of getting a King or a Heart or a Red card.

Example 12.18 A bag contains 10 white, 5 black, 3 green and 2 red balls. One ball is drawn at random. Find the probability that the ball drawn is white or black or green.

Exercise 12.2

3. If ( ) , ( ) , ( ) 1.P A P B P A B21

107 ,= = = Find (i) ( )P A B+ (ii) ( )P A B,l l .

4. If a die is rolled twice, find the probability of getting an even number in the first time or a total of 8 .

5. One number is chosen randomly from the integers 1 to 50. Find the probability that it is divisible by 4 or 6.

6. A bag contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.


8. A basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.


9. In a class, 40% of the students participated in Mathematics-quiz, 30% in Science-quiz and 10% in both the quiz programmes. If a student is selected at random from the class, find the probability that the student participated in Mathematics or Science or both quiz programmes.

10. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that it will be a spade or a king.

14. The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is 0.15. Find the probability that

(i) it will get atleast one of the two awards (ii) it will get only one of the awards.


32

73and respectively.


72 ,


358 .

Find the probability that the problem can be solved by atleast one of them.

Approximations of r

A simple continued fraction for the irrational number r is given by

[3, 7, 15, 1, 292, 1, 1, g ] = 37

151

2921

11

1

g

++

++

+

This includes some interesting approximations for r . (i) [3] = 3, the nearest whole number to r (ii) [3, 7] = 3 + .

71

722 3 142857g= =

This is the value everyone knows from school. It is a good approximation for r .

(iii) [3, 7, 15, 1] = 3 3.14159297

15 11

1133355 g+

++

= =

This an attractive alternative to the familiar 722 .

(iv) [3, 7, 15, 1, 292] = .37

151

2921

11

133102103993 3 141592653g+

++

+

= =

This is over 400 times more accurate than the previous one 133335` j, but the numbers

involved are not easy to remember.


Creative Questions[Creative questions are only sample questions. Similar type of questions

may also be created and asked in the examination]

2. SEQUENCES AND SERIES OF REAL NUMBERS

Objective Type Questions

1. Let , ,b c c a a b1 1 1+ + +

be in A.P. Which one of the following is also in A.P.?

(A) ( )( ), ( )( ), ( )( )a c a b a b b c b c c a+ + + + + +

(B) , ,a b c a b c a b c2 2 2+ + + + + +

(C) , ,a b c a b c b c a- + + - + -

(D) , ,a b c a b c b c a2 2 2- + + - + -

2. If , , , , , ,a a a a a a a1 2 3 4 5 6 7

are in A.P. and if a 104= , then a a

1 7+ is equal to

(A) 5 (B) 10 (C) 15 (D) 20

3. Which one of the following is a part of the Fibonacci sequence?

(A) 14, 21, 35, 56, 91 (B) 11, 19, 30, 49, 79

(C) 13, 21, 34, 55, 89 (D) 13, 20, 33, 53, 86

4. The common difference of an A.P. whose 4th term is 19 and 7th term is 28 is

(A) 2 (B) 3 (C) 4 (D) 5

5. Which one of the following functions is not a sequence?

(A) ( ) 2 1, 1,2,3,4,f p p p g= - =

(B) ( ) , ( , ]f x x x1 0 1!= +

(C) ( ) , , , ,f nn

n n12

3 1 2 3 g= + + =

(D) ( ) , , , ,f t t t2 1 2 3 g= + =

6. Iftheproductoffirst5termsofaG.P.is32,thenthethirdtermis

(A) 41 (B)

21 (C) 2 (D) 4

7. ThefifthtermofaG.P.withfirstterma 5= and common ratio 2 is

(A) 5 2 (B) 10 (C) 10 2 (D) 20

8. When x 2= the value of x x x1 2 9g+ + + + is

(A) 511 (B) 1023 (C) 513 (D) 1025

9. The value of 3 6 9 60g+ + + + is equal to

(A) 510 (B) 570 (C) 600 (D) 630

351

10. If k2 4 6 2 90g+ + + + = , then the value of k is

(A) 8 (B) 9 (C) 10 (D) 11

11. Let t t tm n m n

=+

for all natural numbers m and n and t 31= . Then , ,t t t

1 2 3 are in

(A)A.P. (B)G.P.

(C)bothA.P.andG.P. (D)neitherA.P.norG.P.

12. The nth term of the sequence , , , ,21

43

65

87 g is

(A) n

121+ (B) 1

n21- (C) 1

n 21-+

(D) nn

21

+-

13. If the sequence , , , , ,a a a an1 2 3

g g isinA.P.andalsoinG.P.,thenthe common difference and common ratio are respectively

(A) 0, 0 (B) 0, 1 (C) 1, 0 (D) 1, 1

14. If 2, , , 54x y areinG.P.,thenthecommonratiois

(A) 2 (B) 3 (C) 3 3 (D) 2 2

15. If 2, , 2x 2 are in A.P., then the common difference is

(A) 2 (B) 2 1+ (C) 2 1- (D) 22

16. The sequence , , 5 , 5 , 25 ,11 55 11 55 11 g is

(A)A.P. (B)G.P.

(C)bothA.P.andG.P. (D)neitherA.P.norG.P.

17. If a 1! , then ( ) ( )a a a1 1 1 2+ + + + + =

(A) ( )a

a a a1

3 2 3

-- + + (B) a a3 2 2 2

+ +

(C) a a3 2+ + (D) ( )

aa a

13 1 2

-- + +

18. If , , ca

ac

cab a c2 02

!+ = - , then , ,a b c are in

(A)A.P (B)G.P

(C)bothA.PandG.P (D)neitherA.PnorG.P

19. ThreenumbersareinG.P.Ifwedoublethemiddlenumber,wegetanA.P..ThenthecommonratiooftheG.P.is

(A) 2 3+ (B) 3 2+ (C) 5 3+ (D) 5 3+

20. If , , ,a a a1 2 3

g are in A.P. and a a a a a a 2251 5 10 15 20 24+ + + + + = , then a a

1 24+ is

(A) 150 (B) 100 (C) 75 (D) 125Creative Questions - Sequences and Series of Real Numbers


Two Mark Questions 1. Findthe18thand25thtermsofthesequencedefinedby

( )

,

,

and is even

and is oddC

n nn N n

n

n n N n

42

1

4n

2

!

!=

+

+

Z

[

\

]]

]]. ( Ans: ;C C90

31350

18 25= = )

2. If , ,bc ca ab1 1 1 are in A.P., then prove that , ,a b c are also in A.P.

3. The fourth termof anA.P. is equal to 3 times thefirst term and the seventh termexceedstwicethethirdtermby1.Findthefirsttermandthecommondifference.

( Ans: ,a d3 2= = )

4. A man saves ` 320 in the month of January, ` 360 in the month of February, ` 400 in the month of March. If he continues his savings in this sequence, what will be his savings in the month of November in the same year? ( Ans: ` 720 )

5. Find the value of ( )( )( )a b c b c a c a b2 2 2+ - + - + - if , ,a b c are in A.P. ( Ans: 0 )

6. FindaG.P.whosefirsttermis2andfifthtermis4timesthethirdterm.( )r 0> . ( Ans: , , , ,2 4 8 16 g )

7. IfthefirsttermandseventhtermofaG.P.are24and192respectively( )r 0> . Find the 11th term. ( Ans: 768 )

8. ThethirdtermofaG.P.is31 .Findtheproductofitsfirstfiveterms. (Ans:

2431 )

9. If , ,a b c areinG.P.,provethat , ,a b ab bc b c2 2 2 2+ + + arealsoinG.P.

10. ThefirsttermofaG.P.is1.Thesumofthethirdtermandfifthtermis90.Findthecommon ratio. ( Ans: r 3!= )

11. , , ( )m m72

27 2- - + areinG.P.Findthevalueofm. ( Ans: ,m 2 1= - )

12. Findthesumofthefirst10termsoftheseries:1 2 3 42 2 2 2 g- + - + . ( Ans: – 55 )

13. If , ,a b c are in A.P. and if , ,x y z areinG.P.,thenprovethat 1x y zb c c a a b=- - - .

14. Ifthesumoffirstn terms of a sequence is [ ]n n21 2

+ ,findthe50thterm. (Ans:50 )

15. Ifthesumoffirstn terms of a sequence is [ ( 1) ]n n41 2 2

+ ,findthe5thterm. ( Ans: 125 )

16. Find the sum of 5 156 73 3 3 3g+ + + + . ( Ans: 14300 )

17. Findthesumoffirst7termsoftheseries92

31

94 g+ + + . ( Ans:

935 )

353

18. Find the sum of the series 32

34 2

38

332g+ + + + + . ( Ans: 90

32 )

19. Find the sum of the integers between 50 and 200, which are divisibly 10. ( Ans: 1750 )

20. A student read common difference of an A.P. as – 2 instead of 2 and got the sum of first5termsas–5.Findtheoriginalsum. (Ans:35 )

Five Mark Questions 1. The digits of a 3-digit positive integer, are in A.P. and their sum is 15. The number

obtained by reversing the digits is 594 less than the original number.

Find the number. ( Ans: 852 ) 2. If the mth term of an A.P. is n and the nth term is m, then prove that pth term is m n p+ -

3. An A.P. consists of 21 terms. The sum of the three terms in the middle is 129 and the sumofthelastthreetermsis237.Findthefirsttermandthecommondifference.

( Ans: ,a d3 4= = )

4. The nth termofasequenceisdefinedby 15t an bnn

2= + + . If t 13

2= and t 27

4= ,

Find a and b. ( Ans. 2 ; 5a b= =- ) 5. The pth , qth and rth terms of an A.P. are , ,a b c respectively.

Show that ( ) ( ) ( ) 0q r a r p b p q c- + - + - = .

6. If , ,x y y y z1

21 1

+ + are three consecutive terms of an A.P. , then prove that , ,x y z

are three consecutivetermsofaG.P.

7. If , , ,a b c d areinG.P.,thenprovethat , ,a b b c c d2 2 2 2 2 2+ + + arealsoinG.P.

[Hint: Take , ,b ar c ar d ar2 3= = = ].

8. If the pth , qth and rth termsofaG.P.are ,a b and c respectively, then prove that 1a b cq r r p p q

=- - - .

9. Thesumoffirst3termsofaG.P.is1039 and their product is 1. Find the common ratio

and the geometric sequence. ( Ans: , , ,r52

25

52 1

25G.P.:or= )

10. If 4th , 10th and 16th termsofaG.P.are ,x y and z respectively, then

prove that , ,x y z areinG.P.

11. Thesumoffirstn terms of two arithmetic sequences are in the ratio

( ) : ( )n n3 8 7 15+ + . Find the ratio of their 3rd terms. ( Ans: 23 : 50 )

12. The sum of four consecutive terms in an A.P. is 4 and their product is 385.

Find the four numbers. ( Ans: , , , , , ,735

311 9 9

311

35 7or- - - - )

13. The sum of four consecutive terms in an A.P. is 20 and the sum of their squares is 120. Find the numbers. ( Ans: 2, 4, 6, 8 (or) 8, 6, 4, 2 )

Creative Questions - Sequences and Series of Real Numbers


14. Twocarsstarttogetherinthesamedirectionatthesameplace.Thefirstgoeswithuniformspeedof60km/hr.Thesecondgoesataspeedof50km/hrinthefirsthourand increases the speed by 4 km each succeeding hour. After how many hours will the secondcarovertakethefirstifbothcarsgoonnon-stop? (Ans:6 hrs. )

15. If the mth term of an A.P. is n1 and the nth term is

m1 ,thenprovethatthesumoffirst

mn terms is ( 1) .mn21 +

16. IfthesumoffirstseventermsofanA.P.is105andthefirsttermis6,thenshowthattheratioofthesumoffirstntermsandthesumoffirstn 3- terms is ( ) :n n3 3+ - .

17. Findthesumoffirst50termsofthesequence5, 5.5, 5.55, g . Ans: 4490815

10

149

+; E

18. Ifthesumof4consecutivetermsinaG.P.is60andif , ,t t181 4

are in A.P. ,

thenfindthenumbers. (Ans:4, 8, 16, 32 or 32, 16, 8, 4 )

19. Find the sum of the series ( ) ( ) ( ) ( )2 4 3 6 5 8 7 22 212 2 2 2g+ + + + + .

( Ans: 52107 )

20. Therearefournumberssuchthatthefirstthreeofthemformanarithmeticsequenceandthelastthreeformageometricsequence.Thesumofthefirstandthirdtermis2and that of second and fourth is 26. Find the numbers.

( Ans: , , , , , ,3 1 5 25 7 1 5 25or- - )

3. ALGEBRA

Objective Type Questions 1. If 1

a is a root of the equation x x2 5 7 02

- + = , then the value of 7 52a a- is

(A) 2 (B) – 2 (C) 5 (D) – 5

2. If y x xy

9 4 12+ = , where ,x y0 0> > , then x y3 2- =

(A) 5 (B) 1 (C) 2 (D) 0

3. If (0, – 1) is the unique solution of the equations a x b y c 01 1 1

+ + = and a x b y c 0

2 2 2+ + = , then

(A) a

a

b

b

2

1

2

1= (B) a

a

c

c

2

1

2

1= (C) b

b

c

c

2

1

2

1= (D) b

b

c

c

2

1

2

1!

