ICSE Board Class X Mathematics Board Paper 2013...

ICSE Board

Class X Mathematics

Board Paper – 2013 Solution

Sol.1

(a) A+2X=2B+C

02

6222

20

04

04

23

02

62

2008

04462

02

62

28

422

02

62

28

422

26

1042

13

52

26

104

2

1

(b) P = Rs.4000, C.I. = Rs.1324, n = 3 years

Amount, A = P + C.I. = Rs.4000 + Rs.1324 =Rs. 5324

A=P

nr

1001

5324=4000

3

1001

r

1000

1331

4000

5324

1001

3

r

33

10

11

1001

r

10

11

1001

r

10

11

10

11

100

r

10r %

(C) The observations in ascending order are:

11, 12, 14, (x-2), (x+4), (x+9), 32, 38, 47

Number of observations = 9 (odd)

thn

Median

2

1observation = 5 th observation

∴ x+4=24

x=20

Thus, the observations are:

11, 12, 14, 18, 24, 29, 32, 38, 47

9

473832292418141211 Mean

9

225

=25

Sol.2

(A) Let the number added be x.

xxxx 4320::15:6

x

x

x

x

43

20

15

6

xxxx 1520436

22 1520300436258 xxxxxx

4214 x

3x

Thus, the required number which should be added is 3.

(b) Let p 142 23 bxaxxx

Given, 2x is a factor of p x

02Re pmainder

014222223

ba

0142416 ba

4a + 2b + 2 = 0

2a + b + 1 = 0 …. (1)

Given, when p(x) is divided by (x - 3), it leaves a remainder 52.

523 p

5214333223

ba

052143954 ba

01239 ba

043 ba …. (2)

Subtracting (1) from (2), we get

505 aa

From (1)

110110 bb

(C)

Steps for calculation of mode.

(i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B.

(ii) Mark the inner corner of adjacent rectangles as C and D.

(iii) Join AC and BD to intersect at K. From K, draw KL perpendicular to x-axis.

(iv) The value of L on x- axis represents the mode.

Mode = 13

Sol.3

(a) 0000 31sec59sin210cos.80cos3 ec

000000 31sec3190sin210cos1090cos3 ec

000 31sec,31cos210cos10sin3 ec

sin90cos,cos90sin 00

1213 1sec.cos,1cos.sin c

523

(B) (i) ,ABDin

0180 ADBABDDAB

000 1807065 ADB

0000 456570180 ADB

000 904545, BDCADBADCNow

ADC is the angle of semi-circle so AC is a diameter of the circle.

(ii) ADBACB (angle subtended by the same segment)

045 ACB

(C)

(i) Radius AC 225723

22125

14425

169

=13 units

(ii) Let the coordinates of B be (x, y)

Using mid-point formula, we have

2

32

x

2

75

y

x 34 y 710

7x 17y

Thus, the coordinates of points B are (-7, 17).

Sol.4

(a) x²-5x-10 = 0

Use quadratic formula: a = 1, b = -5, c = -10

a

acbbx

2

42

12

1014552

x

2

40255 x

2

655 x

2

06.85 x

2

06.3,

2

06.13 x

53.1.53.6 x

(b) (i) In ,DECABCand

090 DECABC (perpendiculars to BC)

DCEACB (Common)

ΔABC~ΔDEC (AA criterion)

(ii)Since ΔABC~ΔDEC

CD

AC

DE

AB

CD

15

4

6

606 DC

106

60CD cm

(iii) It is known that the ratio of areas of two similar triangles is equal to the square of the ratio of

their corresponding sides.

4:916:364:6:: 2222 DEABDECarABCar

(c)

(i)

(ii) Co-ordinates of A' = (-6, -4)

Co-ordinates of B' = (0, -4)

(iii) ABA'B' is a parallelogram.

