10th ICSE Guess Paper With Sol

8/9/2019 10th ICSE Guess Paper With Sol

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L.K. Gupta (Mathematics Classes) wwwwww..ppiioonneeeerrmmaat t hheemmaat t iiccss..ccoomm MOBILE: 9815527721, 4617721

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40 –D, CHANDIGARH

SOLUTIONS ON www.pioneermathematics.co m www.pioneermathematics.co m NOTICES

BOARDS(2010)

(10th ICSE GUESS PAPER)

TIME: 2:30 HOURS MAX. MARKS: 80

GENERAL INSTRUCTIONS & MARKING SCHEME

1.

Answers to this paper must be written on the paper provided separately.

2. You will not be allowed to write during the first 15 minutes.This time is to be spent

in reading the question paper.

3. The time given at the head of this paper is the time allowed for writing the answers.

4.

Attempt all questions from Section A and any four questions from Section B.

All working, including rough work, must be clearly shown and must be done on the same

sheet as the rest of the answer. Omission of essential working will result in the loss of mark

5. The intended marks for questions or parts of questions are given in brackets [ ].

6. Mathematical tables are provided.

NAME OF THE CANDIDATE PHONE NUMBER

L.K. Gupta (Mathematics Classes) Pioneer Education (The Best Way To Success)

S.C.O. 320, Sector 40- D, Chandigarh

Ph: - 9815527721, 0172 – 4617721.

http://www.pioneermathematics.com/http://www.pioneermathematics.com/http://www.pioneermathematics.com/


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SECTION – A

1. (a) Determine the value of ‘m’ if x m is a factor of the polynomial 3 2x m 1 x 2 hence find the value of k if 3m 6 k . [3]Sol:

(a)

x m is a factor of 3 2x m 1 x 2 forx m

3 2m m 1 m 2 0

or3 3

m m m 2 0 or m 2 Now 3m 6 k

3 2 6 k

k 0 m 2 and k 0

(b) If 3a 8b:3c 8d ::3a 8b:3c 8d ,then show that a, ,b, c, d,are in Proportion [3Sol:

3a 8b 3a 8b

3c 8d 3c 8d

Using alternendo3a 8b 3c 8d

3a 8b 3c 8d

Using componendo and dividendo3a 8b 3a 8b 3c 8d 3c 8d

3a 8b 3a 8b 3c 8d 3c 8d

or6a 6a

16b 16d

a c

b d

a,b,c and d are in proportion.

(c) On what sum will the difference between the simple interest and compound interest foryears at 5% per annum will be equal to Rs. 50? [4

Sol:

Let Principle = Rs. x

Now , C.I. –S.I =Rs. 50

or

nr p r t

P 1 1 50100 100



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or

25 x 5 2

x 1 1 50100 100

or

221 x

x 1 5020 10

or 441 xx 1 50400 10

441 400 xx 50

400 10

or41x 40x

50400

x 20,000 principle , Rs. x = Rs. 20,000

2.

(a) Mehak deposits Rs. 150 per month in a recurring deposit account for 8 months a

the rate of 8 % per annum. what amount will she get on maturity? [4]

Sol:

Total P for 1 month = Rs 1508 8 1

2

150 36 Rs.5400 5400 8

1 Rs.36100 12

Total amount paid in 8 months 150 8 Rs.1,200 Amount received on maturity

Rs. 1,200 36 Rs.1236

(b) Solve the inequation and represent the solution on the number

Line:2 x 2

1 ,x R3 3 3

. [4]

Sol:

2 x 213 3 3



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Multiplying through out by 3

2 x 3 2 or 2 3 x 2 3 or 5 x 1 or 5 x 1

1 x 5

(c) In the given figureLP 2

PQ||MN andPM 3

calculate the value of :

(i)

ar ΔLPQ

ar ΔLMN

(ii) area of trapeziumPMNQ

area of ΔLMN [3

Sol:

Given : PQ || MN andLP 2

PM 3

L L Common P M [Corresponding angles ]ΔLPQ ΔLMN [ AA similarity ]

Let LP = 2x then PM = 3x

LM = 2x+3x = 5x

(i)

2 22

2

ar ΔLPQ LP LP 2xar ΔLMN LM 5xLM

2

2

4x 44:25

2525x

(ii) Let ar ΔLPQ 4y then ar ΔLMN 25y ar trapeziumPMNQ 25y 4y 21y



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ar trapizium PMNQ 21y 21

21:25ar ΔLMN 25y 25

3.