4. If the roots of x kx3 5 75 02- + = are positive and equal, then the roots are

(A) 3, 3 (B) 6, 6 (C) ,5 5 5 5 (D) ,5 3 5 3

5. If 14a b+ = and 2 3a b- = , then ab =

(A) 42 (B) 44 (C) 46 (D) 48

355

6. If ,xx

x1 23 0>22

+ = , then xx1+ is

(A) 2 (B) 3 (C) 4 (D) 5 7. The equations x x k2 02

+ + = and x x k4 02+ - = have a common root a . Then the

value of k is (A) a (B) a- (C) 3a (D) 3a-

8. The roots of the equation ( )x 3 9 02+ - = are

(A) ( , )0 6- (B) ( , )3 6- - (C) ( , )0 3- (D) ( , )3 3-

9. The value of a for which the roots ,a b of the equation x x a2 02- + = satisfy the

condition 6a b- = is (A) 4 (B) – 4 (C) 8 (D) – 8 10. If a and 2a are roots of the equation x bx 8 02

- + = , then the value of b is (A) 2 (B) 4 (C) 6 (D) 8 11. The value of a for which the system 12, 2 6 15ax y x y6- = - = has no solutions is (A) 1 (B) 2 (C) 3 (D) 4 12. If the system of equations x y2 6 m- = and x y

32 2 5- = has no solution, then !m

(A) 5 (B) 9 (C) 12 (D) 15 13. Which one of the following is a root of the equation x x x x2 5 3 13 9 04 3 2

- - + + = ? (A) 1 (B) – 1 (C) 2 (D) 0 14. If x 1= is a zero of the polynomial ( ) 2 3p x x x x x44 3 2 m= - + - + , then the value of

m is (A) 2 (B) – 2 (C) 3 (D) – 3

15. The value of ( ) ( ) ( )x x x1 2 32 2 2- - - when x 4= is

(A) 3 (B) – 3 (C) 6 (D) – 6 16. The number of zeros (real numbers) of the polynomial ( )p x x 12

= + is

(A) 0 (B) 1 (C) 2 (D) 3

17. If x ax bx 63 2+ - - is the quotient on dividing x x x x2 14 19 64 3 2

+ - + + by x2 1+ , then the values of a and b are respectively

(A) 6, 7 (B) 0, 7 (C) 7, 0 (D) 7, 6 18. The LCM of , ,x y x yz x y z6 9 122 2 2 2 is

(A) xy z36 2 2 (B) x y z36 2 2 (C) 6x y z3 2 2 2 (D) xy z36 2

19. If x2 isG.C.D.ofthepolynomials ( )p x and ( )q x , then the L.C.M. of ( )p x and ( )q x is

(A) ( ) ( )x p x q x2 (B) ( ) ( )

x

p x q x2

(C) ( ) ( )

x

p x q x4

(D) ( ) ( )x p x q x4

20. When ,a b2 3= = , the value of ab b

a b a b2

2 3 3 2

+

+ is

(A) 12 (B) 18 (C) 24 (D) 36

Creative Questions - Algebra


Two Mark Questions

1. Solve: ( ) ( ) ( )x y x2 2 3 1 2 4 8+ = + = + . ( Ans: ,2 1- - )

2. Solve: y x y2

3 13

7 15

8 1+= + =

+ . ( Ans: ,2 3 )

3. Solve: 0.3 0.1 0.9, 0.2 0.1 0.4x y x y- =- + = . ( Ans: ,1 6- )

4. Solve: x y x y x y8 2 14 3 2 10+ - = + - = + - . ( Ans: ,2 6- )

5. The sum of two numbers is 55 and their difference is 7. Find the numbers. ( Ans: 31, 24 )

6. The cost of a pen and a note book is ` 60. If the cost of a pen is ` 10 less than the cost ofanotebook,thenfindtheircost. (Ans:` 25, ` 35 )

7. Find the zeros of the quadratic polynomial 1x x x3

2 5 22+ = + . ( Ans: ,

23 1- )

8. Find the zeros of the polynomial xx3

4 12 1

2 1+ -+

= . ( Ans: – 1, 1 )

9. Find a quadratic polynomial each with the given numbers as a sum and product of its

zeros respectively, (i) 9, 14 (ii) ,3

1 3 (iii) ,21

23- . (each subdivision carries

two marks) ( Ans: (i) x x9 142- + , (ii) x x

3

1 32- + , (iii) x x

2 232

+ + )

10. Find the quotient and the remainder when x x x3 17 31 123 2- + - is divided by x3 2- .

( Ans: Quotient: x x5 72- + , Remainder = 2 )

11. Find the quotient and the remainder when x x x x2 7 7 34 3 2+ + - - is divided by 2 1.x +

( Ans: Quotient: x x x3 33 2+ - - , Remainder = 0 )

12. Check whether ( )x3 1+ is a factor of x x x3 8 3 23 2+ + - . ( Ans: Not a factor )

13. FindtheG.C.D.ofthefollowing:(i) , ,a bc ab c a b c25 75 1253 3 2 2 4 2 2 , (ii) ,x y x y3 3 4 4+ -

(iii) ,a a a a2 13 2 2- - + . ( Ans: (i) abc25

2 , (ii) x y+ , (iii) a 1- )

14. Find the L.C.M. of the following: (i) , ,xy z x y x yz10 5 22 2 4 3 4 , (ii) ,a b a b2 2 3 3- +

(iii) ( )( ), ( )( ), ( )( )a b b c b c c a c a a b+ + + + + + . (each subdivision carries two

marks) ( Ans: (i) x y z103 4 4 , (ii) ( )( )a b a b

3 3- + , (iii) ( )( )( )a b b c c a+ + + )

15. The L.C.M. and G.C.D. of two polynomials are x y z5 4 7 and x z2 3 . If one of the polynomial is x z2 3 ,findtheotherpolynomial. (Ans:x y z

5 4 7 )

16. Simplify: (i) a

x

x x

a a

1

4

22

2

3 2

3

#-

-

+

- , (ii) x

x x

x x

x x

9

4 3

6 5

4 32

2

2

2

#-

- +

+ +

+ + .

(each subdivision carries two marks) ( Ans: (i) ( )

( )

x a

a x

1

22

2

+

- , (ii) xx

51

+- )

17. Simplify: (i) xx

xx

12

132 2

++ +

+- , (ii)

a ba

b ab3 3

-+

-, (iii)

x x11

12

--

+.

(each subdivision carries two marks) Ans: (i) xx

12 1

2

+- , (ii) a ab b

2 2+ + , (iii)

x

x

1

32-

-

357

18. Find the square root of ( ) 4x x4 22+ + - . ( Ans: | 2 |x + )

19. Solve: x x2 8 3 02+ + = . ( Ans:

4,

44 10 4 10- - - + )

20. Find a positive number whose square is added to it gives 30. ( Ans: – 6 and 5 )

21. Find the value of a b ab+ - if a and b are the roots of the equation 0.cx bx a2+ + =

( Ans: ( )ca b- + )

22. Find the common root of the equations x x 6 02+ - = and x x5 6 02

+ + = . ( Ans: 3- )

Five Mark Questions

1. Solve the following system of equations using cross multiplication method:

(i) ;y x xy y x xy4 1 5 2 3 13- = + = , (ii) , , , .

x y x yx y5 2 3 3 4 7 0 0! !+ =- - =-

(each subdivision carries two marks) ( Ans: (i) (2, 3), (ii) (– 1, 1) )

2. The perimeter of the rectangle is 60 cm. If the length is increased by 3cm and breadth is decreased by 3cm, then their sides are in the ratio 2 : 1. Find the dimension of the rectangle. ( Ans: 17 cm, 13 cm )

3. The sum of the numerator and denominator of a fraction is 12. If 3 is added to the denominator, it is 2 times of the numerator. Find the fraction. ( Ans:

75 )

4. Factorise: ( )x x x6 13 2+ + - . (Hint: Add and subtract x

2 ) ( Ans: ( )( )( 3)x x x1 2+ + + )

5. If x x 12+ + is the GCD. of the polynomials x x x x3 6 5 34 3 2

+ + + + and x x x2 24 2

+ + + findtheirLCM. (Ans:( )( )x x x x x2 3 2 22 4 2+ + + + + )

6. TheGCD.andLCM.oftwopolynomialsare( )x2 3- and (2 3)(3 4)(5 4) .x x x- + + If one of the polynomials is x x6 122

- - ,findtheotherpolynomial. ( Ans: ( )( )x x2 3 5 4- + )

7. Simplify: a

a

a a

a a

a a

a a

8

16

2 9 4

2 3 2

2 4

3 11 43

2

2

2

2

2

# '-

-

+ +

- -

+ +

- - . ( Ans: a3 11+

)

8. Simplify: x

x x

x x

x x

9

4 3

6 5

2 152

2

2

2

#-

- +

+ +

- - . ( Ans: ( )( )( )( )x xx x

5 11 5

+ +- - )

9. Simplify: m

m m

m m

m m

16

12

8 16

62

2

2

2

'-

- -

+ +

+ - . ( Ans: mm

24

-+ )

10. Simplify: a a a a a a7 12

1

5 6

1

6 8

22 2 2+ +

++ +

-+ +

. ( Ans: 0 )

11. Simplify: a b a b a b

a1 1 22 2+

+-

--

( Ans: 0 )

12. Simplify: ( ) ( )

a a

a

a a

a

3 2

4 1

6

4 32 2+ +

++

- -

- . ( Ans: a 28+

)

Creative Questions - Algebra


13. Simplify: a a a

a14

13

1

72+

--

--

. ( Ans: a 1

7+- )

14. Find the square root of the polynomial x x x x4 10 12 94 3 2+ + + + .

( Ans: x x2 32+ + )

15. Find the square root of the polynomial x x x x1 4 10 12 92 3 4+ + + + .

( Ans: 3 2 1x x2+ + )

16. Find the values of a and b if x x x ax b4 12 254 3 2+ + + + is a perfect square.

( Ans: ,a b24 16= = )

17. Find the values of a and b if a bx x x x25 24 162 3 4+ + - + is a perfect square.

( Ans: ,a b4 12= =- ) 18. Solve:

xx

xx

11

3061

++ + = . ( Ans: ,6 5- )

19. Solve by factorization method: x a x a b2 02 2 4 4- + - = .

( Ans: ( , )b a a b2 2 2 2- + )

20. The sum of a number and its reciprocal is 750 . Find the number. ( Ans: ,

71 7 )

21. If a and b are the roots of the equation x x3 1 02- + = , then form a quadratic equation

whose roots are 1a b+

and 1ab

. ( Ans: x x3 4 1 02- + = )

22. Sum of two numbers is 24 and sum of their reciprocal is 61 . Find the numbers..

( Ans: 12, 12 )

23. A man bought a certain number of articles for `6000. If he had bought superior kind costing each `10 more, he would have bought 50 articles less. Find the number of articles bought and the cost of one article. ( Ans: 200, ` 30 )

24. The sides of a right angled triangle are 2, 2 1x x+ - and x2 1+ . Find the length of the sides and also its area. ( Ans: Sides are 4, 3, 5 units and Area = 6 sq. units )

25. A motorist reduced the speed of a car by 10 km/hr. than the usual speed. He takes 32 minutes more than the usual time to travel a distance of 160 km. Find the usual speed.

( Ans: 60 km/hr. )

26. If one root of the equation x x m7 02+ + = exceedstheotherbyunity,thenfindm.

( Ans: m = 12 )

27. If the roots of the equation x px q 02- + = differ by unity, prove that p q4 12

= + .

28. Find the values of a and b if x x ax x b4 16 724 3 2- + - + is a perfect square.

( Ans: ,a b52 81= = )

29. Simplify: 2x x

x x

x x

x x

3 2

2

2 5 3

2 32

2

2

2

- +

- - ++ +

+ - - . ( Ans: x 1

42-

)

359

5. COORDINATE GEOMETRY


1. If ( , 2)x is the midpoint of the line segment joining (3, 4) and ( , )y1 , then the value of x and y are respectively

(A) 1, 2 (B) 2, 0 (C) ,2 2- (D) ,1 2-

2. The point of intersection of the lines y 5=- and x y 1 0+ + = is

(A) (– 4, – 5) (B) (6, – 5) (C) (4, – 5) (D) (– 5, 6)

3. The centroid of a triangle is the origin. If ( , )1 2- and ( , )3 5- are two vertices, then the third vertex is

(A) (– 2, 3) (B) (2, 3) (C) (– 2, – 3) (D) (2, – 3)

4. Let ( , )P x y be a point on the line y 5= . Then the distance of P from the x -axis is

(A) 3 (B) 4 (C) 5 (D) 6

5. If the distance between the points ( , )y21

and ( , )y22

is 7 units, then | |y y1 2- is equal

to

(A) 7 (B) 7 (C) 4 (D) 0

6. If the equations x y2 3 9+ = and ax y 3+ = represent the same line, then the value of a is equal to

(A) 2 (B) 31 (C)

32 (D) 3

7. The centre of a circle is at ( , )3 4 . If the circle touches the x -axis, then the radius of the circle is

(A) 3 (B) 4 (C) 5 (D) 7

8. The point ,P 037-` j divides a line segment joining two points A and B internally in

the ratio :2 1 . If A is ( , )2 3- , then the point B is

(A) (2, 1) (B) (– 2, – 1) (C) (2, – 1) (D) (– 1, – 2)

9. The midpoint of the segment joining the point P and the origin is 21 , 1-` j. Then the

point P is at

(A) (1, – 2) (B) (4, – 8) (C) (– 2, 1) (D) (– 8, 4)

10. The area of the triangle formed by the points ( , ), ,0 0746 0` j and ,0

2321` j is

(A) 1 (B) 2 (C) 3 (D) 4

11. The centre of a circle is ( , )4 2- . If a diamter of the circle has its one end at the origin, then the other end of the diameter is

(A) (2, – 4) (B) (– 8, 4) (C) (4, – 8) (D) (– 4, 8)

Creative Questions - Coordinate Geometry


12. The slope of a straight line parallel to x -axis is

(A)0 (B)1 (C)–1 (D)notdefined

13. The slope of a straight line parallel to the line x y2 4 5 0+ + = is

(A) 2 (B) 21 (C)

21- (D) – 2

14. The equation of the line passing through the origin and parallel to the line x y4 1 0- + = is

(A) x y4 0- = (B) x y4 0+ = (C) x y4 2 0- + = (D) y4 4 0+ =

15. The equation of the line passing through origin and perpendicular to the line x y5 10- = is

(A) x y5 10+ = (B) x y5 0- = (C) x y5 0- = (D) 5 0x y+ =

16. The angle of inclination of a straight line whose slope is 3 is

(A) 0c (B) 30c (C) 60c (D) 90c

17. The slope of the straight line whose angle of inclination is 90c , is

(A)1 (B)–1 (C)0 (D)notdefined

18. The equation of a straight line passing through ( , )2 3- and parallel to y -axis is

(A) 0x 2- = (B) 0x 2+ = (C) 0y 2+ = (D) 0y 2- =

19. The equations of the straight line parallel to x -axis which are at a distance of 5 units from the x -axis are

(A) ,y y5 5= =- (B) ,x x5 5= =-

(C) ,x y0 0= = (D) ,x c y k= =

20. The equation of a straight line with slope 3- and y intercept 4 is

(A) x y3 4 0+ + = (B) x y3 4 0- + =

(C) x y3 4 0+ - = (D) x y3 4 0- - =

21. The y -intercept of the line x y5 3= is

(A) 35 (B)

53 (C) 0 (D)

35-

22. The equation of the straight line whose x and y intercepts are 2 and 3 respectively is

(A) x y2 3 6+ = (B) x y3 2 6+ = (C) x y2 3 0+ = (D) x y3 2 0+ =

23. The lines y 7=- and x 4= meet at the point

(A) ( , )7 4- (B) (7, 4) (C) (4, – 7) (D) (4, 7)

24. The angle of inclination of the straight line y x5 5 10= + is

(A) 0c (B) 30c (C) 60c (D) 45c

25. If the slope of the line joining the points ( , )2 1- - and ( , )k 0 is 61 , then the value of

k is (A) – 2 (B) 1 (C) 0 (D) 4

361

Two Mark Questions

1. The coordinates of the midpoint of the line segment joining the points ( , )a2 2 3+ and ( , )b4 2 1+ are ( , )a b2 2 . Find the values of a and b . (Ans: a = 3, b = 2 )

