CBSE-XII-2017 EXAMINATION CBSE-X-2011 …...CBSE-X-2011 EXAMINATION 2 / 24 20 15 20 15 Ap 3 19 19. 3...

CBSE-X-2011 EXAMINATION CBSE-XII-2017 EXAMINATION

MATHEMATICS

Paper & Solution

Time: 3 Hrs. Max. Marks: 80

www.vedantu.com 1 / 24

Code: 30/1 Series: RHB

General Instuctions :

(i) All questions are compulsory.

(ii) The question paper consists of 34 questions divided into four sections — A, B, C and D.

(iii) Section A contains 10 questions of 1 mark each, which are multiple choke type questions, Section B

contains 8 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D

contains 6 questions of 4 marks each.

(iv) There is no overall choke in the paper. However, internal choice is provided in one question of 2 marks,

three questions of 3 marks and two questions of 4 monks.

(v) Use of calculators is not permitted.

SECTION – A

Question numbers 1 to 10 carry 1 mark each. For each of the question numbers 1 to 10, four alternative

choices have been provided, of which only one is correct. Select the correct choke.

1. The roots of the equation x2 – 3x – m (m + 3) = 0, where m is a constant, are

(A) m, m + 3

(B) –m, m + 3

(C) m, – (m + 3)

(D) – m, – (m + 3)

Solution:

Given a quqelaric eqution 2

2

3 ( 3) 0

( 3) ( 3) 0

( ( 3)) ( ( 3)) 0

( ( 3))( ) 0

, 3

x x m m

x m x mx m m

x x m m x m

x m x m

x m m

2. If the common difference of an A.P. is 3, then a20 - a15 is

(A) 5

(B) 3

(C) 15

(D) 20

Solution:

Given common difference -cl ,B

lc ks

CBSE-X-2011 EXAMINATION


20

15

20 15

Ap 3

19 19. 3

57

14 14 3

42

lets say the list term

d

a

a a d a x

a

a a d a x

a

a a a 57 a 42

15

3. In Figure 1, 0 is the centre of a circle, PQ is a chord and PT is the tangent at P. If POQ = 70°, then

TPQ is equal to

(A) 55°

(B) 70°

(C) 45°

(D) 35°

Solution:

Given POQ = 700

In

0

POQ OP OQ ( )

POQ OQP

POQ

POQ OPQ OPQ 180

rad i

It is an isosules triangle

In



0

0

0

0 0

0

POQ 2 180

OPQ 55

OP PT

OPT 90

OPT TPQ OPQ

90 TPQ 55

TPQ 35

we know that

4. In Figure 2, AB and AC are tangents to the circle with centre O such that BAC = 40°. Then BOC is

equal to

(A) 40°

(B) 50°

(C) 140°

(D) 160°

Solution:

0

0

OB AB OBA 90

AC OCA 90

AB and AC are the tangents

drown from external pointA to the circle.

OC



0

0

0

0

0

ABCD

180

; OBA OCA 180

ABCD

BAC BOC 180

BOC 180 40

BOC 120

is a quadrilateral in which

sum of opposite angles is

ie

is a cyclic quadrilateral

5. The perimeter (in cm) of a square circumscribing a circle of radius a cm, is

(A) 8 a

(B) 4 a

(C) 2 a

(D) 16 a

Solution:

Given OE =OF = a

Side of the square circomscrcling the circle =2a

∴ perimeter of square =4×2a =8a

6. The radius (in cm) of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is

(A) 4.2 (

(B) 2.1

(C) 8.4

(D) 1.05

Solution:

largest cone that can be cube from a cube has the

Diameter = side of cube

Height = side of cube

∴ radius =4.2

2 = 2.1cm



7. A tower stands vertically on the ground. From a point on the ground which is 25 in away from the foot of

the tower, the angle of elevation of the top of the tower is found to be 46°. Then the height (in meters) of

the tower is

(A) 25 2

(B) 25 3

(C) 25

(D) 12.5

Solution:

Given AB=25m

And angle elevation of the lower (BC) from A=450

∴∠BAC =450

In ∆ABC tan450 =BC

AB

⟹BC=25m

∴ height of the lower=25m

8. If , 42

aP

is the mid-point of the line-segment joining the points A(- 6 5) and B(- 2, 3), then the value of

a is

(A) - 8

(B) 3

(C) - 4

(D) 4

Solution:

Given a line segment joining

A(-6,5) and B(-2,3)



Midpoint of A&B is P , 42

a

6 2 5 3,4 ,

2 2 2

A B

2 2

B

a

a

9. If A and B are the points (- 6, 7) and (-1, -5) respectively, then the distance 2AB is equal to

(A) 13

(B) 26

(C) 169

(D) 238

Solution:

Given 2 points

A(-6,7) and B(-1,-5)

Distance between the

points=AB

2 2( 6 1) (7 5)

35 144

AB 13

2AB 26

10. A card is drawn from a well-shuffled deck of 52 playing cards. The probability that the card will not be

an ace is

(A) 1

13

(B) 1

4

(C) 12

13

(D) 3

4

Solution:

No.of aces in deck of cards=4 probability that card drown is

An Ace=. 4

52

No of Ace

Totalcards

Probability that the card is not an Ace

4 121

52 13

SECTION B



Question numbers II to 18 cony 2 marks each.

11. Find the value of in so that the quadratic equation ins (x — 7) + 49 = 0 has two equal roots.

Solution:

Given;mx(x-7)+49=0

⟹mx2-7mx+49=0

2

2

D (7 ) 4 49

49 4 49 0

49

m mx

m mx

2 4 49m mx

0, 2m

12. Find how many two-digit numbers are divisible by 6.

Solution:

Divisibility test for 6:

A number is divisible by 6 if it divisible by both 2 and 3

Test for 2: Any even number is divisible by 2

Test for 3: Any number whose sure of the digits is divisible by 3

∴ List of 2 digit numbers divisible by 6 are 12,18,24,30,36,42,48,54,60,66,72,78,84,90 and 96.

13. In Figure 3, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB et 6 cm, BC = 9

cm and CD = 8 cm. Find the length of side AD.

Solution:



AB=6 cm

BC=9 cm

CD=8 cm AD=?

AB,BC,CD,AD, are tangents to the circle

And AP=PS RD=DS

BP=BQ AS=AP

CQ=CR

Also AB=AP+PQ (1)

BC=BQ+QC (2)

CD=RC+DR (3)

AD=AS+DS (4)

Adding (1),(2),(3),(4)

6+9+8+AD=AP+AS+BP+BQ+CQ+RC+RD+DS

23+AD=2(AP)+2(BP)+2(RC)+2(RD)

23+AD=2(AB)+2(CD)

AD 5cm

14. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that

3.

5

AP

AB

Solution:

Given a line segment AB= 7 cm

Given

AP 3 AP 3 2 2PB AP AB.

PB 5 AP+ PB 5 3 5

AP 3

PB 2

:Steps 1) Draw the line segment AB=7cm

2) Draw a ray AX such that ∠BAX is an acute angle with AB

3) Along AX, mark 5 (3+2) points A1,A2,A3,A4,A5 such that

AA1=A1A2=A2A3=A3A4=A4A5



4) join A5 to B

5) through A3 draw a line parallel to A5B intersecting AB at P

∴ the desired point which divides AB in 3:2 is P

15. Find the perimeter of the shaded region in Figure 4, if ABCD is a square of side 14 cm and APB and CPD

are semicircles. (Use 22

7 )

Solution:

Given a square ABCD with side =14cm

AB=CD =BC=AD =14cm

Semicircles APB and CPD with diameter =14cm

Perimeter of shaded region =AD+BC + CD AB

Length of CD are =0

0

180 14

360 2

Length of AB CD 3.5cm

Perimeter of =14+14+3.5+3.5

shaded region =35cm

16. Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the

resulting cuboid.

OR

A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the

form of a sphere. Find the diameter of the sphere.