(a)

(i) Point P(k, m) is reflected in the x-axis to P' 5, 2 .Write down the values of k and m .

(ii) P’’ is the reflection of P when reflected in the y-axis .Write down the coordinates of P’’.(iii) Name a single transformation that maps P’ to P’’. [4]Sol:

(a)

(i) xM k,m 5, 2

k 5,m 2 xM x,y x, y Co –ordinates of P are 5,2

(ii) y yM 5,2 5,2 M x,y x,y Co –ordinates of P’’ are 5,2

(iii) Reflection of p’ in the origin

0M x,y x, y

(b) Find the mean, mode and median of the following data : 25 , 27 , 19, 29, 21 , 23 , 25 , 30

28 , 20 [3]

Sol:

(i) mean

x 247

24.7n 10

(ii) Ascending order : 19, 20, 21 , 23, 25, 27, 28, 29, 30n 10(even)

Mediann/2T Next term

2

10/2 5 6T Next term T T

2 2

25 25 5025

2 2

(iii) Mode =25 as it occurs maximum number of times in the data.

(c) The area enclosed between two concentric circles is 7702

cm . If the radius

of the outer circle is 21cm. Calculate the radius of the inner circle22

Use π7

[3

Sol:



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Radius of outer circle (R) = 21 cm

Let radius of inner circle = r cm

or2 2 2πR πr 770cm

or 2 2π R r 770

or 2 222 21 r 7707

or 2770 7

441 r22

or2

441 r 245 or

2r 441 245

or2

r 196 or r 14cm Radius of the inner circle is r 14cm .

4. (a) Prove that :1 1

secθ tanθ cosθ

1 1

cosθ secθ tanθ

[3

Sol:

(a)

L.H.S =1 1

secθ tanθ cosθ

secθ tanθ1

secθsecθ tanθ secθ tanθ

2 2

secθ tanθ secθsec θ tan θ

secθ tanθ secθ secθ secθ tanθ

secθ tan θ

secθ secθ tan θsecθ tan θ

2 2sec θ tan θsecθ

secθ tanθ

1 1

cosθ secθ tanθ

RHS. Hence proved.

(b) If the mean of the distribution is 62.8 and sum of frequencies is 50, find P and Q. [4]



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Sol:

Class X F Fx

0-20 10 5 50

20-40 30 P 30p

40-60 50 10 500

60-80 70 Q 70q

80-100 90 7 630

100-120 110 8 88050 2060+30p+70q

Now, 5 p 10 q 7 8 50 or p q 50 30 or p q 20................(i)

fxx

f

or2060 30p 70q

62.8

50

or10 206 3p 7q

62.850

or 314 206 3p 7q or 3p 7q 108......................(ii) (i) × 3 3p + 3q = 60 ………….(iii) Subtracting (iii) from (ii)

4q 48 q 12 p 12 20 p 8 from (i)

p 8 andq 12

(c) List Price of a washing machine is Rs. 17,658.The rate of sales tax is 8% .The customer

requests the shopkeeper to allow a discount in the Price of the washing machine to such a

extent that the Price remains Rs. 17,658 inclusive of sales tax. Find the discount in the pri

of the washing machine. [3]

Sol:

Let new marked price of the washing machine= Rs. x

Class Frequency

0-20 5

20-40 P

40-60 10

60-80 Q

80-100 7

100-120 8



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No total amount to be paid = M.P +S.T% of M.P17658 = x 8%of x

17658 =8

x x100

217,658 x x

25

27x17658

25

17,658 25x

27

x 16,350 Discount in the price of the washing machine

17,658 16,350 Rs.1,308

SECTION-B

(Attempt any four question from this Section.}

5. (a) Solve the following equation and give your answer up to two decimal places:

73x 1

x [3]

Sol:

273x 1 0 3x x 7 0

x

Comparing this to the equation2

ax bx c 0 we get a 3,b 1,c 7

2b b 4acx

2a

21 1 4 3 72 3

1 1 84

6

1 85 1 9.219

6 6

1 9.219 1 9.219,

6 6

10.219 8.219,

6 6



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1.703, 1.369 1.70, 1.37

(b) In the given figure , a circle touches all the four sides of a quadrilateral ABCD

whose sides are AB = 6cm , BC = 7cm and CD = 4 cm. Find AD . [3]