2. Find the coordinates of the point dividing internally the line segment joining ( , )A 3 7 and ( , )B 8 2 in the ratio :2 3 . ( Ans: (5, 5) )

3. If ( , ), ( , ), ( , )A B a C b2 1 0 4- - and ( , )D 1 2 are the vertices of a parallelogram ABCD ,findthevaluesofa and b . ( Ans: a = 1, b = 3 )

4. Find the coordinates of the point dividing externally the line segment joining ( , )P 1 2

and ( , )Q 4 5 in the ratio :5 3 . ( Ans: ,217

219` j )

5. Find the point P on the line segment joining the points ( , )A 3 1- and ( , )B 6 5- such that AP PB2 = . ( Ans: (0, 1) )

6. P and Q trisect the line segment joining the points ( , )2 1 and ( , )5 8- . If the point P lies on x y k2 0- + = ,thenfindthevalueofk . ( Ans: k = – 8 )

7. If the three vertices of a rhombus, taken in order, are ( , ), ( , ), ( , )4 1 6 0 7 2- ,findthefourth vertex. ( Ans: (5, 1) )

8. The coordinates of two vertices of a ABCT are A ( , )1 4 and B ( , )5 3 . The centroid of the ABCT is (3, 3). Find the coordinates of the third vertex C . ( Ans: (3, 2) )

9. Find the area of the triangle whose vertices are ( 4, 6), ( , )A B 2 1- - - and ( , )C 1 2 . ( Ans: 11.5 sq. units )

10. The vertices of ABCT are ( 5, 1), (3, 5)A B- - - and (5, ) .C k If the area of ABCT is32sq.units,findthevalueofk. ( Ans: k = 2) )

11. Show that the points ( , ), ( , )2 5 3 4- - and ( , )8 1 are collinear.

12. If the points ( , )p 02 and ( , )q0 2 and ( , )1 1 are collinear, prove that 1.p q

1 12 2+ =

13. Let ( , )2 1- be the centre of a circle and AB be a diameter. If A is ( , )3 5- ,

thenfindB . ( Ans: (7, – 7) )

14. Show that the points ( , ), ( , )4 3 5 1 and ( , )1 9 are collinear using the concept of slope.

15. The points ( 5, ), ( , 7)P a Q b- and ( , )R 1 3- are collinear such that PQ QR= . Find the values of a and b . ( Ans: a = 17, b = – 2 )

16. If the points ( , ), ( , )A B p1 3 2- and ( , )C 5 1- arecollinear,findthevalueofp. ( Ans: p = 1 )

17. Find the equation of the straight line whose angle of inclination is 60° and y intercept

is 3

1 . ( Ans: x y3 3 1 0- + = )



18. Find the equation of the straight line passing through ( , )3 5- with slope 52 .

( Ans: x y2 5 31 0- - = )

19. Find the equation of the straight line whose x and y intercepts are 72 and

31-

respectively. ( Ans: 7 6 2 0x y- - = )

20. Find the equation of the line passing through ( , )2 3 and parallel to the line x y3 7 21 0+ - = . ( Ans: x y3 7 27 0+ - = )

21. Show that the line joining ( , )4 4- and ( , )10 12- - and the line joining ( , )8 16- and ( , )14 0 are parallel.

22. Find the equation of the line through the point ( , )5 2- and parallel to the line joining the points ( , )A 12 2- and ( , )B 4 10- - . ( Ans: x y2 9 0- + = )

23. Show that the straight lines x y2 3 6 0+ + = and x y3 2 12 0- - = are perpendicular to each other.

24. Examine whether the lines x y2 4

1+ = and x y2 5+ = are parallel or not. ( Ans: Parallel )

25. If kx y3 14 0- - = and x y3 4 10 0+ + = areperpendiculartoeachother,findk .

( Ans: k = 4 )

Five Mark Questions

1. Prove that the points ( , ), ( , ), ( , )A B C4 1 2 4 4 0- - - - and ( , )D 2 3 are the vertices of a rectangle. (Using the concept of slope)

2. Prove that ( , ), ( , ), ( , )0 5 2 2 5 0- - and ( , )7 7 are the vertices of a rhombus. (Using the concept of slope)

3. If M is the midpoint of the line segment joining ( , )A a2 0 and ( , )B b0 2 and if O is the origin, then show that M is equidistant from the points ,O A and B .

4. The line joining the points ( , )5 4- and ( , )3 2- is trisected. Find the coordinates of

the points of trisection. ( Ans: ,37 2-` j and ,

31 0-` j )

5. Find the coordinates of the points which divide the line segment joining the points

( 2, 0)- and ( , )0 8 into four equal parts. ( Ans: , , ( , ) ,and23 2 1 4

21 6- - -` `j j )

6. If ( , )P a 2- and ,Q b35` j trisect the line segment joining the points ( , )3 4- and

( , )1 2 ,findthevaluesofa and b . ( Ans: , ba37 0= = )

7. Find the ratio in which the point ( , 6)m divides the line segment joining the points ( , )A 2 2- and ( , )B 3 7 .Alsofindthevalueofm . ( Ans: 4 : 1, m = 2 )

363

8. Find the ratio in which the line segment joining ( , )2 3- and ( , )5 6 is divided by x -axis.Also,findthecoordinatesofthepointofdivision. (Ans: 1 : 2, (3, 0) )

9. In what ratio is the segment joining the points ( 2, 3)- - and ( , )3 7 divided by the y -axis?.Also,findthepointofdivision. (Ans: 2 : 3, (0, 1) )

10. The vertices of a triangle are ( , ), ( , )A B1 3 1 1- - and ( , )C 5 1 . Find the length of medians through the vertices ( , )1 3- and ( , )5 1 . ( Ans: 5, 5 )

11. Find the area of the quadrilateral whose vertices are ( , ), ( , ), ( , )1 6 3 9 5 8- - - - - and ( , )3 9 . ( Ans: 60 sq. units)

12. If the area of the quadrilateral whose vertices, taken in order, are (1,2), ( 3,4),- , , (4, )k5 6- -^ h is43sq.units,findk . (Ans. – 1)

13. The vertices of a triangle are ( , ), ( , )2 5 4 1- - and ( , )6 3 . Find the slopes of the

medians. ( Ans: , ,74

25

51- - )

14. The line x y4 3 12 0+ - = intersect the ,x y -axes at A and B respectively. Find the area of AOBT . ( Ans: 6 sq. units )

15. Find the equation of the perpendicular bisector of the straight line segment joining the points ( , )A 2 4- and ( , )B 6 8- . ( Ans: 2 0x y3 1 0- - = )

16. The vertices of a ABCT are ( , ), ( , )A B1 7 0 2- and ( , )C 3 3 . Find the slope and equation of the median through A. ( Ans: – 13, 13 20 0x y+ - = )

17. Find the equation of the straight line joining the point ( , )4 5 and the point of intersection of the straight lines x y5 3 8- = and x y2 3 5- = . ( Ans: x y2 3 0- - = )

18. Find the equation of the straight line which passes through the intersection of the lines x y 2 0+ - = , x y2 3 0+ - = and bisects the line joining the points ( , )4 2 and ( , )6 4- . ( Ans: x y 2 0+ - = )

19. Find the equation of the straight line joining the point of intersection of the lines x y 2 0- - = and x y3 4 15 0+ + = and the point of intersection of the lines x y3 3 0- + = and 2 8 0x y+ - = . ( Ans: (– 1, – 3), (3, 2), x y5 4 7 0- - = )

20. Find the equation of the straight line passing through the point of intersection of the lines x y 5 0+ - = and x y3 1 0- + = and parallel to the line joining the points ( , )3 1 and (2,3) . ( Ans: x y2 6 0+ - = )

21. Find the equation of the straight line passing through the point of intersection of the lines x y5 8 23 0- + = and 7 6 7 0x y 1+ - = and is perpendicular to the line joining the points ( , )5 1 and ( , )2 2- . ( Ans: (5, 6), x y7 29 0- - = )



6. GEOMETRY


1. Inthefigure,ifDE AB< and : 3:2AD DC = .

Thenarea ABCT^ h:area DECT^ h =

(A) 4 : 25 (B) 4 : 9

(C) 9 : 4 (D) 25 : 4

2. IfABCisanisosceles,righttriangle,rightangledatC,then (A) AB AC22 2

= (B) AC AB22 2=

(C) BC CA22 2= (D) AC BC22 2

=

3. If ABC PQRT T+ andarea 4PQRT =^ h area ABCT^ h,then :AB PQ is

(A) 2 : 1 (B) 4 : 1 (C) 1 : 2 (D) 1 : 4

4. IntrapeziumABCD,ifAB DC< and AB = 2DC, thenarea AOBT^ h:area CODT^ his

(A) 1 : 2 (B) 4 : 1

(C) 1 : 4 (D) 2 : 1

5. Intheadjointfigure,AB = AC .Whichofthefollowingistrue?

(A) TABD +TCFE

(B) TABD +TFCE

(C) TABD +TECF

(D) TABD +TEFC

6. Inthefigure,ifDE AC< and DF AE< ,thenFEBF =

(A) ECFE (B)

ECBE

(C) BABD (D)

BEEC

7. Inthefigure,Cisthecentreoftheconcentriccircles,thechordPQ touchesthesmallercircleofradius3cmatR.IfPQ=8cm,thentheradiusofthelargercircleis

(A)3cm (B)4cm

(C)5cm (D)2cm

8. In TABC,AB=6cmandADistheanglebisectorof A+ .

IfBD : DC=3:2,thenAC =

(A)4cm (B)6cm (C)6cm (D)8cm

365

9. Inthefigure, ,AB QRAPPQ

2< = and QR = 8 cm, then AB =

(A) 10 cm (B) 8 cm

(C) 6 cm (D) 4 cm

10. Inthefigure,thevalueofx is

(A) 3 (B) 4

(C) 5 (D) 6

11. Inthefigure,T PTl is tangent to the circle at P.

If 130QPT+ =l c, then PRQ+ =

(A) 65c (B) 50c

(C) 130c (D) 40c

12. ABCD is a parallelogram and 110B+ = c. If E is a point on BD such that ,ADAB

EDEB=

then EAD+ =

(A) 35c (B) 45c

(C) 60c (D) 70c

13. Inthefigure,ifPT is tangent at P and PQ = QR, then QPT+ =

(A) 30c (B) 60c

(C) 45c (D) 90c

14. Inthefigure,LM NQ< and LN PQ< . Which one of the following is true?

(A) TLMN +TNQP

(B) TQNP +TMNL

(C) TLNM +TQNP

(D) TLMN +TQNP

15. The distance between the tops of two vertical poles of height 6 m and 11 m standing on the same plane at a distance of 12 m is

(A) 7 m (B) 13 m (C) 9 m (D) 10 m

Creative Questions - Geometry


7. TRIGONOMETRY


1. cos tan sin cot1 12 2i i i i+ + + =

(A) 2 (B) 1 (C) cosec i (D) sec i

2. ( ) ( )cos cosec sin sec9 900 i i i i- - - =c c

(A) 1 (B) 0 (C) 2 (D) 1-

3. (90 ) (90 )tan tan cot coti i i i- + - =c c

(A) tan 45c (B) 60cosec c (C) sec 60c (D) cot 90c

4. (90 ) (90 )cos cos sin sini i i i- - - =c c

(A) sin90c (B) cot45c (C) cosec45c (D) cos90c

5. If A + B = 90c, then cos A sin B + sin A cos B is equal to

(A) 2 (B) 1 (C) 0 (D) 45c

6. cosec tan67 232 2- =c c

(A) 0 (B) 1 (C) – 1 (D) 2

7. IntheadjointfigureAB BC3= , then i =

(A) 30c (B) 45c

(C) 60c (D) 90c

8. cosecsin

seccos

ii

ii+ =

(A) sin cos2 2i i- (B) cosec cot2 2i i-

(C) sec cosec2 2i i+ (D) 0

9. If 1 (90 ) 45tan cosec2 2i+ - =c c, then i =

(A) 30c (B) 90c (C) 60c (D) 45c

10. If , ( )sec cotx y5 5 90i i= = -c , then y x2 2- =

(A) 0 (B) 5 (C) 25 (D) – 25

11. cossin sin cos

3

ii i i+ =

(A) sini (B) cosi

(C) tani (D) coti

12. Intheadjointfigure,thevalueofx is

(A) 5 cm (B) 20 cm

(C) 10 cm (D) 40 cm10 cm

367

13. If the length of the shadow of a tower is 3

1 times that of its height then the angle of elevation of the sun is

(A) 30c (B) 45c (C) 60c (D) 90c

14. The value of 5 85tan tanc c is

(A) 3

1 (B) 3 (C) 1 (D) 0

15. cossin1ii+ =

(A) sin

cos1 i

i-

(B) cos

sin1 i

i+

(C) sinsin

11

ii

-+ (D)

sinsin

11

ii

+-

16. sec tan

sec tan2 2

4 4

i i

i i

+

- =

(A) 1 (B) sec tan2 2i i (C) 2 (D) sin cos2 2i i

17. tan cot

1i i+

=

(A) sin cosi i+ (B) sin cosi i (C) sin cosi i- (D) cosec coti i+

18. sin sin tan20 70 452 2+ - =c c c

(A) 1 (B) 0 (C) 2 (D) – 1

19. In a right triangle ABC, 90 , sinB A31+ = =c and BC 11= , then AC =

(A) 311 (B) 120 (C) 33 (D) 118

20. cos cos

sin sin3

3

i i

i i

-

- =

(A) tan2i (B) cot2i (C) tan i (D) cot i

Creative Questions - Trigonometry


8. MENSURATION

Two Mark Questions

1. A solid right circular cylinder has base circumference 132 cm and height 10 cm. Find its total surface area. ( Ans. 4092 cm2 )

2. The total surface area of a solid right circular cylinder is 600 rcm2. If its heightis13cm,findtheradius. (Ans.12 cm )

3. The radius of a circular well is 5 m .Find the cost of cementing the surface to the depth of 14m. at the rate of `2 per square metre. ( Ans. ` 880 )

4. The radius and the total surface area of a right circular cone are 6 cm and 96rcm2, respectively. Find the height of the cone. ( Ans. 8 cm )

5. Total surface area of a right circular cone is 770 cm2. If the slant height of the cone is 4timestheradiusofitsbase,thenfindthediameterofthebase. (Ans.14 cm )

6. The bucket is in the form of a frustum of a cone. The radii of the circular ends of the bucket of height 24 cm are 12 cm and 4 cm. Find its volume. ( Ans. 1664r cm3 )

7. Find the total surface area of a hollow cylinder of length 14 cm, whose external and internal radii are 4 cm and 3 cm respectively. ( Ans. 660 cm2 )

8. If the total surface area of a sphere is 616 cm2,thenfinditsradius. (Ans.7 cm )

9. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of the surface area of the moon to that of the earth. ( Ans. 1 : 16 )

10. Volume of a right circular solid cylinder is 744 cm3.Ifitsheightis8cm,thenfindthe

diameter of the cylinder. ( Ans. 1 cm )

11. The base radius and the slant height of a solid right circular cone are 5 cm and 13 cm

respectively. Find the volume of the cone. ( Ans. 31472 cm3 )

12. Radius and height of a right circular cone are 8 cm and 12 cm respectively. How many times the volume of a sphere of radius 4 cm is equal to the volume of the right circular cone?. ( Ans. 3 times )

13. The largest sphere (sphere with maximum volume) is carved out of a cube of sides 14 cm. Find the volume of the sphere. ( Ans. 1437

31 cm3 )

14. A barrel (drum) is to be painted inside and outside of its curved surface. The radius of its base and the height of the drum are 1.4 m and 3 m respectively. Find the cost of painting at the rate of `10 per square metre. ( Ans. `528 )

15. Volume of a solid sphere is 17932 cm3. Find its curved surface area. ( Ans. 154cm2 )

369

Five Mark Questions

1. Find the number of coins each of diameter 1.5 cm and thickness 0.2 cm to be melted to make a right circular solid cylinder of height 10 cm and diameter 4.5 cm. ( Ans. 450 coins )

2. The length, breadth and height of a solid metallic rectangular parallelopiped are 44cm, 21cm and 12cm respectively. It is melted and a solid cone is made out of it.