Solution:



Choice-2

Cone: height =20cm

Base radius =5cm

Cone is reshaped into a sphere

∴ volume of cone = volume of sphere

1

37 5( 30 2 4

)(5) 4

7 3

3 3

( )

5

5

r

r

r cm

17. Find the value of y for which the distance between the points A(3, -1) and 8(11, y) is 10 units.

Solution:

Distance between the points A(3,-1) and B(11,y) is 10 units

2 2

2

2

AB 10

(3 11) ( 1 ) 10

64 ( 1) 100

( 1) 36

1 6 1 6

7,5

y

y

y

y or y

y

18. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability

that the selected ticket has a number which is a multiple of 5.

Solution:

A ticket is drawn at random from 40 tickets

Total outcomes=40

Out of the tickets numbered from 1 to 40 the number of tickets with number

By 5 = 5,10,15,20,25,30,35,40

= 8 tickets

∴ favorable outcomes = 8

∴ probability

8

40

1

5

SECTION C

Question numbers 19 to 28 carry 3 marks each.

19. Find the roots of the following quadratic equation :

x2 - 3 5 x + 10 = 0



Solution:

2 3 5 10 0x x

For any quadratic equation 2

2 40

2

b b acax bx c x

a

∴ for the given equation

3 5 45 40

2

3 5 55,2 5

2

x

x x

20. Find an A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.

Solution:

Given an A.P

say first term = a

common difference =d

Given T4=9

A+3d=9 -(1)

Also T6+T13=40

a+5d+a+12d=40

2a+17d=40 (2)

Solving (1) and (2)

a=3 d=2

∴A.P will be 3,5,7,9,……….

21. In Figure 5, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT

and TR into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the

area of PQR = 189 cm2, then find the lengths of sides PQ and PR.

Solution:



PQ PN+ QN PQ & PR

PR = PM+ RM PN = PM,RM = RT QN = QT

As are tangenst to circle

PQ+PR=PN+PM+QN+RM

=2(PN)+QT+RT

42=2(PN)+21

PN=21

2cm

PQ=PN+QN

=PN+QT

=21

2+12

PQ=45

2cm

PR=21

2+9

PR=39

2cm

22. Draw a pair of tangents to a circle of radius 3 cm, which are inclined to each other at an angle of 60°.

OR

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then

construct another triangle whose sides are 3

5 times the corresponding sides of the given triangle.

Solution:

Choice-1

Pair of to a circle with radius=3cm inclined to each other with angle 600



If ∠APB=600 AOBPAs is a cyclic quadrilateral

Then ∠AOB=180-600

=1200

Tee tangents can be constructed in the following manner:

Step 1

Draw a circle 2cm and with center as O.

Step 2

Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.

Step 3

Draw a radius OB, making an angle of 120 (180 - 60 ) with OA.

Step 4

Draw a perpendicular to OB at point B. Let both the perpendicular intersect at point P. PA and PB are the

required tangents at an angle of 60 .

Justification

The construction can be justified by proving that 60APB

By our construction

90

90

and 120

OAP

OBP

AOB

We know that the sum of all interior angles of a quadrilateral = 360

360

90 120 90 360

60

OAP AOB OBP APB

APB

APB



This justifies the construction

Choice-2

Steps:

1) Draw BC=4cm

2) Draw a ray BX such that ×BC=900

3) take compass with radius 3cm and draw an are from B cutting B× at A

4) join A and C to from △ABC

5) Draw a ray BY such that ∠CBY is acute angle

6) Along BY mark 5 equidistant points

B1,B2,B3,B4,B5 such that B B1=B1,B2=B2B3=B3B4=B4B5

7) join B5 to C and draw a line parallel to B5C from B3 such that it cuts BC at C1

8) from C1 draw a line parallel to AC such that it cuts AC at A1 thus ∆A1BC1 is the repaired triangle 1 1 1 5

3

A B A B BC

AB AC BC

Instification

B5C || C3C1

∆BB5C~∆BB3C1

5

1 1

3

B CBC 5= =

BC B C 3

23. A chord of a circle of radius 14 cm subtends an angle of 120° at the centre. Find the area of the

corresponding minor segment of the circle. [Use 22

7 and , 3 = 1.73]



Solution:

Area of minor segment =Area of sector AOB –Area of ∆AOB

Given

∠AOB = 1200

OA=OB=14cm

Find area of shaded region

Area of minor segment = Area of sector AOB – Area of AOB

Area of triangle AOB=1

2×14×14× sin 1200

=49 3 cm2

Area of sector AOB=0

2

0

120 22(14)

360 7

2 2616205.33

3cm cm

Area of minor segment=205.33-49 3 = 120.46cm2

24. An open metal bucket is in the shape of a frustum of a cone of height 21 cm with radii of its lower and

upper ends as 10 cm and 20 cm respectively. Find the cost of milk which can completely fill the bucket at

Rs. 30 per litre. [Use 22

7 ]

Solution:

Volume of frustum = 2 2( )3

h R r rR

2 2 0 )22

21 (10 20 207 3



3

315400 15.4

22(700)

cm

cm

Cost of milk:

= 15.4 × 30

=462 Rs

25. Point P(x, 4) lies on the line segment joining the points A(— 5, 8) and B(4, —10). Find the ratio in which

point P divides the line segment AB. Also find the value of x.

Solution:

Lets say ratio =m:n

P(x,4)=4 5 10 8m n n

,m n m n

10 84

4 4 10 8

m m

m nm n m n

14 4

2

7

4 4

4 5

1

8 35

9

3

3 4

we know

m n

m

n

m nx

m n

m

nx

m

n

x

x

P ( , )

26. Find the area of the quadrilateral ABCD, whose vertices are A(-3, -1), B(- 2, - 4), C(4, -1) and D(3, 4).

OR

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are

A(2, 1), B(4, 3) and C(2, 5).

Solution:



Area of quadrilateral ABCD

= ar (∆ABC)+ar(∆ADC)

Area of quadrilateral ABCD = Area of ABC + Area of ADC

We know that,

Area of triangle

We know that,

Area of triangle u3 1 2 3 2 3 1 3 1 2

1( ) ( ) ( )

2x y y x y y x y y

Thus,

Area of 1

( 3)( 4 1) ( 2)( 1 1) 4( 1 4)2

ABC

1[9 0 12]

221

.2

sq units

Area of 21

.2

ABC sq unitss

Area of 1

( 3)(4 1) 3( 1 1) 4( 1 4)2

ADC

1[ 15 0 20]

235

.2

sq units

Area can’t be negative, So,

Area of 35

.2

ABC sq units

Substitute these values in equation (1), we have,

Area of quadrilateral ABCD = 21 35 56

282 2 2

.sq units

Hence, area of quadrilateral is 28 square units.



P,Q,R are the midpoints to the sides of the ∆ABC

4 2 33 2

2 2

3 4 2 3

HP , ,

Q , ,R ,

Area of ∆PQR

1

3 4 3 3 3 2 2 2 42

13 3 4

2

1sq.unit

27. From the top of a vertical tower, the angles of depression of two cars, in the same straight line with the

base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 in apart and are on the

same side of the tower, find the height of the tower. (Use 3 .= 1.73]

Solution:

Given CD=100m, AB=?

In ∆ABC Tan600=AB

BC



0

3

45

1003

3 1100

3

ABBC

ABIn ABD Tan

BD

BD AB

BD BC CD

ABAB

AB

100 3

3 1

236 98

237

AB

AB .

AB mt

28. Two dice arc rolled once. Find the probability of getting such numbers on the two dice, whose product

is 12.

OR

A box contains 80 discs which are numbered from 1 to 80. If one disc is drawn at random from the box,

find the probability that it bears a perfect square number.

Solution:

tow dice are rolled total possible outcomes =6×6=36

Product of outcomes will be 12 for

(2,6),(6,2),(3,4)(4,3)

No. of favorable cases =4

Probability =4 1

36 9

28) (or)

A disc drawn from a box containing 80

Total possible outcomes =80

No. of cases where the disc will be numbers perfect square =8

Perfect squares less then 80

=1,4,9,16,25,36,49,64

Probability =8 1

80 10

SECTION D

Question numbers 29 to 34 cony 4 marks each.



29. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of

contact.

Solution:

Given: A circle C (0, r) and a tangent /at point A

To prove : OA l

Construction : Take a point B, other roan A. on the tangent I. Join OB. Suppose OB meets the circle in C.