Sol:

Given that : AB = 6cm , BC =7cm , CD = 4cm . let the circle touches the sides AB , BC, CD

and DA at points P,Q,R and S respectively.

then AP = AS , BP =BQ ,CR =CQ , DR = DS

(tangents from external point are of equal length)

Adding we get :

AP +BP+CR+DR =AS +DS+BQ+QC

AP PB CR RD BQ QC DS SA AB CD BC DA 6 4 7 AD

AD = (10-7) cm = 3 cm.

(c) Find the equation of the straight line that passes through the point (3,4) and

Perpendicular to the line 3x 2y 5 0 . [4]Sol:

slope of 1Coefficient of x 3

AB mCoefficient of y 2

Slope of CD 21

1 1m 2/3

3m

2



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then equation of CD , 1 1y y m x x

N 3,4 is the passing point and slope 2/3

2y 4 x 3 3y 12 2x 63

2x 3y 6 0 2x 3y 6 0

6.

(a) Manan, Produces an item for Rs. 216, which he sells to Rohan , Rohan sells it to

Sohan and Sohan sells it to Mohan. The tax rate is 10 %. The Profit Rs. 20 at each

stage of the selling chain. Find the amount of VAT. [4]

(b) If2 2

A3 4

, then find2

A 6A [3]

(c) A bag contain 3 red, 5 black and 6 white balls. A ball is drawn at a random. Find the

Probability that the ball drawn is (i) black (ii) not red (iii) either red or white . [3]

Sol:

(a) The selling price for Manan = Rs. 216 +Rs. 20

= Rs. 236

Tax charged = Rs.236 10

Rs.23.60100

So VAT = Rs. 23. 60

The value of invoice = Rs. 236 +Rs.23.60Rs. 259.60

cost price for Rohan = 236

The selling price for Rohan = Rs. 236+Rs, 20

= Rs. 256


Rs.25.60100

So, VAT = Rs. 25.60 –Rs. 23.60=Rs. 2.00

The value of invoice = Rs. 256 +Rs. 25.60

=Rs. 281.60

Cost Price for Sohan = Rs. 256

The selling price of Sohan = Rs. 256 + Rs. 20

= Rs. 276


100

=Rs. 27.60

VAT = Rs. 27.60 – Rs. 25.60 = Rs. 2.00The value of invoice = Rs. 276 + Rs. 27.60

=Rs. 303.60



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Cost price for Mohan = Rs. 276

The selling price for Mohan = Rs. 276 + Rs. 20

=Rs. 296


Rs.29.60100

VAT =Rs. 29.60 –Rs. 27.60 =Rs. 2.00The value of invoice = Rs. 296+Rs. 29.60=Rs. 325.60

So Total VAT =Rs. 23.60 3 2.00 Rs.29.60

(b)

2 2 A

3 4

2 A A A 2 2 2 2

3 4 3 4

2 2 2 3 2 2 2 4

3 2 4 3 3 2 4 4

4 6 4 8 10 12

6 12 6 16 18 22

2 10 12 A18 22

2 2 12 126A 6

3 4 18 24

210 12 12 12

A 6A18 22 18 24

2 0

0 2

(c)

No.of favaurable outcomesP ETotal no.of possible outcomes

Total numbers of balls 3 5 6 14 (i) No. of black balls = 5



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P(black ball) =5

14

(ii) No. of balls which are not redB w

5 6 11

P ( not red ) =11

14

or

P(not red)3 11

1 p(red) 114 14

(iii) No. of favorable outcomes = 3 6 9

P(either red or while)9

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7. (a) Anu sold Rs. 100 shares at 10% discount and invested in 15 % Rs . 50 shares

at Rs. 33 . If she sold her shares at 10 % premium instead of 10 % discount, she would have

earned Rs. 450 more. Find the number of shares sold by her. [3]Sol:

(a)

Let the number of shares sold = x

S.P. of x shares = Rs. 90x (Sold at 10% discount)

S.P. of x shares sold at 10% premium = Rs. 110x

No. of shares (each of Rs. 50 ) that can be purchased wit h Rs. 90x =90x

33

Total face value =90

Rs.50 x

33

Annual income90

15% of 50 x33

15 9050 x

100 33

225Rs. x

11 .