Iftheheightoftheconeis24cm,thenfindthediameterofitsbase.(Ans.d = 42 cm )

3. Ahemisphericalbowlofradius9cmcontainsfullofaliquid.Thisliquidistobefilledinto cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are needed to empty out the bowl? ( Ans. 54 bottles )

4. The diameter of a small iron sphere is 6cm. 8 such small iron spheres are dropped into a cylindrical vessel of diameter 24 cm containing some water. Find the rise in the level of water when the spheres are completely immersed in the water. ( Ans. 2 cm )

5. How many metres of cloth of 4m wide will be required to make 5 identical conical tents each of radius 14m and slant height 50 m . ( Ans. 2750 m )

6. Waterisflowingattherateof20km/hrthroughacylindricalpipeofradius20cmintoahemisphericaltankofdiameter10m.Findthetimerequiredtofillthetank.

( Ans. 6 min 15 sec. )

7. A cricket stump is in the shape of a cylinder surmounted by a cone. The diameter and the total height of the stump are 10 cm and 80 cm respectively. If the height of the

conicalpartis12cm,thenfinditstotalsurfacearea. (Ans. 234173 cm2 )

8. A spherical wooden solid of diameter 12 cm is cut off from a solid cylinderical wood of

diameter 14 cm and height of 10 cm. Find the volume of the wood wasted.

( Ans. 63476 cm3 )

9. The total surface area of a right circular solid cone is 704 cm2. If its radius is 7 cm, thenfindthevolumeofthecone. (Ans.1232 cm3 )

10. The curved surface area and total surface area of a solid circular cylinder are

880 sq.cm and 1188 sq.cm respectively. Find its volume. ( Ans. 3080 cm3 )

11. The total surface area of a solid hemispherical ball is 942 cm2 . It is melted and made into small solid identical spherical balls of diameter 2 cm each. How many balls can be made? ( 3.14r = ) ( Ans. 500 spheres)

12. The surface area of a given sphere is 154 cm2 . Find the volume of a sphere whose diameter is twice that of given sphere. ( Ans. 3 cm14 7

31 3 )

13. The total surface area of a right circular solid cylinder is 96 cm2

r . If its height is 4 cm morethanitsradius,findthevolumeofthecylinder. (Ans.402 cm

72 3 )

Creative Questions - Mensuration


Objective Type Questions - Answers

2. Sequences and Series of Real Numbers

1 2 3 4 5 6 7 8 9 10A D C B B C D B D B11 12 13 14 15 16 17 18 19 20B B B B B D A D A C

3. Algebra

1 2 3 4 5 6 7 8 9 10B D C B C D A A D C11 12 13 14 15 16 17 18 19 20B D B A C A B B B A

5. Coordinate Geometry

1 2 3 4 5 6 7 8 9 10B C D C A C B D A C11 12 13 14 15 16 17 18 19 20B A C A D C D A A C21 22 23 24 25C B C D D

6. Geometry

1 2 3 4 5 6 7 8 9 10D A C B B B C A D C11 12 13 14 15C A C C B

7. Trigonometry

1 2 3 4 5 6 7 8 9 10A B C D B B C B D C11 12 13 14 15 16 17 18 19 20C B C C A A B B C D

Model Question Papers 371

MODEL QUESTION PAPERS

GENERAL GUIDELINES

(i) Model Question Papers are based on the Blue-Print given in the

Text Book.

(ii) 2 mark, 5 mark and 10 mark Questions are taken according to the

classification of questions given in the SCORE Book.

(iii) Creative Questions in all the categories are framed based on the

syllabus and content of the Text Book. (Please refer to the Sample

Questions in the SCORE Book)

(iv) Diagrams are given along with the questions wherever required.


Model Question Paper - 1Time: 2.30 Hrs.] [Maximum marks: 100

General instructions:(i) This question paper consists of four sections. Read the note carefully under each section before answering them.(ii) The rough work should be shown at the bottom of the pages of the answer book.(iii) Calculator and other electronic devices are not permitted.

Section – ANote: (i) Answer all the 15 questions. (ii) Each question contains four options. Choose the most suitable answer from the four alternatives.

(iii) Each question carries 1 mark 15 × 1 = 15

1. Given ( )f x = 1 x-^ h is a function from N to Z . Then the range of f is

(A) { 1} (B) N (C) { 1, – 1 } (D) Z

2. The sequence –3, –3, –3,g is

(A) an A.P. only (B) a G.P. only

(C) neither A.P. nor G.P (D) both A.P. and G.P.

3. The 108th term of the sequence 1, – 1, 0, 1, –1, 0 ... is

(A) 1 (B) –1 (C) 0 (D) 108

4. If anda b are the roots of ax bx c 02+ + = , then one of the quadratic equations whose

roots are 1 1anda b

, is

(A) ax bx c 02+ + = (B) 0bx ax c2

+ + =

(C) 0cx bx a2+ + = (D) 0cx ax b2

+ + =

5. The remainder when x x7 2 12- + is divided by x 3- is

(A) 58 (B) 70 (C) 0 (D) 3

6. A is of order m n# and B is of order p q# , addition of A and B is possible only if

(A) m p= (B) n = q (C) n = p (D) m = p, n = q

7. The value of k if the straight lines 3x + 6y + 7 = 0 and 2x + ky = 5 are perpendicular is

(A) 1 (B) –1 (C) 2 (D) 21

8. The length of a diagonal of the quadrilateral whose vertices are (1,0), (0,1), (–1,0) and (0,–1) is ______ units.

(A) 1 (B) –1 (C) 2 (D) 2


9. In the figure, PA and PB are tangents to the circle drawn from an external point P. Also CD is a tangent to the circle at Q. If PA = 8 cm and CQ = 3 cm, then PC is equal to

(A) 11 cm (B) 5 cm

(C) 24 cm (D) 38 cm

10. The areas of two similar triangles are 16 cm2 and 36cm2 respectively. If the altitudes are in the ratio 2 : x , then x is

(A) 2 (B) 3 (C) 4 (D) 6

11. cos sinx x4 4

- =

(A) 2 1sin x2

- (B) 2 1cos x2

- (C) 1 2sin x2

+ (D) 1 2 .cos x2

-

12. sin cos sec tan cosec cot2 2 2 2 2 2i i i i i i+ + - - + =

(A) 0 (B) 1 (C) 2 (D) 3

13. If the surface area of a sphere is 36r cm2, then the volume of the sphere is equal to

(A) 12r cm3 (B) 36r cm3 (C) 72r cm3 (D) 108r cm3.

14. Variance of the first 11 natural numbers is

(A) 5 (B) 10 (C) 5 2 (D) 10

15. If ( ) 0.25, ( ) 0.50, ( ) 0.14 ( )thenP A P B P A B P A Bneither nor+= = = =

(A) 0.39 (B) 0.25 (C) 0.11 (D) 0.24

Section – BNote: (i) Answer 10 questions. (ii) Answer any 9 questions from the first 14 questions. Question no. 30 is compulsory. (iii) Each question carries 2 marks. 10 × 2 = 20

16. Let { , , }, { , , , } { , , , }P a b c Q g h x y R a e f sand= = = . Find \R P Q+^ h.

17. Does each of the following arrow diagrams represent a function? Explain.

18. Find the sum of the first 25 terms of the geometric series 16 48 144 432 g- + - +

19. What rational expression should be added to x

x

2

12

3

+

- to get x

x x

2

2 32

3 2

+

- + ?

A

BC

D

PQ


20. Form a quadratic equation whose roots are ,2

4 72

4 7+ -


= 6 @ whose elements are given by a i j2 3ij= -

22. If A B4

5

2

9

8

1

2

3and=

-

-=

- -e eo o find A B6 3- .

23. If , , , ,7 3 6 1^ ^h h ,8 2^ h and ,p 4^ h are the vertices of a parallelogram taken in order, then find the value of p.

24. In ABC3 , the internal bisector AD of A+ meets the side BC at D. If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, then find DC.

25. A girl of height 150 cm stands in front of a lamp-post and casts a shadow of length 150 3 cm on the ground. Find the angle of elevation of the top of the lamp-post .

26. Prove: sinsin sec tan

11

ii i i

+- = -


28. The smallest value of a collection of data is 12 and the range is 59. Find the largest value of the collection of data.

29. A ticket is drawn from a bag containing 100 tickets. The tickets are numbered from one to hundred. What is the probability of getting a ticket with a number divisible by 10?

30. (a) Find the x and y intercepts of the straight line y x xy10 9 8+ =- .

(OR) (b) If the total surface area of a solid right circular cylinder is thrice its curved surface

area, then find the height in terms of its radius.

Section – CNote: (i) Answer 9 questions. (ii) Answer any 8 questions from the first 14 questions. Question no. 45 is compulsory. (iii) Each question carries 5 marks 9 × 5 = 45

31. Using venn diagram, verify \A B C,^ h = \ \A B A C+^ ^h h

32. Let A= { 0, 1, 2, 3 } and B = { 1, 3, 5, 7, 9 } be two sets. Let :f A B" be a function given by ( )f x x2 1= + . Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph.

33. Find the sum of first n terms of the series. 7 77 777 g+ + + .

34. Factorize the following polynomial. 2 5 6x x x3 2- - +

35. Find the square root of the polynomial 9 6 7 2 1x x x x4 3 2- + - + by division method.

36. If A B

2

4

5

1 3 6and=

-

= -f ^p h , then verify that ( )AB B AT T T= .


37. Find the area of the quadrilateral whose vertices are (6,9), (7,4), (4,2) and (3,7)

38. The vertices of a 3ABC are A(1 , 2), B(-4 , 5) and C(0 , 1). Find the slopes of the altitudes of the triangle.


40. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30° with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree.


42. A solid wooden toy is in the form of a cone surmounted on a hemisphere. If the radii of the hemisphere and the base of the cone are 3.5 cm each and the total height of the toy is 17.5 cm, then find the volume of wood used in the toy. ( Take r =

722 )

43. For a collection of data, if xR = 35, n = 5, 82x 9 2R - =^ h , then find ( )andx x x2 2/ / -


32

73and respectively.


72 ,


358 . Find

the probability that the problem can be solved by atleast one of them.

45. (a) Find the sum of all natural numbers between 400 and 600 which are divisible by 11.(OR)

(b) Prove that ( )xx

xx

xx x x

11

11

12 14 2 33

-- +

+- =

++ + +

Section – DNote: (i) This section contains 2 questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries 10 marks 2 ×10 = 20

46. (a). Draw a circle of radius 3.2 cm. At a point P on it, draw a tangent to the circle using the tangent-chord theorem.

(OR) (b) Construct a cyclic quadrilateral ABCD, where AB = 6.5 cm, 110ABC+ = c,

BC = 5.5 cm and AB || CD.

47. (a) Draw the graph of 2 6y x x2

= + - and hence solve 2 10 0x x2+ - = .

(OR) (b) The cost of the milk per litre is `15. Draw the graph for the relation between the

quantity and cost . Hence find (i) the proportionality constant. (ii) the cost of 3 litres of milk.



General instructions:

(i) This question paper consists of four sections. Read the note carefully under each section before answering them.(ii) The rough work should be shown at the bottom of the pages of the answer book.(iii) Calculator and other electronic devices are not permitted.



1. If { , , , }A p q r s= , { , , , }B r s t u= , then \A B is

(A) { p, q } (B) { t, u } (C) { r, s } (D) {p, q, r, s }

2. If the nth term of a sequence is 100 n +10, then the sequence is

(A) an A.P. (B) a G.P.

(C) a constant sequence (D) neither A.P. nor G.P.

3. General term of the sequence 52 ,

256 , ,

12518 g is

(A) 53 (B)

52 n 1-

` j (C) 52

53 n 1-

` `j j (D) 53

52 n 1-

` `j j

4. If ax bx c 02+ + = has equal roots, then c is equal

(A) ab2

2

(B) ab4

2

(C) ab2

2

- (D) ab4

2

-

5. (x – a) is a factor of p(x) if an only if ...

(A) ( ) ( )P a p x= (B) ( )p a 0! (C) ( )p a 0= (D) ( )p a 0- =

6. If A and B are square matrices such that AB = I and BA = I , then B is

(A) Unit matrix (B) Null matrix

(C) Multiplicative inverse matrix of A (D) A-

7. The centroid of the triangle with vertices at ,2 5- -^ h, ,2 12-^ h and ,10 1-^ h is

(A) ,6 6^ h (B) ,4 4^ h (C) ,3 3^ h (D) ,2 2^ h

8. The angle of inclination of the line passing through the points (1, 2), and (2, 3) is

(A) 30c (B) 45c (C) 60c (D) 90c

9. The sides of two similar triangles are in the ratio 2:3, then their areas are in the ratio

(A) 9:4 (B) 4:9 (C) 2:3 (D) 3:2


10. In ABCT a straight line DE BC< , intersects AB at D and AC at E, then

(A)ADAB

AEAC= (B)

AEAB

ADAC= (C)

ECAB

DBAC= (D) AB AC=

11. cos cot1 12 2i i- +^ ^h h =

(A) sin2i (B) 0 (C) 1 (D) tan2i

12. In the adjoining figure 60CAB+ = c. AB = 3.5m, then AC =

(A) 7 m (B) 3.5 m

(C) 1.75 m (D) 1 m

13. If the surface area of a sphere is 100r cm2, then its radius is equal to

(A) 25 cm (B) 100 cm (C) 5 cm (D) 10 cm .