(iv) AB = A'B' = 6 units

In ΔOBO'

OO' = 3 units

OB = 4 units

BO’ 2,

2 00 OB

22 34

25

= 5 units

Since BO' = 5 units

BA' = 10 units = AB'

Perimeter of ABA'B' = (6 + 10 + 6 + 10) units = 32 units

SECTION – B (40 MARKS)

Sol.5

(a) The given inequation is Rxxx

,6

1

3

11

23

3

11

23

xx

3

4

23

xx

6

1

3

11

2

x

3

4

6

32

xx and

3

4

6

1

2

x

3

4

6

5

x

6

81

2

x

85 x 6

9

2

x

5

8x

6

18x

6.1x 3x

The solution set is {x: x R and 1.6 x < 3}

It can be represented on a number line as follows:

(b) Maturity amount = Rs 8088

Period (n) = 3 yrs = 36 months

Rate = 8% p.a

Let x be the monthly deposit.

S.I. =

100122

1 rnnp

xp 44.4100

8

24

3736

Total amount of maturity xxx 44.4044.436

Now,

40.44x=8088

200 x

Thus, the value of monthly installment is Rs 200

(c)

(i) No. of shares = Rs 50

Market value of one share = Rs 132

Salman's investment = (132 x 50) = Rs 6600

(ii) Dividend on one share = 7.5% of Rs 100 = Rs 7.50

His annual income = 50 Rs7.50 = Rs375

(iii) Salman wants to increase his income by 150.

Income on one share = Rs 7.50

No. of extra shares he buys 2050.7

150

So.6

(a)

LHS

CosA

CosA

1

1

CosACosA

CosACosA

11

11

2

2

1

1

CosA

ACos

2

2

1 CosA

ASin

RHSCosA

SinA

1

(b) In cyclic quadrilateral ABCD,

0180 DB (opp. angles of cyclic quadrilateral are supplementary)

00 180100 ADC

080 ADC

Now, in ∆ACD,

0180 ADCCADACD

40 000 18080 CAD

000 60120180 CAD

Now CADDCT (angles in the alternate segment)

∴ DCT = 60 0

(c)

On completing the given table, we get:

Principal for the month of Feb = Rs 4500

Principal for the month of March = Rs 4500

Principal for the month of April = Rs 4500

Principal for the month of May = Rs 6730

Principal for the month of June = Rs 1730

Principal for the month of July = Rs 7730

Total Principal for 1 month

= Rs (4500+4500+4500+6730+1730+7730) Rs = 29690

P = Rs 29690, ,12

1yearsT R=6%

∴ Interest 100

TRP

12100

1629690

= Rs 148.45

Sol. 7

(a) The vertices of ABC are A (3, 5), B (7, 8) and C (1, -10).

Coordinates of the mid-point D of BC are

2,

2

2121 yyxx

2

)10(8,

2

17

2

2,

2

8

1,4

Slope of AD12

12

xx

yy

34

51

61

6

Now, the equation of median is given by:

11 xxmyy

365 xy

1865 xy

0236 yx

(b) Since, the shopkeeper sells the article for Rs 1500 and charges sales-tax at the

rate of 12%.

Tax charged by the shopkeeper = 12% of Rs 1500 = Rs 180

VAT = Tax charges - Tax paid

Rs 36 = Rs 180 - Tax paid

Tax paid = Rs 144

If the shopkeeper buys the article Rs x

Tax on it = 12% on Rs x = Rs 144

x Rs 144 12

100Rs 1200

Thus, the price (inclusive of tax) paid by the shopkeeper = Rs 1200 + Rs 144 = Rs 1344.

(c)

(i) In ∆ AEC,

Tan 30EC

AE0

x

y

3

1

)1.......(3

xy

In ∆ DBA,

Cot 60EC

AE0

603

1 x

mx 64.343203

60

Therefore, the horizontal distance between AB and CD = 34.64 m.