No. of shares (each of Rs. 50) that can be purchased with Rs. 110x110x

33

Total face value = Rs.110

50 x33

Annual income15 110

50 x100 33

15 110xRs.

2 33

Rs.25x



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225x25x 450

11

275 225x450

11

50x 450 11 450 11

x 50

x 90 Number of shares sold by Anu = 99

(b) From a window (60m high above the ground ) of a house in a street,the angles of

elevation and depression of the top and the foot of another house on the opposite side of

the street are0

60 and0

45 respectively . Show that the height of the opposite house

is60 1 3 m . [4]

Sol: Let P denote the position of the window of a house and AB denote the opposite house .In

right triangle PQA

0 AQ tan45 1PQ

0Each angle of quadrilatral OPQA is 90

OPQA is a rectangle

AQ 60m

PQ AQ

PQ 60 AQ OP 60m In right triangle PQB

0BQ tan60PQ

BQ 3 60

BQ 60 3 The height of the opposite house AB AQ BQ

60 60 3 60 1 3 m



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(c) In the given figure , find CEB and ADB , where E is the Point of intersection of chords

AC and BD of circle. [3]

Sol: (c) BAC BDC [ angle of same segment ]

0BAC 35 in AEB

0EAB AEB EBA 180 0 0 0

35 AEB 50 180 0 0 0 0

AEB 180 35 50 95 0

CEB AEB 180

[Linear pair angles]0 0

CEB 95 180 0 0 0

CEB 180 95 85 In ΔADB

0 ADB DAB ABD 180

0 0 0 0 ADB 55 35 50 180 0 0

ADB 140 180



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0 0 ADB 180 140

0 ADB 40

8. (a) Prove that if the bisector of any angle of a triangle and the perpendicular bisector o

its apposite side intersect, they will intersect on the circumcircle of the triangle. [3

Sol:

(a)

Let ABC be the given triangle and AD be the angle bisector of A and PR be right bisectorof side BC, intersecting each other at P.

As P lies on right bisector of BC0PB PC and PQB 90

with BP and PC as diameters draw circles

these circles will pass through Q.0BQP CQP 90

and will touch the sides AB and ACNow ABL is a tangent and BQ is a chord.

ABD BPA [ angle in the alt. segment ]Similarly,

ACD CPA ABD ACD BPA CPA BPC

Adding A to the both sides, we get ABD ACD A A BPC

or0

180 A BPC [ angle sum property]Now, ABPC is a quadrilateral in which

0 A P 180 A,B,P and C are concylic .

(b) The radii of the internal and external surface of a hollow spherical shell are 3cm and 5c



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respectively. If it is melted and recast in to a solid cylinder of height2

23

cm, find the

diameter and the curved surface area of the cylinder. [4]

Sol:

r 3cm R 5cm

Volume of sphere (shell ) 3 3 3 34 4 22π R r 5 33 3 7

4 22 125 273 7

4 2298

3 7 3

88 14cm

3

For cylinder

2 8h 2 cm cm

3 3

Volume of shell = volume of cylinder

288 14 πr h3

288 14 22 8r3 7 3

2 88 14 7 3r3 22 8

2r 49 r 7m Diameter of the cylinder 2r 2 7 14cm Surface area of cylinder 2πrh

22 82 7

7 3

2117.33cm

(c) From the given fig. find the value of x.

[3]



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PA × PB = PC × PD x × x = 1 × 8 x² = 8 x = √8 = 2√2

9. (a) Draw ΔABC , having A 2,0 ,B 6,0 and C 2,8 (i) Draw the line of symmetry of ΔABC .(ii) Find the coordinates of the point D, if the line (i) and BC are both lines of

symmetry of the quadrilateral ABCD.

(iii) Assign, special name to the quadrilateral ABCD. [3

Sol: (a) (i)

(ii) D(6,8) (iii) Square

(b) The Point A 5, 1 on reflection in x-axis is mapped as A’. Also A on reflection in y-axis

mapped as A’’. Write the coordinates of A’ and A’’ also calculate the distance AA’: Sol:

A' 5,1 and A'' 5, 1

So, 2 22 1 1 2 AA' x x y y

2 25 5 1 1



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2 210 2 100 4 104

2 26

(c) A page from Pooja’s saving bank account is given below.