14. The variance of 10, 10, 10, 10, 10 is

(A) 10 (B) 10 (C) 5 (D) 0

15. If A and B are two events such that ( ) 0.25, ( ) 0.05P A P B= = and ( ) 0.14,P A B+ = then ( )P A B, =

(A) 0.61 (B) 0.16 (C) 0.14 (D) 0.6

Section – BNote: (i) Answer 10 questions (ii) Answer any 9 questions from the first 14 questions. Question no. 30 is compulsory. (iii) Each question carries two marks 10 × 2 = 20

16. If ,A B1 then find A B+ and \A B (use Venn diagram).

17. Let A = { 1, 2, 3, 4, 5 }, B = N and :f A B" be defined by ( )f x x2

= .

Find the range of f . Identify the type of function.

18. Find the sum of the series. 1 2 3 203 3 3 3

g+ + + +

19. Multiply the following and write your answer in lowest terms.x

x

x x

x x

4

81

5 36

6 82

2

2

2

#-

-

- -

+ +


= 6 @ whose elements are given by ai ji j

ij=

+-

21. If A B

8

2

0

7

4

3

9

6

3

1

2

5and= -

-

=-

- -f ep o , then find BA if it exist.

22. Find the coordinates of the point which divides the line segment joining (-3, 5) and(4, -9) in the ratio 1 : 6 internally.

23. Find the equation of the straight line passing through the point ,2 3-^ h with slope 31 .


24. In ABC3 , AE is the external bisector of A+ , meeting BC produced at E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then find CE.

25. Prove: 1sec sin sec tan1i i i i- + =^ ^h h

26. Find the angular elevation (angle of elevation from the ground level) of the Sun when the length of the shadow of a 30 m long pole is 10 3 m.

27. If the circumference of the base of a solid right circular cone is 236 cm and its slant height is 12 cm, find its curved surface area.

28. If the coefficient of variation of a collection of data is 57 and its S.D is 6.84, then find the mean.

29. Three coins are tossed simultaneously. Find the probability of getting at least two heads.

30. (a) Simplify. x

x x x2 18

4 122

3 2

-

- -

(OR)

(b) The surface area of a sphere is 616 sq.cm. Find its diameter.

Section – CNote: (i) Answer 9 questions (ii) Answer any 8 questions from the first 14 questions. Question no. 45 is compulsory. (iii) Each question carries five marks 9 × 5 = 45

31. A radio station surveyed 190 students to determine the types of music they liked. The survey revealed that 114 liked rock music, 50 liked folk music, and 41 liked classical music, 14 liked rock music and folk music, 15 liked rock music and classical music, 11 liked classical music and folk music. 5 liked all the three types of music.

Find (i) how many did not like any of the 3 types? (ii) how many liked any two types only? (iii) how many liked folk music but not rock music?

32. Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If :f A B" is defined by f x x21 1= +^ h


33. Find the sum to n terms of the series 6 + 66 + 666 +g

34. The GCD of 3 5 26 56x x x x4 3 2+ + + + and 2 4 28x x x x

4 3 2+ - - + is 5 7x x

2+ + .

Find their LCM.

35. Find the values of a and b if the polynomial is perfect squares.

4 12 37x x x ax b4 3 2- + + +

36. If a and b are the roots of 5 1x px2- + = 0 and a b- = 1, then find p.


37. If A 1

2

1

3=


2

2- + = .


39. ABCD is a quadrilateral with AB parallel to CD. A line drawn parallel to AB meets AD at

P and BC at Q. Prove that PDAP

QCBQ

= .

40. From the top and foot of a 40 m high tower, the angles of elevation of the top of a lighthouse are found to be 30cand 60c respectively. Find the height of the lighthouse. Also find the distance of the top of the lighthouse from the foot of the tower.

41. If the total surface area of a solid right circular cylinder is 880 sq.cm and its radius is

10 cm, find its curved surface area. ( Take 722r = )


43. Find the standard deviation of the numbers 62, 58, 53, 50, 63, 52, 55.

44. A die is thrown twice. Find the probability that at least one of the two throws comes up with the number 5 (use addition theorem).

45. (a) The sum of first 10 terms of an A.P. is 25 and the common difference is twice the first term. Find the 10th term.

(OR)

(b) Find the area of the quadrilateral whose vertices are (–1, 6), (–3, –9), (5, –8) and (3, 9)

Section – DNote: (i) This section contains two questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries ten marks 2 ×10 = 20

46. (a) Construct a cyclic quadrilateral ABCD where AB = 6 cm, AD = 4.8 cm, BD = 8 cm and CD = 5.5 cm.

(OR) (b) Construct a DPQR in which the base PQ = 6 cm, 60R+ = c and the altitude from R

to PQ is 4 cm.

47. (a) Draw the graph of 2y x2

= and hence solve 2 6 0x x2+ - = .

(OR) (b) Draw the Graph of xy = 20, x , y > 0. Use the graph to find y when x 5= , and to find x when y 10= .






1. Which one of the following is not true ?

(A) \A B = A B+ l (B) \A B A B+=

(C) \ ( )A B A B B, += l (D) \ ( ) \A B A B B,=

2. If a, b, c are in G.P, then b ca b

-- is equal to

(A) ba (B)

ab (C)

ca (D)

bc

3. The 17th term of the A.P 19, 14, 9, ... is

(A) 84 (B) –61 (C) –84 (D) –51

4. If one zero of the polynomial p x^ h = ( )k x x k4 13 32+ + + is reciprocal of the other,

then k is equal to

(A) 2 (B) 3 (C) 4 (D) 5

5. The quadratic equation x kx 4 02- + = has equal roots, then the value of k is/are

(A) 4! (B) 2 (C) 3 (D) 5!

6. If A aij 2 2

=#

6 @ and ,a i jij= + then A =

(A) 1

3

2

4c m (B) 2

3

3

4c m (C) 2

4

3

5c m (D) 4

6

5

7c m

7. If the points (2, 5), (4, 6) and ,a a^ h are collinear, then the value of a is equal to

(A) -8 (B) 4 (C) -4 (D) 8

8. The angle of inclination of the straight line x y3= is

(A) 0c (B) 60c (C) 30c (D) 45c


9. In 9ABC, DE is || to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm , then AC is equal to

(A) 6.5 cm (B) 4.5 cm (C) 3.5 cm (D) 5.5 cm

10. In figure andDE BC ABC ADET T< + if AD = 1 cm, BD = 2 cm, then the ratio of the area of ABCT to the area of ADET is

(A) 1 : 9 (B) 1 : 2 (C) 9 : 1 (D) 2 : 1

11. In the adjoining figure ABC+ =

(A) 45c (B) 30c

(C) 60c (D) 05 c

12. In the figure the height of the CE is

(A) 15 cm (B) 12 cm

(C) 45 cm (D) 18 cm

13. The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio

(A) 81 : 625 (B) 729 : 15625 (C) 27 : 75 (D) 27 : 125

14. If t is the standard deviation of , ,x y z , then the standard deviation of x + 5, y + 5, z + 5 is

(A) t3

(B) t 5+ (C) t (D) x y z

15. The probabilities of three mutually exclusive events A, B and C are given by ,31

41 and

125 . Then P A B C, ,^ h is

(A) 1219 (B)

1211 (C)

127 (D) 1

Section – BNote: (i) Answer 10 questions (ii) Answer any 9 questions from the first 14 questions. Question no. 30 is compulsory. (iii) Each question carries two marks 10 × 2 = 20

16. Let U = { , , , , , , }4 8 12 16 20 24 28 , A= { , , }8 16 24 and B= { , , , }4 16 20 28 .

Find A B, l^ h .


17. Verify whether the relation f = { (1, 2), (4, 5), (9, 4), (16, 5) } is a function from A = { 1, 4, 9, 16 } to B = { – 1, 2, – 3, – 4, 5, 6 }. In case of a function, write down its range.

18. Find the quotient and remainder using synthetic division when. ( 3 5x x x3 2+ - + ) is

divided by (x 1- ). 19. If the sum and product of the roots of the quadratic equation 5 0ax x c

2- + = are both

equal to 10, then find the values of a and c.

20. If A1

2

3

2

4

5

3

5

6

=

-

-f p, then verify that ( )A AT T

= .

21. Prove that 3

1

5

2

2

1

5

3and

-

-c em o are multiplicative inverses to each other.

22. Find the equation of the straight line perpendicular to the straight line x y2 3 0- + = and passing through the point (1, -2).

23. AB and CD are two chords of a circle which intersect each other externally at P. If AB = 4 cm, BP = 5 cm and PD = 3 cm, then find CD.

24. Prove the following identity 1sin cos

cos sin cot1

2

i ii i i+

+ - =^ h

.

25. The angle of elevation of the top of a tower as seen by an observer is 30c. The observer is at a distance of 30 3 m from the tower. If the eye level of the observer is 1.5 m above the ground level, then find the height of the tower.

26. A solid right circular cylinder has raidus 7 cm and height 20cm. Find total surface area. (Take

722r = )

27. The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.

28. Find the standard deviation of the first 10 natural numbers.


52= = and ( )P A B

21, = ,then find

( )P A B+ .

30. (a) Find the sum of 1 5 52 g+ + + upto 8 terms (OR)

(b) Find the equation of the straight line whose x and y intercepts are and72

53- .


31. Let A = { , , , , , , , , , }a b c d e f g x y z , B = { , , , , }c d e1 2 and C = { , , , , , }d e f g y2 . Verify \ \ \A B C A B A C, +=^ ^ ^h h h.


32. Let A= { 5, 6, 7, 8 }; B = { –11, 4, 7, –10,–7, –9,–13 } and f = {( ,x y) : y = x3 2- , x A! , y B! }

(i) Write down the elements of f . (ii) What is the co-domain? (iii) What is the range ? (iv) Identify the type of function.

33. If the product of three consecutive terms in G.P. is 216 and sum of their products in pairs is 156, find them.

34. Find the total area of 14 squares whose sides are 11 cm, 12 cm, g , 24 cm, respectively.

35. The LCM and GCD of two polynomials are x x x4 5 13- +^ ^h h and x x5

2+^ h respectively.

If one of the polynomial p x^ h = x x x5 9 23 2- -^ h , then find the other polynomial q x^ h .

36. Solve the equation x x11

22

++

+ =

x 44+

, where x 1 0!+ , x 2 0!+ and x 4 0!+

using quadratic formula.

37. If A a

c

b

dI

1

0

0

1and

2= =c cm m , then show that ( ) ( )A a d A bc ad I2

2- + = - .

38. In what ratio is the line joining the points (-5, 1) and (2 , 3) divided by the y-axis? Also, find the point of intersection .

39. The vertices of TABC are A (1, 8), B (-2, 4), C (8, -5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN is parallel to BC.

40. The image of a tree on the film of a camera is of length 35 mm, the distance from the lens to the film is 42 mm and the distance from the lens to the tree is 6 m. How tall is the portion of the tree being photographed?

41. If tan tanni a= and ,sin sinmi a= then prove that cosn

m

1

12

22i =

-

- .

42. Radius and slant height of a solid right circular cone are in the ratio 3 : 5. If the curved surface area is 60r sq.cm, then find its total surface area.

43. The mean and the standard deviation of a group of 20 items was found to be 40 and 15 respectively. While checking it was found that an item 43 was wrongly written as 53. Calculate the correct mean and standard deviation.


45. (a) A solid sphere of diameter 42 cm is melted and recast into a number of smaller identical cones, each of diameter 7 cm and height 3 cm. Find the number of cones so formed

(OR)

(b) Find the square root of 81 72 70 24 9x x x x4 3 2- + - + .


Section – DNote: (i) This section contains 2 questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries 10 marks 2 × 10 = 20

46. (a) Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 6 cm. Also, measure the lengths of the tangents.

(OR)

(b) Construct a cyclic quadrilateral PQRS given PQ = 5 cm, QR = 4 cm, 35QPR+ = c and 70PRS+ = c.

47. (a) A cyclist travels from a place A to a place B along the same route at a uniform speed on different days. The following table gives the speed of his travel and the corresponding time he took to cover the distance.

Speed in km/hr x 2 4 6 10 12Time in hrs y 60 30 20 12 10

Draw the speed-time graph and use it to find

(i) the number of hours he will take if he travels at a speed of 5 km/hr

(ii) the speed with which he should travel if he has to cover the distance in 40 hrs.

(OR)

(b) Draw the graph of 12y x x2

= + - and hence solve 2 2 0x x2+ + = .






1. If A = { 5, 6, 7 }, B = { 1, 2, 3, 4, 5 }and :f A B" is defined by ( )f x x 2= - , then the range of f is

(A) { 1, 4, 5 } (B) { 1, 2, 3, 4, 5 } (C) { 2, 3, 4 } (D) { 3, 4, 5 }

2. The next term of 201 in the sequence , , , ,

21

61

121

201 g is

(A) 241 (B)

221 (C)

301 (D)

181

3. If a, b and c are the three consecutive terms of a G.P., then

(A)ac

cb= (B)

ac

ab= (C)

ac

ab 2

= ` j (D) ca

ab=

4. If a ba3

- is added with

b ab3

-, then the new expression is

(A) a ab b2 2+ + (B) a ab b

2 2- + (C) a b

3 3+ (D) a b

3 3-

5. a ba b

b aa b

-+ -

-- =

(A) 1 (B) a b

b2-

(C) b a

b2-

(D) ( )a ba b2-+

6. If x

y

1

2

2

1

2

4=c c cm m m, then the values of x and y respectively, are

(A) 2 , 0 (B) 0 , 2 (C) 0 , 2- (D) 1 , 1

7. Slope of the straight line which is perpendicular to the straight line joining the points ,2 6-^ h and ,4 8^ h is equal to

(A) 31 (B) 3 (C) -3 (D)

31-

8. Angle of inclination of the line joining the two points (2, ) (5, 2 )and3 3 is

(A) 30c (B) 45c (C) 60c (D) 90c


9. In figure, if ACAB

DCBD= , ,B 40c+ = and ,C 60c+ = then BAD+ =

(A) 30c (B) 50c

(C) 80c (D) 40c

10. In , andABC DE BCDBAD

32T < = . If AE = 6 then BC is

(A) 9 (B) 18 (C) 15 (D) 12

11. If secx a i= , tany b i= , then the value of ax

b

y2

2

2

2

- =

(A) 1 (B) –1 (C) tan2i (D) cosec2i

12. cosec cot

tan sec2 2

2 2

i i

i i

-

- =

(A) 1 (B) –1 (C) sini (D) cosi

13. The ratios of the respective heights and the respective radii of two cylinders are 1:2 and 2:1 respectively. Then their respective volumes are in the ratio

(A) 4 : 1 (B) 1 : 4 (C) 2 : 1 (D) 1 : 2

14. Standard deviation of a collection of data is 2 2 . If each value is multiplied by 3, then the standard deviation of the new data is

(A) 12 (B) 4 2 (C) 6 2 (D) 9 2

15. The probability that a leap year will have 53 Fridays or 53 Saturdays is

(A) 72 (B)

71 (C)

74 (D) 7

3

Section – BNote: (i) Answer 10 questions (ii) Answer any 9 questions from the first 14 questions. Question no. 30 is compulsory. (iii) Each question carries 2 marks 10 × 2 = 20 16. Draw the Venn diagram \B C A,^ h .