Proof: We know that, among all line segment joining the point 0 to a point on I, the perpendicular is

shortest to l

OA = OC (Radius of the same circle)

NOW, OB = OC + BC.

OB > OC

OB > OA

OA < OB

B is an arbitrary point on the tangent I. Thus, OA is shorter than any other line segment joining O to any

point on l

Here, OA l

30. The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how

many terms are there and what is their sum ?

OR

How many multiples of 4 lie between 10 and 250 ? Also find their sum.

Solution:

Given an A.P with first (a)=8

Last term =350

Common difference (d)=9

Tn =a+(n-1)d

=a+(n-1)d=350

⟹8+(n-1) 9=350

39n

No. of terms =39



Sum of the terms

2

398 350

2

6981

na

OR

Multiples of 4 between 10 and 250 first multiple of 4 after 10 =12

Last multiple of 4 less then 350=248

We now have an A.P with first term=12 and last term =248

Common difference=4

∴248=12+(n-1)4

60n multiples of 4 between 10 and 250

31. A train travels 180 km at a uniform speed. If the speed had been 9 km/hour more, it would have taken

1 hour less for the same journey. Find the speed of the train.

OR

Find the roots of the equation 2 1 3

1, , 5.2 3 5 5

xx x

Solution:

Distance travelled by train =180 km lets say speed = s km|hr

180

Time taken t s

It is given if speed had been (s+9) km | hr

Train would have travelled AB M (t-1) hrs

∴t-1=180

9s

1801

9t

s

2

2

180 1801

9

189 180 1520

189 180 1620

9 1620 0

36

s s

s s s

s s s

s s

s km | hr



OR

1 1 31 5

2 3 5 2,x ,

x x

Taking L.C.M on left side of equality

2

2

5 2 31

2 3 5

3 8 2 3 10 15

2 16 23 0

x x

x x

x x x x

x x

256 2 4 2

4

16 72

4

16 6 2

4

8 3 2 8 3 2

2 2

16; ;

x

,

x

x

x

32. In Figure 6, three circles each of radius 3.5 cm are drawn in such a way that each of them touches the

other two. Find the area enclosed between these throe circles (shaded region). [Use 22

7 ]

Solution:

Given that all circles have radii =3.5cm

∴AB=BC =AC=7cm

∆ABC is an equilateral triangle area of ∆ABC= 3

4 49 cm2



Area of sector BPQ

0

2

0

2

60 223 5

7360

77

12

.

cm

Similarly areas of other sectors PCR and RAQ = 277

12cm

Area of shaded regin

=ar(∆ABC)-3(area of BPQ)

249 3 772

4 12cm

33. Water is flowing at the rate of 15 km/hour through a pipe of diameter 14 cm into a cuboidal pond

which is 50 in long and 44 m wide. In what time will the level of water in the pond rise by 21 cm ?

Solution:

Let the level of water in the pond rises by 21 cm in /hours.

Speed of water = 15 km/hr

Diameter of pipe =14 cm = 14

100m

Radius of the pipe, r = 7

100m

Volume of water flowing out of the pipe In 1 hour

2

2

22 715000

7 100

r h

m m

= 231 m3

Volume of water flowing out of the pipe in t hours = 231 t m3

Volume of water in the cuboidal pond

= 50 m 44m 21

100m (Volume of cuboid = lbh)

= 462 m3

Volume of water flowing out of the pipe in t hours = Volume of water In the cuboidal pond

231 t = 462

4622

231t

Thus, the water in the pond rise by 21 cm in 2 hours.

34. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another

point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower.

Solution:



Height of the lower (AB) =h

Given CD =10m & BC=ED

BE=CD= 10m

In ∆ABC tan600=h

BC

BC=3

h

In ∆ADE

Tan300=10h

ED

10 3ED ( h )

10 33

210

3

15

hh

h

h mt

CBSE-XII-2017 EXAMINATION CBSE-X-2011 …...CBSE-X-2011 EXAMINATION 2 / 24 20 15 20 15 Ap 3 19 19. 3...

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