Date Particulars Withdrawals

Rs. P

Deposits

Rs, P

Balance

Rs, P.

01.01.2000 B/f - - - - 2,800.00

08.01.2000 By cash - - 2,200.00 5,000.00

18.02.2000 To cheque 2,700.00 - - 2,300.00

19.05.2000 By cash - - 1,800.00 4,100.00

Calculate the total interest earned by her upto 30 -06-2000, the rate of interest are as follow

(i) 4.5% p.a. from 01. 10-99 to 31 -03. 2000.

(ii) 4 % p.a. from 01. 04.2000 to date . [4]

Sol:

Month Qualifying amount (in Rs.)

January 5,000

February 2,300

March 2,300 =9600

April 2,300

May 2,300

June 4,100 = 8700

(i) Interest earned by pooja at 4.5%P R T 9600 4.5 1

Rs.36100 100 12

(ii) Interest earned by pooja at 4 %8,700 4 1

Rs.29100 12

Total interest =Rs. 36 29 Rs.65 .



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10.

(a) Using a ruler and compass only construct :

(i) A triangle ABC in which AB = 9 cm, BC = 10 cm and0

ABC 45 .(ii) Also, Construct a circle of radius 2 cm to touch the arms of ABC . [4Sol:

(a)

(b) The Marks of 200 students in an exam were recorded as follows:

Marks % 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No. of

students

7 11 20 46 57 37 15 7

Draw a cumulative frequency table and hence, draw the Ogive and use it to find:

(i) The median, and

(ii) The number of students who score more than 40 % marks. [6]

Sol:

(b)

Marks % No. of

students(frequency)

Cumulative

(frequency)

10-20 7 7

20-30 11 18

30-40 20 3840-50 46 84

50-60 57 141

60-70 37 178

70-80 15 193

80-90 7 200



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(i) Median

n n1

2 2

2

th observation

200 2001

2 22

th observation

100 101

2

the observation

100.5 th observation = 52.5%

(ii) Number of students getting more than 40%

Marks = 200 – 38 = 162

11.

(a) Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is al

cyclic. [3Sol:

(a)

PQRS is a quadrilateral



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0P Q R S 360

01 2 3 4 5 6 7 8 360

02 1 2 4 2 5 2 8 360

1 2, 3 4, 5 6, 7 8 02 1 4 5 8 360

01 4 5 8 180

01 8 4 5 180 0 0 0180 9 180 10 180

0 0360 9 10 180

So,0 0

9 10 360 180 09 10 180

Hence XUYW is a cyclic quadrilatral.

(b) Without using mathematical tables evaluate :

0 0 0 0 0

2 0 2 0

tan2 tan3 tan45 tan87 tan88

2 sec 20 cot 70 [3

Sol:

0 0 0 0 0

2 0 2 0

tan2 tan3 tan45 tan87 tan88

2 sec 20 cot 70

0 0 0 0 0 0

2 0 2 0

tan2 .tan3 .tan45 .tan 90 3 tan 90 2

2 sec 20 cot 90 20

0 0 0 0 0

2 0 2 0

tan2 tan3 tan45 cot3 cot2

2 sec 20 tan 20



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Using , 0 2 2tan 90 θ cot θ & sec θ tan θ 1

1 1

2 1 2

(c) A train covers a distance of 90km at a uniform speed. Had the speed been 15 km per ho

more, it would have taken 30 minute less for the journey . Find the original speed of the tra[4]

Sol:

Total distance = 90km

Let original speed of the train be = x km/hr

DistenceTime

Speed

90hr

x

Now the increased speed of the trainx 15 km/ hr

DistanceTime

speed

90hr

x 15

According to the Problem

90 90 1

x x 15 2

90 x 15 90x 1

x x 15 2

2

90x 1350 90x 1

2x 15x

22 1350 x 15x

2x 15x 2700 0

x x 60 45 x 60 0 Either x 60 0 or x 45 0

x 60 or x 45 Rejecting x 60 [as speed cannot be negative]Hence original speed of the train = 45 km / hr

10th ICSE Guess Paper With Sol

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