17. Let ifx

x x

x x

0

0if 1

$=

-' , where .x Rd Does the relation { ( ,x y) | y = |x |, x R! }

define a function? Find its range.

18. Find the zeros of the quadratic polynomial 6 3 7x x2- - and verify the basic relationships

between the zeros and the coefficients.

19. If a and b are the roots of the equation 0x x3 5 22- + = , find the value of

2 2

ba

ab

+

20. Let A 3

5

2

1= c m and B 8

4

1

3=

-c m. Find the matrix C if C A B2= + .

21. Solve : y

x

x

y3

6 2

31 4=

-

+c em o.


22. The side BC of an equilateral ABCT if parallel to x-axis. Find the slope of AB and the slope of BC

23. If the area of the ABCT is 12 sq.units and the vertices are A (a, –3), B (3, a) and C(–1, 5) taken in order, then find the value of a.

24. In a MNO3 , MP is the external bisector of M+ meeting NO produced at P. If MN = 10 cm, MO = 6 cm, NO = 12 cm, then find OP.

25. Prove that secsec

cossin1

1

2

ii

ii+ =

-.

26. A pendulum of length 40 cm subtends 60c at the vertex in one full oscillation. What will be the shortest distance between the initial position and the final position of the bob?

27. The radii of two circular ends of a frustum shaped bucket are 15 cm and 8 cm. If its depth is 63 cm, find the capacity of the bucket in litres. ( Take

722r = )

28. The weight (in kg) of 13 students in a class are 42.5, 47.5, 48.6, 50.5, 49, 46.2, 49.8, 45.8, 43.2, 48, 44.7, 46.9, 42.4. Find the range and coefficient of range.

29. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of drawing a red ball, then find the number of blue balls in the bag.

30. (a) Find the sum of all odd natural number between 6 and 40(OR)

(b) Find the curved surface area of a cone whose volume and height are 120 cm3r and 10 cm respectively.

Section – CNote: (i) Answer 9 questions (ii) Answer any 8 questions from the first 14 questions. Question no. 45 is compulsory. (iii) Each question carries 5 marks 9 × 5 = 45

31. Use Venn diagrams to verify A B A B+ ,=l l l^ h . 32. Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and :f A B" be defined by

f x^ h = x33- . Represent f by

(i) an arrow diagram (ii) a set of ordered pairs (iii) a table (iv) a graph . 33. Find the sum of first 20 terms of the arithmetic series in which 3rd term is 7 and 7th term

is 2 more than three times its 3rd term. 34. The first term of a geometric series is 375 and the fourth term is 192. Find the common

ratio and the sum of the first 14 terms.

35. Factorize 2 3 3 2x x x3 2- - + into linear factors.

36. Simplify xx

x

xxx

12 5

1

11

3 22

2

++ +

-

+ ---` j= G in the simplest form.


37. If ,A B C3

7

3

6

8

0

7

9

2

4

3

6and= = =


Is ( )A B C AC BC+ = + ? 38. The vertices of a 3ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the

median through the vertex A. 39. State and prove the converse of Thales Theorem. 40. From the top of a tower of height 60 m, the angles of depression of the top and the bottom

of a building are observed to be 30 60andc crespectively. Find the height of the building. 41. The total surface area of a solid right circular cylinder is 231 cm2 . Its curved surface area

is two thirds of the total surface area. Find the radius and height of the cylinder. 42. A circus tent is to be erected in the form of a cone surmounted on a cylinder. The

total height of the tent is 49 m. Diameter of the base is 42 m and height of the cylinder is 21 m. Find the cost of canvas needed to make the tent, if the cost of canvas is `12.50/m2 . ( Take r =

722 )

43. Find the standard deviation of the following distribution.

x 70 74 78 82 86 90f 1 3 5 7 8 12


45. (a) If 25x x x ax b30 114 3 2- - + - is a perfect square, then find a and b

(OR)

(b) Find the equation of a straight line which is passing through the point of intersection

of the straight lines x y2 3 1 0+ - = and x y3 2 4+ = , and the midpoint of the straight

line joining the two points ,83

107-` j and ,

87

103- -` j

Section – DNote: (i) This section contains 2 questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries10 marks. 2 ×10 = 20

46. (a) Draw a circle of radius 3 cm. From an external point 7 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

(OR) (b) Construct a ABCD in which the base BC = 5 cm, 40BAC+ = c and the median from

A to BC is 6 cm. Also, measure the length of the altitude from A. 47. (a) Solve the equation graphically. x x2 1 3 0+ - =^ ^h h .

(OR) (b) A bus travels at a speed of 40 km / hr. Write the distance-time formula and draw the

graph of it. Hence, find the distance travelled in 3 hours.






1. If ( )f x x 52= + , then ( )f 4- =

(A) 26 (B) 21 (C) 20 (D) –20

2. If the nth term of an A.P. is 3 5t nn= - , then the sum of the first n terms is

(A) n n21 5-6 @ (B) n n1 5-^ h (C) n n

21 5+^ h (D) n n

21 +^ h

3. Find the common ratio of the sequence 4, –2, +1, –7, .... (A) 4 (B) –2 (C)

21 (D)

21-

4. The common root of the equations 0x bx c2- + = and x bx a 0

2+ - = is

(A) b

c a2+ (B)

bc a2- (C)

ac b2+ (D)

ca b2+

5. w s

x y z

64

8112 14

4 6 8

=

(A) w s

x y z

8

912 14

4 6 8

(B) w s

x y z

8

912 14

2 3 4

(C) w s

x y z

8

96 7

2 3 4

(D) x y z

w s

9

82 3 4

6 7

6. If 20x5 1

2

1

3

- =^ f ^h p h, then the value of x is

(A) 7 (B) 7- (C) 71 (D) 0

7. The slope of the straight line y x7 2 11- = is equal to (A)

27- (B)

27 (C)

72 (D)

72-

8. Mid point of the line segment joining the points (1, –1) and (–5, 3) is

(A) (–2, 1) (B) (2, –1) (C) (–2, –1) (D) (–1, –2)

9. In the adjoining figure, chords AB and CD intersect at P. If AB = 16 cm, PD = 8 cm, PC = 6 and AP > PB, then AP =

(A) 8 cm (B) 4 cm (C) 12 cm (D) 6 cm


10. In the figure PQ is a tangent, BAQ 62+ = c and BAC 52+ = c, then ACB+ =

(A) 64c (B) 90c

(C) 54c (D) 62c

11. 11

cottan

2

2

i

i

++ =

(A) cos2i (B) tan2i (C) sin2i (D) cot2i

12. sin cos tan coti i i i+ =^ h

(A) 0 (B) 2 (C) 1 (D) tani 13. If the radius of a sphere is half of the radius of another sphere, then their respective

volumes are in the ratio (A) 1 : 8 (B) 2: 1 (C) 1 : 2 (D) 8 : 1 14. For any collection of n items, ( )x xR - = (A) nx (B) ( 2)n x- (C) ( 1)n x- (D) 0 15. A fair die is thrown once. The probability of getting a prime or composite number is

(A) 1 (B) 0 (C) 65 (D)

61

Section – BNote: (i) Answer 10 questions (ii) Answer any 9 questions from the first 14 questions. Question no. 30 is compulsory. (iii) Each question carries 2 marks 10 × 2 = 20

16. For the given sets { 10,0,1, 9, 2, 4, 5} { 1, 2, 5, 6, 2,3,4}A Band= - = - - , verify that the set intersection is commutative


f x^ h = ;

;

;

x x

x x

x x

4 1 3 2

3 2 2 4

2 3 4 7

2 1

1 1

#

# #

- -

-

-

* . Find f f1 3- -^ ^h h

18. If a, b, c are in A.P. then prove that ( ) 4( )a c b ac

2 2- = - .

19. Solve the system of equations by elimination method. x y2 7+ = , x y2 1- =

20. Find the product of the matrices, if exists, 3

5

2

1

4

2

1

7

-c cm m

21. Matrix A shows the weight of four boys and four girls in kg at the beginning of a diet programme to lose weight. Matrix B shows the corresponding weights after the diet

programme. ,A B35

42

40

38

28

41

45

30

32

40

35

30

27

34

41

27

BoysGirls

BoysGirls= =c cm m Find the

weight loss of the Boys and Girls.

22. Find the point which divides the line segment joining the points (3 , 5) and (8 , 10) internally in the ratio 2 : 3.


23. In PQRT , AB ;;QR. If AB is 3 cm, PB is 2cm and PR is 6 cm, then find the length of QR.

24. A ramp for unloading a moving truck, has an angle of elevation of 30c. If the top of the ramp is 0.9 m above the ground level, then find the length of the ramp.

25. Prove the following identity cos

sin cosec cot1 i

i i i-

= + .


27. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.

28. Find the range and the coefficient of range of 43, 24, 38, 56, 22, 39, 45. 29. An integer is chosen from the first twenty natural numbers. What is the probability that it

is a prime number? 30. (a) Find the square root of 1

12x

x

166

+ ++

+^^

hh

. (OR)

(b) Find the equation of the straight line whose angle of inclination is 60° and y-intercept is

3

1 .


31. An advertising agency finds that, of its 170 clients, 115 use Television, 110 use Radio and 130 use Magazines. Also, 85 use Television and Magazines, 75 use Television and Radio, 95 use Radio and Magazines, 70 use all the three. Draw Venn diagram to represent these data. Find

(i) how many use only Radio? (ii) how many use only Television? (iii) how many use Television and magazine but not radio?

32. A function : [1, 6)f R$ is defined as follows,

,

,

f x

x x

x x

x x

1 1 2

2 1 2 4

3 10 4 62

1

1

1

#

#

#

=

+

-

-

^ h *

Here, [1 , 6) = { x Re : 1# x 1 6} ) Find the value of

(i) ( )f 5 (ii) f 3^ h (iii) f 1^ h (iv) f f2 4-^ ^h h (v) 2 3f f5 1-^ ^h h

33. The ratio of the sums of first m and first n terms of an arithmetic series is :m n2 2 show

that the ratio of the mth and nth terms is :m n2 1 2 1- -^ ^h h

34. Solve x y3 2 +^ h = xy7 ; x y3 3+^ h = xy11 using elimination method

35. Find the square root of the following: x x x x x x6 5 6 6 2 4 8 32 2 2+ - - - + +^ ^ ^h h h

36. A car left 30 minutes later than the scheduled time. In order to reach its destination 150 km away in time, it has to increase its speed by 25 km/hr from its usual speed. Find its usual speed.


37. Find X and Y if 2 3X Y2

4

3

0+ = c m and 3 2X Y

2

1

2

5+ =

-

-e o.

38. Find the area of the quadrilateral formed by the points (-4, -2), (-3, -5), (3, -2) and (2 , 3).

39. Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5.

40. State and prove Angle Bisector Theorem. 41. A girl standing on a lighthouse built on a cliff near the seashore, observes two boats due

East of the lighthouse. The angles of depression of the two boats are 30° and 60°. The distance between the boats is 300 m. Find the distance of the top of the lighthouse from the sea level.

42. A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out soil is evenly spread to form a cuboid-platform with base dimension 20 m × 14 m. Find the height of the platform.

43. The time (in seconds) taken by a group of people to walk across a pedestrian crossing is given in the table below.

Time (in sec) 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30No. of people 4 8 15 12 11

Calculate the variance and standard deviation of the data.

44. The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is 0.15. Find the probability that

(i) it will get atleast one of the two awards (ii) it will get only one of the awards. 45. (a) Find the sum upto n terms of the series . . .0 7 0 97 0 997 g+ + + . (OR) (b) A solid metalic cylinder of diameter 4cm and height 45cm. is melted and recast into

identical spherical shots of radius 3cm each. Find the number of spherical shots.

Section – DNote: (i) This section contains two questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries ten marks 2 ×10 = 20 46. (a) Construct a ABCD in which BC = 5.5 cm., 60A+ = c and the median AM from

the vertex A is 4.5 cm.(OR)

(b) Construct a cyclic quadrilateral ABCD, given AB = 6cm, 70ABC+ = c, BC = 5 cm and 30ACD+ = c.

47. (a) Draw the graph of 2 3y x x2

= + - and hence find the roots of 6 0x x2- - = .

(OR) (b) The following table gives the cost and number of notebooks bought.

No. of note books x 2 4 6 8 10 12Cost `y 30 60 90 120 150 180

Draw the graph and hence (i) Find the cost of seven note books. (ii) How many note books can be bought for ` 165.

Evaluation - Deptl. Model Question Paper 393

scheme of evaluation - mathematics

GENERAL GUIDELINES

(i) The answers given in the Scheme of Evaluation are based on the Text Book and SCORE Book.

(ii) Full Credit (marks) should be awarded to a student if his / her approach in giving solution to a problem is correct and leading to required answer. This approach may be Text Book oriented / SCORE Book oriented / Mathematically correct.

(iii) Marks should be allotted according to the different stages required to arrive at the appropriate answer.

(iv) Stage marks are meant for only when a student gives partial answer/commits mistakes/furnishes irrelevant information in the course of his/her answer to a particular question.

(v) While answering a question, if a student starts from a stage with correct step but reaches the next stage with a wrong result, then suitable credits should be given to the correct steps instead of denying the entire marks allotted for that stage.

(vi) If the solution to a particular question, other than questions under Section D, requires a diagram, then a rough sketch of the diagram is enough. Full Credit must be given for such a rough diagram.

(vii) Full credit must be given for an equivalent answer wherever possible.

(viii) There is no separate marks allotted for formula. If a particular stage is wrong and if the student writes the appropriate formula, then suitable mark, which is attached with that stage, should be awarded for the formula. Mark should not be deducted for not writing the formula if the student arrives at the correct answer.