(ii) Substituting the value of x in (1),

3

320 y m20

Height of the lamp post = CD = (60 - 20) m = 40 m

Sol. 8

(a)

yy

xx

4

3

1

2=

12

5

yy

xx

42

32=

12

5

1532 xxx

and 21242 yyy

(b) Sphere : R = 15 cm

Cone : r = 2.5 cm, h = 8 cm

Let the number of cones recasted be n

n Volume of one cone = Volume of solid sphere

32

3

4

3

1Rhrn

3215485.2 xn

85.25.2

1515154

n

270n

Thus, 270 cones were recasted.

(c) x 2 + (p - 3) x + p = 0

Here, A = 1, B = (p - 3), C = p

Since, the roots are real and equal, D = 0

042 acb

01432

pp

04692 ppp

09102 pp

0)9)(1( pp

1 p or 9p

Sol. 9

(a) Area of the quadrant OACB =

2

4

1r

5.35.37

22

4

1

2625.9 cm

Area of the triangle OAD =

heightbase2

1

25.32

1

2,5.3 cm

Shaded Area = Area of quadrant OACB - area of triangle OAD area of triangle OAD

= 9.625 – 3.5 cm 2

= 6.125 cm 2

(b) Number of white balls in the bag = 30

Let the number of black balls in the box be x

Total number of balls = x + 30

P (drawing a black ball) 30

x

x

P (drawing a white ball) 30

30

x

It is given that:

P (drawing a black ball) x5

2 P(drawing a white ball)

30

30

5

2

30

xx

x

30

12

30

xx

x

12 x

Therefore, number of black balls in the box is 12.

(c)

Here, A = 55, h = 10

Mean hf

fdA

1050

2755

4.555

6.49

Sol. 10

(i) Steps of constructions:

(a) Draw a line segment BC = 6 cm.

(b) At B, draw a ray BX making an angle of 1200 with BC.

(c) From point B cut an arc of radius 3.5 cm to meet ray BX at C

(d) Join AC

ABC is the required triangle

(ii)

(a) Bisect BC and draw a circle with BC as diameter.

(b) Draw perpendicular bisectors of AB. Let the two bisectors meet the ray of angle

bisector of ABC at point P. P is equidistant from AB and BC

(iii) On measuring BCP = 30 0

(b)

602

120n

(i) Through marks 60, draw a line segment parallel to x-axis which meets the curve

at A. From A, draw a line perpendicular to x-axis meeting at B.

Median = 43

(ii) Through marks 75, draw a line segment parallel to y-axis which meets the curve

at D. From D, draw a line perpendicular to y-axis which meets y-axis at 110.

Number of students getting more than 75% = 120 - 110 = 10 students

(iii) Through marks 40, draw a line segment parallel to y-axis which meets the curve

at C. From C, draw a line perpendicular to y-axis which meets y-axis at 52.

Number of students who did not pass = 52.

Sol. 11

(a) Let the coordinates of A and B be (x, 0) and (0, y) respectively.

Given P divides AB is the ratio 2:3,

Using section formula, we have:

32

3023

x

32

0324

y

5

33

x

5

24

y

x315 20=2y

5x 10y

Thus, the coordinates of A and B are (-5, 0) and (0, 10) respectively

(b) 8

17

2

12

4

x

x

Using componendo and dividendo,

817

817

21

214

24

xx

xx

9

25

1

122

22

x

x

2

2

22

3

5

1

1

x

x

3

5

1

12

2

x

x

35

35

11

1122

22

xx

xx (Using componendo and dividend)

2

8

2

2 2

x

42 x

2 x

(c) Original cost of each book = Rs x

Number of books brought for Rs x

960960

In 2 nd case:

The cost of each book = (x - 8)

Number of books bought for Rs 8

960960

x

From the given information, we have:

4960

8

960

xx

48

8960960960

xx

xx

19204

8960)8(

xx

0192082 xx

04048 xx

48 x or 40

But x can't be negative,

x = 48

Thus, the original cost of each book is Rs 48