X Std. Mathematics

Departmental Question Paper - Distribution of Questions

section-a section-B section-cQ.No. Chapter Exercise Creative Q.No. Chapter Exercise Example Q.No. Chapter Exercise Example

1 Set ü 16 Set ü 31 Set ü

2 Seq. ü 17 Set ü 32 Set ü

3 Seq. ü 18 Seq. ü 33 Seq. ü

4 Alg. ü 19 Alg. ü 34 Alg. ü

5 Alg. ü 20 Mat. ü 35 Alg. ü

6 Mat. ü 21 Mat. ü 36 Alg. ü

7 Co-or. ü 22 Co-or. ü 37 Mat. ü

8 Co-or. ü 23 Geo. ü 38 Co-or. ü

9 Geo. ü 24 Tri. ü 39 Co-or. ü

10 Geo. ü 25 Tri. ü 40 Geo. ü

11 Tri. ü 26 Men. ü 41 Tri. ü

12 Tri. ü 27 Men. ü 42 Men. ü

13 Men. ü 28 Stat. ü 43 Stat. ü

14 Stat. ü 29 Prob. ü 44 Prob. ü

15 Prob. ü Creative Questions Creative Questions

30 a Alg. ü 45 a Seq. ü

30 b Co-or. ü 45 b Men. üsection-D

Q.No. Chapter Exercise Example Q.No. Chapter Exercise Example

46 a Prac.Geo

ü 47 aGraph

ü

46 b ü 47 b ü

Distribution of Questions : consolidated table

Type of Questions Exercise Example CreativeObjectiveType Questions 10 - 52 Mark Questions 8 6 25 Mark Questions 8 6 210 Mark Questions 3 1 -


Departmental Model Question Paper – EvaluationTime: 2.30 Hours] [Maximum marks: 100




1. Let A = { 1, 3, 4, 7, 11 }, B = {–1, 1, 2, 5, 7, 9 } and :f A B" be given byf = { (1, –1), (3, 2), (4, 1), (7, 5), (11, 9) }. Then f is

(A) one-one (B) onto (C) bijective (D) not a function

( Ans. (A) )

2. The common ratio of the G.P. , , ,52

256

12518

62554 g is

(A) 52 (B) 5 (C)

53 (D)

54 ( Ans. (C) )

3. If , , ,a a a1 2 3

g are in A.P. such that ,a

a

23

7

4 = then the 13th term of the A.P. is

(A) 23 (B) 0 (C) a12 1 (D) a14 1

Solution: 2( a + 3d ) = 3( a + 6d ) & 3a + 18d – 2a – 6d = 0 & a + 12d = 0 t 0

13& = ( Ans (B) )

4. The LCM of , ,x y x yz x y z6 9 122 2 2 2 is

(A) x y z36 2 2 (B) xy z36 2 2 (C) 6x y z3 2 2 2 (D) xy z36 2

Solution: 6x y2 = x y2 3 2# LCM = 2 3 x y z2 2 2 2

#

x yz9 2 = x yz32 2 = 4 9x y z2 2#

x y z12 2 2 = x y z2 32 2 2# = x y z36 2 2 (Ans (A))

5. Let b = a + c . Then the equation 0ax bx c2+ + = has equal roots, if

(A) a = c (B) a = – c (C) a = 2c (D) a = – 2cSolution: 4 4 0 .b ac a c ac a c a c2 2 2& & &= + = - = =^ ^h h (Ans. (A) )


6. If A 1

0

1

21 2# =c ^m h, then the order of A is

(A) 2 1# (B) 2 2# (C) 1 2# (D) 3 2#

Solution: A 1

0

1

21 2# =c ^m h, ( )A

1

0

1

21 2

m n2 2

1 2=

##

#c m

2 1. ,Son m Aand& = = is of order 1 2# (Ans. (C) )

7. The slope of the straight line y x7 2 11- = is equal to

(A) 27- (B)

27 (C)

72 (D) 7

2-

Solution: Slope .mba

72

72= - =- - =` j (Ans. (C) )

8. The perimeter of a triangle formed by the points (0, 0), (1, 0), (0,1) is

(A) 2 (B) 2 (C) 2 2+ (D) 2 2-

Solution: AB = 1, BC = 2 , AC = 1

Perimeter = AB + BC + CA = 1 2 1+ + = 2 2+ (Ans. (C) )

9. In 9PQR, RS is the bisector of R+ . If PQ = 6 cm, QR = 8 cm, RP = 4 cm, then PS is equal to

(A) 2 cm (B) 4 cm

(C) 3 cm (D) 6 cm

Solution: Let cmPS x= . (6 ) cmSQ x= -

RS is the bisector of PRQ+ , we have

QRPQ =

SQPS

84

21= =

2 6 2x

x x x x6 2

1& & &-

= = - = (Ans. (A) )

10. Chords AB and CD of a circle intersect at P inside the circle. If AB = 7, AP = 4, CP = 2, then CD =

(A) 4 (B) 8 (C) 6 (D) 10

Solution: AP × PB = CP × PD & 4×3 = 2 × PD& PD212= = 6

2 6 8CD CP PD` = + = + = (Ans. (B) )

11. A man is 28.5 m away from a tower. His eye level above the ground is 1.5 m. The angle of elevation of the tower from his eyes is 45c. Then the height of the tower is

(A) 30 m (B) 27.5 m (C) 28.5 m (D) 27 m

Solution: Height of the tower = tanx y i+

P

S

Q R8cm

4cm6cm


= 1.5 28.5 45tan o#+

= 1.5 28.5 30m+ = (Ans. (A) )

12. tan cot

1i i+

=

(A) sin cosi i+ (B) sin cosi i (C) sin cosi i- (D) cosec coti i+

Solution: tan cot

cossin

sincos sin cos

sin cos1 12 2i i

ii

ii i i

i i+

=+

=+

= sin cosi i (Ans. (B) )

13. If the total surface area of a solid hemisphere is 12 cm2r , then its curved surface area is equal to

(A) 6r cm2 (B) 24r cm2 (C) 36r cm2 (D) 8r cm2

Solution: TSA , r3 2r = 12r cm2 r 42

& =

CSA, r2 2r = 8r cm2 (Ans. (D) )

14. Meanandstandarddeviationofadataare48and12respectively.Thecoefficientofvariation is

(A) 42 (B) 25 (C) 28 (D) 48

Solution: Given that v = 12 and x = 48


= 1004812 25# = . (Ans. (B) )

15. If S is the sample space A and B are mutually exclusive events and such that

( ) ( )P A P B31= and A B S, = , then ( )P A =

(A) 41 (B)

21 (C)

43 (D) 8

3

Solution: ( ) ( ) ( )P A B P A P B, = + (A , B are mutually exclusive events )

( ) ( ) ( )P S P A P A3= + 4 ( ) .P A 1& =

Thus, ( )P A41= . (Ans. (A) )

Section – BNote: (i) Answer 10 questions. (ii) Answer any 9 questions from the first 14 questions. Question no. 30 is compulsory. (iii) Each question carries 2 marks. 10 × 2 = 20

16. If {4,6,7,8,9}, {2,4,6} {1,2,3,4,5,6}A B Cand= = = ,thenfindA B C, +^ h Solution: B C+ = {2, 4, 6} + {1, 2, 3, 4, 5, 6} = {2, 4, 6}. ... 1 mark

( )A B C, + = {4, 6, 7, 8, 9} , {2, 4, 6} = {2, 4, 6, 7, 8, 9}. ... 1 mark


17. Let X = { 1, 2, 3, 4 }. Examine whether the relation

g = { (3, 1), (4, 2), (2, 1) } is a function from X to X or not. Explain.

Solution: The element 1 does not have a image. Domain of {2, 3, 4}g X!= . ... 1 mark

Hence, the relation g = { (3, 1), (4, 2), (2, 1)} is not a function. ... 1 mark

18. Threenumbersareintheratio2:5:7.Ifthefirstnumber,theresultingnumberon subtraction of 7 from the second number and the third number form an arithmetic sequence, thenfindthenumbers.

Solution: Let the numbers be 2 ,5 7x x xand for some unknown x,(x 0! ).

By the given information, we have 2 , 5 7, 7x x x- are in A.P.

2 ( )x x x x5 7 7 5 7- - = - -^ h ( 3 7x x2 7- = + and so x = 14. ... 1 mark

Thus, the required numbers are 28, 70, 98. ... 1 mark

19. If a and b are the roots of the equation 2 3 1 0x x2- - = , find the value of

a b- if >a b .

Solution: Given equation is x x2 3 1 02- - = .

Comparing the given equation with 0ax bx c2+ + = , we get

a 2= , b 3=- , c 1=- . Given a and b are the roots of the equation.

` a b+ = ab- =

2

3- -^ h = 23 and

ac

21ab = =- ... 1 mark

a b- = 423 4

21

2172

2a b ab+ - = - - =^ ` `h j j ... 1 mark

20. If A 2

9

3

5

1

7

5

1=

--

-e eo o,thenfindtheadditiveinverseofA.

Solution: A = 2

9

3

5

1

7

5

1--

-e eo o 2

9

3

5

1

7

5

1=

-+

-

-

-e eo o

= 1

16

2

6-

-e o ... 1 mark

The additive inverse of A is –A. Hence, the additive inverse is

A- = 1

16

2

6

1

16

2

6-

-

-=

-

-e eo o ... 1 mark

21. Find the product of the matrices, if exists, 2

4

9

1

3

0

4

6

2

2

7

1-

--

-

e fo p.

Solution: Let A = 2

4

9

1

3

0-

-e o and B =

4

6

2

2

7

1

-

-

f p.

The Order of A is 2 # 3 and the order B is 3 # 2. Thus, the product ABisdefined ... 1 mark


AB = 2

4

9

1

3

0

4

6

2

2

7

1-

--

-

e fo p = 8 54 6

16 6 0

4 63 3

8 7 0

- +

+ -

+ -

- +e o

= 40

22

64

1

-c m (or) ( For correct answer award 2 marks.) ... 1 mark

22. The centre of a circle is at (-6, 4). If one end of a diameter of the circle is at the origin, thenfindtheotherend.Solution: Let ( , )x y be the other end of the diameter.Centre of the circle is the midpoint of the diameter.

Thus, ,xy

20

20+ +

c m = (– 6, 4) ... 1 mark

Equating x and y-coordinates,

we get 6 12x x2

0 &+ =- =- and 4 8.y

y2

0&

+= =

Thus, the other end of the diameter is ( , )12 8- . ... 1 mark

23. In ABCT , DE BC< and DBAD

32= . If AE=3.7cm,findEC.

Solution: In ABC3 , DE BC<

` DBAD = EC

AE (Thales theorem) ... 1 mark

& EC = AD

AE DB#

Thus, EC = .2

3 7 3# = 5.55 cm ... 1 mark

24. A ladder leaning against a vertical wall, makes an angle of 60c with the ground. The foot of the ladder is 3.5 m away from the wall. Find the length of the ladder.

Solution: Let AC denote the ladder and B be the foot of the wall.

Given that CAB 60+ = c and AB =3.5 m.

In the right 3ABC, cos60c = ACAB ... 1 mark

& AC = 60cos

ABc

= 2 3.5 7# = m ... 1 mark

Thus, the length of the ladder is 7 m..

25. Prove the identity 1cosecsin

seccos

ii

ii+ = .

Solution: Now, cosecsin

seccos

ii

ii+ =

sin

sin

cos

cos1 1i

i

i

i+` `j j

... 1 mark

= sin cos2 2i i+ = 1. ... 1 mark

B

D3.7cm

C

E

A

m


26. A solid right circular cylinder has radius of 14 cm and height of 8 cm . Find its curved surface area and total surface area.Solution: Given that radius, r = 14 cm and height, h = 8 cm Curved Surface Area = 2 rhr

= 2 × × ×722 14 8= 704 sq.cm ... 1 mark

Total Surface Area = 2 ( )r h rr +

= 2 × 722 ×14 (8 + 14)

Thus, the total surface area = 88 × 22 = 1936 sq.cm. ... 1 mark

27. The circumference of the base of a 12 cm high wooden solid cone is 44 cm.

Find the volume.Solution: Let r and h be the radius and height of the wodden solid cone respectively.Given that h = 12 cm Base circumference of the wodden solid, 2 rr = 44

& r = 244r

= 2 2244 7## = 7 cm ... 1 mark

Volumeofthewoodensolid = r h31 2r

= 7 131

722 22

# # # cm3

= 166 cm3. ... 1 mark

28. Calculatethestandarddeviationofthefirst13naturalnumbers.Solution:

TheSDofthefirstn natural numbers v = n12

12-

` SDofthefirst13naturalnumbersv = n12

12- =

1213 1

2- ... 1 mark

= .12168 14 3 74= = ... 1 mark

29. Two coins are tossed together. What is the probability of getting at most one head?.

Solution: The sample space is { , , , } ; ( ) 4.S HH HT TH TT n S= = ... 1 mark

Let A be the event of getting atmost one head.

Thus, A = , ,HT TH TT" , and 3n A =^ h .

Hence, ( )P A = n S

n A

43=

^^

hh . ... 1 mark

30. (a) Simplify x x

x

7 12

6 542

2

+ +

-

(OR) (b) Show that the lines y x2 4 3= + and x y2 10+ = are perpendicular.


Solution: (a) x x

x

7 12

6 542

2

+ +

- = x x

x4 3

6 92

+ +-

^ ^^

h hh

= x x

x x

4 3

6 3 3

+ +

+ -

^ ^^ ^

h hh h ... 1 mark

= x

x

4

6 3

+

-^ h ... 1 mark

(b) 2y = 4x + 3 & 4x – 2y + 3 = 0.

Slope m1 = coefficient of

coefficient ofyx- = 2

2

4

-

-=

^ h

For x y2 10 0+ - = , slope m2 =

21- ... 1 mark

Now, 2m m21

1 2#= - = – 1 ... 1 mark

Hence, the given lines are perpendicular.

Section – CNote: (i) Answer 9 questions. (ii) Answer any 8 questions from the first 14 questions. Question no. 45 is compulsory. (iii) Each question carries 5 marks 9 × 5 = 45

31. UseVenndiagramstoverifyDeMorgan’slawforsetdifference\ \ \A B C A B A C+ ,=^ ^ ^h h h.

Solution: (each diagram carries 1 mark)

From (2) and (5), we have \ \ \A B C A B A C+ ,=^ ^ ^h h h.


32. A function f : ,7 6- h6 " R isdefinedasfollows

( )f x = ;

;

; .

x x x

x x

x x

2 1 7 5

5 5 2

1 2 6

2 1

1 1

#

# #

+ + - -

+ -

-

* Find ( ) ( )( ) ( )

f ff f

6 3 14 3 2 4

- -- + .

Solution: When 6x =- , ( ) 2 1f x x x2= + +

( )f 6- = ( ) ( )6 2 6 1 36 12 1 252- + - + = - + = . ... (1 mark)

When 3 1, ( ) 5x f x xand the function is=- = +

Thus, ( )f 3- = 2 and (1) 6f = . ... (2 marks)

When 4, ( ) 1.x f x xthe function is= = - So, (4) 3.f = ... (1 mark)

( ) ( )( ) ( )

f ff f

6 3 14 3 2 4

- -- + =

25 3 64 2 2 3

25 188 6

714 2

## #

-+ =

-+ = = . ... (1 mark)

33. Findthesumofthefirst2n terms of the series 1 2 3 4 ...2 2 2 2- + - + .

Solution: 1 2 3 42 2 2 2

g- + - + to n2 terms

= 1 4 9 16 25 g- + - + - to n2 terms

= 1 4 9 16 25 36 g- + - + - +^ ^ ^h h h to n terms. (after grouping)

= 3 7 11 g- + - + - +^ ^h h n terms ... (1 mark)

Now,theaboveseriesisinA.P.withfirstterm 3a =- and common difference d 4=-

. ... (1 mark)

Therefore, the required sum = n a n d2

2 1+ -^ h6 @

= n n2

2 3 1 4- + - -^ ^ ^h h h6 @ ... (2 marks)

= n n2

6 4 4- - +6 @ = n n2

4 2- -6 @

= n n22 2 1- +^ h = n n2 1- +^ h ... (1 mark)

34. Factorize the polynomial 2 5 6x x x3 2- - + .


Sumofthecoefficientsof ( )p x is 1 5 2 24 0!- - + .Thus, ( )x 1- is not a factor of ( )p x .Also, ( )p 1- = 1 5 2 24 0!- - + + . Thus, 1x + is not a factor.By trial and error, we see that ( ) .p 2 0- = Thus, ( )x 2+ is a factor. ... (1 mark)

– 2 1 – 5 – 2 240 – 2 14 – 241 – 7 12 0 " Remainder ... (2marks)

The other factor is ( )( )x x x x7 12 3 42- + = - - . ... (1 mark)


Thus, 5 2 24x x x3 2- - + = ( )( )( )x x x2 3 4+ - - ... (1 mark)

35. If 28 12 9m nx x x x2 3 4

- + + + isaperfectsquare,thenfindthevaluesofm and n.

Solution: Arrange the polynomial in descending powers of x. 9 12 28x x x nx m

4 3 2+ + - +

3 2 4x x2+ +

3 x2

9 12 28x x x nx m4 3 2+ + - +

9x4

... (2 marks)

6 2x x2+ 12 28x x

3 2+

12 4x x3 2+

... (1 mark)

6 4 4x x2+ + 24x nx m

2- +

24 16 16x x2+ + ... (1 mark)

0

Since the given polynomial is a perfect square, we must have n = –16 and m = 16. ... (1 mark)

36. The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and return downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream.Solution: Let the speed of the stream be x km/hr.Given that the speed of the boat in still water is 15 km/hr.Thus, the speed of the boat in the downstream and in the upstream are ( )x15 + km/hr and ( )x15 - km/hr respectively .... (1 mark)Let T

1 be the time taken to cover the distance of 30 km in the downstream.

Let T2

be the time taken to cover the same distance in the upstream.


1 =

x1530+

and T2

= x15

30-

. .... (1 mark)

Now, T1 + T

2 = 4 hrs. 30 min. = 4

21 hrs.

& x x15

301530

-+

+ =

29 .... (1 mark)

& ( )( )

( ) ( )x x

x x15 15

30 15 30 15- +

+ + - = 29

& ( )x9 225 2- = 1800

& x225 2- = 200 .... (1 mark)

& x = 5!

Since the speed of the stream cannot be negative, we take .x 5= Thus, the speed of the stream is 5 km/hr. .... (1 mark)

37. If A B5

7

2

3

2

1

1

1and= =

-

-c em o verify that ( )AB B A

T T T= .


Solution: A = 5

7

2

3c m, B = 2

1

1

1-

-e o

A is of order 2 #2 and B is of order 2 #2. Hence, AB is of order 2 #2

AB = 5

7

2

3

2

1

1

1-

-c em o

= 10 2

14 3

5 2

7 3

-

-

- +

- +e o = 8

11

3

4

-

-e o .... (1 mark)

(AB)T = 8

3

11

4- -e o ... (1) .... (1 mark)

Now, BT = 2

1

1

1-

-e o .... (1 mark)

AT = 5

2

7

3c m .... (1 mark)

BT AT = 2

1

1

1

5

2

7

3-

-e co m = 10 2

5 2

14 3

7 3

-

- +

-

- +e o

= 8

3

11

4- -e o ... (2) .... (1 mark)

From (1) and (2), we have (AB)T = BT AT

38. Find the area of the quadrilateral formed by the points (-4, -2), (-3, -5), (3, -2) and (2 , 3).

Solution: Let us plot the points roughly and take the vertices in counter clock-wise direction.

Let the vertices be A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3).

21 4

2

3

5

3

2

2

3

4

2

-

-

-

- -

-

-) 3 .... (2 marks)


= 21 20 6 9 4 6 15 4 12+ + - - - - -^ ^h h" , .... (2 marks)

= 21 31 25+" , = 28 sq.units. .... (1 mark)

39. The vertices of ABCT are A (2, 1), B (6, –1) and C (4, 11). Find the equation of the straight line along the altitude from the vertex A.

Solution: Slope of BC = 4 611 1-+ = – 6 .... (1 mark)


Since the line AD is perpendicular to the line BC, slope of AD = 61 .... (1 mark)

Equation of AD is y y1

- = m x x1

-^ h .... (1 mark)

y 1- = x61 2-^ h ( y6 6- = x 2- .... (1 mark)

Thus, equation of the required straight line is x y6 4- + = 0. .... (1 mark) 40. Aboyisdesigningadiamondshapedkite,asshowninthefigure

where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?

Solution: We know that if a perpendicular is drawn from the vertex of a right angled triangle to its hypotenuse then the triangles on each side of the perpendicular are similar.

EAD EDC` +D D EDEA& =

ECED .... (1 mark)

ED2 = 16 81EA EC# #= .... (1 mark)

ED` = 4 9 3616 81# #= = .... (1 mark)

Now, ABDT is an isosceles triangle and AE BD=

` BE = ED .... (1 mark)

BD& = ED2 = 2 × 36 = 72 cm .... (1 mark)

41. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30c with it. If the top of the tree touches the ground 30 m away from its foot, then findtheactualheightofthetree.

Solution: Let C be the point at which the tree is broken and let the top of the tree touch the ground at A. Let B denote the foot of the tree. AB =30 m and 30CAB+ = c. In the right angled ABCT ,Now, tan 30c =

ABBC .....(1 mark)

& BC = 30tanAB c

Hence, BC = 3

30 = 01 3 m .... (1 mark)

Also, 30cos c = ACAB .....(1 mark)

& AC = cosAB30c

So, AC = 10 2 203

30 2 3 3 m# #= = .... (1 mark)

Thus, the height of the tree = 10 20BC AC 3 3+ = +

= 03 3 m. ..... (1 mark)


42. Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere.Solution: Let r1 and h be the radius and height of a right circular cone.

Let r2 be the radius of the spherical shaped clay.

Given that r1 = 12cm, h = 48cm.

After the conical clay reshaped into a spherical shaped clay.

Now,Volumeofthesphere = Volumeofthecone ..... (1 mark)

r34

2

3r = r h31

1

2r

r2

3 = 31 12 48

432

# # # #r

r = 123 ..... (3 marks)

Hence, the radius of the spherical clay = 12cm. ..... (1 mark)

43. Calculate the standard deviation of the following data.

x 3 8 13 18 23

f 7 10 15 10 8

Solution:LetusfindtheStandardDeviationbyusingassumedmeanmethod.

Let us take A = 13 be the assumed mean. Then, d x A x 13= - = -

x f d = x – 13 d2 fd fd2

38

131823

71015108

-10-5 0 5

10

100250

25100

-70-50 0 50 80

700250

0250800

fR =50 fdR =10 fd2

R =2000

Standard deviation v = f

fd

f

fd2 2

- e o//

//

= 50

20005010 2

- ` j .... (2 marks)

= 40251- =

25999 = .

531 61

` v - 6.321 .... (1 mark)

44. The probability that a new car will get an award for its design is 0.25, the probability that itwillgetanawardforefficientuseoffuelis0.35andtheprobabilitythatitwillgetboththe awards is 0.15. Find the probability that

(i) it will get atleast one of the two awards (ii) it will get only one of the awards.

.... (2 marks)


Solution: Let A be the event of getting award for design and B be the event of getting awardforefficientuseoffuel.

( ) .P A 0 25= , ( ) .P B 0 35= and ( ) .P A B 0 15+ = .... (1 mark)

(i) Probability of getting at least one award,

( )P A B, = ( ) ( ) ( )P A P B P A B++ -

= .25 0.35 0.150 + -

= 0.45 .... (2 marks)

(ii) Probability of getting only one of the awards,

( ) ( )P A B P A B+ ++ = [ ( ) ( )] [ ( ) ( )]P A P A B P B P A B+ +- + -

= (0.25 0.15) (0.35 0.15)- + -

= 0.10 0.20+

= 0.3 .... (2 marks) 45. (a) The sum of three consecutive terms in an A.P is – 6 and their product is 90. Find the

three numbers. (OR)

(b) A cylindrical jar of diameter 14cm and depth 20cm is half-full of water. 300 lead shots of same size are dropped into the jar and the level of water raises by 2.8 cm. Find the diameter of each lead shots.

Solution: (a) Let a – d, a, a + d be the three consecutive terms of the A.P. (1 mark)

Then, a – b + a + a + d = – 6 & a = – 2 .... (1 mark) (a – d) a (a + d) = 90

( ) [( ) ]d2 22 2

- - - = 90 & d 492= & d = 7! .... (2 marks)

When ,a d2 7=- = , the three numbers are , ,2 7 2 2 7- - - - + & , ,9 2 5- - When ,a d2 7=- =- , the three numbers are ( ), ,2 7 2 2 7- - - - - - & , ,5 2 9- -

Thus, the required three numbers are , ,9 2 5- - or , ,5 2 9- - . .... (1 mark)

(OR)

(b) Cylindrical jar Diameter 2r = 14 & r = 7 cm, h = 2.8 cm (water level raised)

Let r1 be the radius of the leadshot.

Volumeofthewaterlevelraised

in the cylindrical jar = 300 (volume of the lead shot) .... (1 mark)

r h2r = 300 r34

1

3# r

.7 7 2 8# # = 300 r34

1

3#


r1

3 = .100 4

7 7 2 8#

# # = . . .0 7 0 7 0 7# #

r1 = 0.7 cm .... (3 marks)

Diameter = r21 = 2 × 0.7 cm = 1.4 cm

Thus, diameter of the lead shot = 1.4 cm .... (1 mark)

Section – DNote: (i) This section contains 2 questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries 10 marks 2 ×10 = 20

46. (a) Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 6 cm. Also, measure the lengths of the tangents.

Rough diagram .... (1 mark)First circle .... (3 marks)Line segment OP .... (1 mark)Perpendicular bisector .... (1 mark)Second circle .... (2 marks)Two tangent lines .... (1 mark)Measuring the length .... (1 mark)


Construction: (i) With O as the centre draw a circle of radius 6 cm. (ii) Mark a point P at a distance of 10 cm from O and join OP. (iii) Draw the perpendicular bisector of OP. Let it meet OP at M. (iv) With M as centre and MO as radius, draw another circle. (v) Let the two circles intersect at A and B. (vi) Join PA and PB . They are the required tangents. Length of the tangent, PA = 8cmVerification: In the right angled OPAT ,

PA = OP OA2 2- = 10 62 2

-

= 100 36- = 64 8=

` PA = 8 cm.(OR)

(b) Construct a ABCT in which the base BC = 5 cm, 40BAC+ = c and the median from A to BC is 6 cm. Also, measure the length of the altitude from A.

Rough Diagram .... (1 mark)Line Segment BC .... (1 mark)Circle .... (5 marks)Triangle .... (2 marks)Altitude .... (1 mark)


Construction:

(i) Draw a line segment BC = 5 cm.


(iii) Draw BY=BX.





(viii) ABCT or A BCT l is the required triangle.

(ix) Produce CB to CZ.

(x) Draw AE CZ= .

(xi) Length of the altitude AE is 3.8 cm.

47. (a) Draw the graph of 8y x x2

= - - andhencefindtherootsof 2 15 0x x2- - = .

Solution: 8y x x2

= - -

x – 4 – 3 – 2 – 1 0 1 2 3 4 5

x2 16 9 4 1 0 1 4 9 16 25

x- 4 3 2 1 0 – 1 – 2 – 3 – 4 – 5– 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8 – 8y 12 4 – 2 – 6 – 8 – 8 – 6 – 2 4 12

Points: ( , ), ( , ), ( , ), ( , ), ( , )4 12 3 4 2 2 1 6 0 8- - - - - - -

( , ), ( , ), ( , ), ( , ), ( , )1 8 2 6 3 2 4 4 5 12- - -

Solve: y = x x 82- -

0 = x x2 152- -

y = x 7+

Let us draw the graph of the straight line y x 7= + .

Now, form the table for the line y x 7= + .

x – 3 – 2 – 1 0 1 2 3 4 5y x 7= + 4 5 6 7 8 9 10 11 12

Points: ( 3,4), ( 2,5), ( 1,6), (0,7), (1,8), (2,9), (3,10), (4,11), (5,12)- - -


The straight line and parabola intersects at the points (– 3, 4) and (5, 12).

The x coordinates of the points are – 3 and 5. Hence, the solution set is {– 3, 5}.

First Table .... (2 marks)Solving the equation .... (1 mark)Second Table .... (1 mark)Drawing x and y axis, Scale

.... (2 marks)

Plotting the points .... (3 marks)Solution Set .... (1 mark)

(OR)


(b) A cyclist travels from a place A to a place B along the same route at a uniform speed on different days. The following table gives the speed of his travel and the corresponding time he took to cover the distance.

Speed in km/ hr x 2 4 6 10 12Time in hrs y 60 30 20 12 10

Draw the speed-timegraphanduseittofind (i) the number of hours he will take if he travels at a speed of 5 km / hr (ii) the speed with which he should travel if he has to cover the distance in 40 hrs

Solution: From the table, we observe that as x increases, y decreases.This type of variation is called indirect variation.

Here, xy = 120. Thus, y = x

120 .

Plot the points (2 , 60), (4 , 30), (6 , 20), (10 , 12) and (12 , 10).

Join these points by a smooth curve.

From the graph, we have (i) The number of hours he needed to travel at a speed of 5 km/hr is 24 hrs.(ii) The required speed to cover the distance in 40 hrs, is 3 km / hr.

Formation of Equation .... (1 mark)Plotting the points and Drawing the curve

.... (5 marks)

Drawing x and y axes, Scale .... (2 marks)Solution Set .... (2 marks)


Namma Kalvi - The No.1 Educational Website for 9th, 10th ......11th, 12th, TRB TET & TNPSC Materials...

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