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MAHESH TUTORIALS I.C.S.E.

Model Answer Paper

Marks : 80

Time : 2½ hrs.

SUBJECT : MATHEMATICSICSE X

Exam No. : MT/ICSE/Semi Prelim I-Set-B-003

T17 I SP A003 Turn over

SECTION – A (40 Marks)

Attempt all questions from this Section.A.1(a) Sum deposited per month (P) = ` 640

No. of months (n) = 54 monthsRate of interest (r) = 12%Maturity value = ?

Interest earned = P ×( +1)2×12

n n × 100

r

I = 640 ×54(54 1)

2 12

×

12100

I = 640 ×54 552 12

×12

100 I = ` 9504 Maturity value= (640 × 54) + 9504 Maturity value= ` 44,064 [3]

(b) Cards are marked as 1, 2, 3, 4, ....... 29, 30 Total possible outcomes = 30(i) A multiple of 4 or 6

4, 6, 8, 12, 16, 18, 20, 24, 28, 30i.e. number of favourable outcomes = 10

P (a multiple of 4 or 6) =1030

=13

(ii) Multiple of 3 and 5 are 15, 30

P (a multiple of 3 and 5) =2

30 =

115

(iii) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30 Number of favourable outcomes = 14

P (a multiples of 3 or 5) =1430

=7

15 [3]

(c) Let A (–2, 4)B (4, 8)C (10, 7)

and D (11, –5)

Turn overT17 I SP A003

Set B

Let M be the mid-point of AB, By Mid-point formula,

M –2 + 4 4 + 8

,2 2

2 12

,2 2

(1, 6)Let N be the mid-point of CD

By Mid-point formula,

N 11 + 10 –5 + 7

,2 2

21 2

,2 2

21, 1

2Let P be the mid-point of AD

By Mid-point formula

P –2 + 11 4 + (–5)

,2 2

9 –1

,2 2

Let Q be the mid-point of BC By Mid-point formula

Q 4 + 10 8 + 7

,2 2

14 15

,2 2

157,

2 M (1, 6)

N

21, 1

2

P 9 –1

,2 2

and Q 15

7,2

Slope of QM = 2 1

2 1

y yx x

15 627 1

... 2 ...

(11, –5)

B

C

A

D(10, 7)

(–2, 4) (4, 8)

P

M

N

Q


Set B

3

2 6

14

....(i)

Slope of QN = 2 1

2 1

y yx x

=

1512

21 72

=

13272

=137

....(ii)

Slope of NP = 2 1

2 1

y yx x

=

1 12

9 212 2

=

32122

=3

12

=14

....(iii)

Slope of PM = 2 1

2 1

y yx x

=

162

912

=

13272

=137

....(iv)

Slope of QM = Slope of NP [From (i) and (iii)]

... 3 ...


Set B... 4 ...

QM || NP ....(v)Slope of QN = Slope of MP [From (ii) and (iv)]

QN || MP ....(vi) QNPM is a parallelogram [A quadrilateral is said to be a

parallelogram if its both pair of opposite sides are parallel] [4]

A.2.(a) Diameter of cylinder = 20 cm

its radius (r) = 10 cm(i) Curved surface area of the cylinder = 100 cm2

2rh = 100

2 ×227× 10 × h = 100

h =100×720×22

=5 7

22 =

3522

h = 1.6 The height of the cylinder is 1.6 cm

(ii) Volume of the cylinder = r2h

=227

× 10 × 10 × 1.6

=3520

7= 502.85= 502.9 cm3

Volume of the cylinder is 502.9 cm3 [3]

(b) A =–1 1

a b

A2 = A.A

=–1 1

a b

–1 1

a b

=

–1 × –1 + 1 × a –1 × 1 + 1 × b

a× –1 + b × a a × 1 + b × b

=

2

1 + a –1 + b

–a + ab a + b

Since A2 = I (Given)

2

1 + a –1 + b

–a + ab a + b

=

1 0

0 1


Set B... 5 ...

1 + a = 1 a = 1 – 1 = 0

– 1 + b = 0 b = 1 a = 0; b = 1 [3]

(c) (i) Marked price of article = ` 500S.P. of wholesaler or C.P. of retailer at 20% discount

= 500 ×

100 – 20100

= 500 ×80

100 = ` 400

Sales tax of 12.5% charged by wholesaler

=12.5100

× 400 = ` 50

S.P. of retailer or C.P. of customer at marked price = ` 500

Sales tax charged by retailer =12.5100

× 500 = ` 62.50

Also, sales tax paid by retailer = Sales tax chargedby wholesaler = ` 50

(i) Price paid by the customer = 500 + 62.50= ` 562.50

(ii) VAT paid by the retailer = Tax charged by retailer – Tax paid by retailer

= 62.50 – 50= ` 12.50 [4]

A.3.(a) Let the ratio be k : 1

Let A (–1 , 4) (x1 , y1)

B (4 , –1) (x2 , y2)

Now, x =1 2 2 1

1 2

++

m x m xm m

1 =(4) + 1 (–1)

+ 1k

k

1 =4 – 1

+ 1kk

k + 1 = 4k – 1 2 = 4k – k 2 = 3k

k =23

m1 : m2 = 2 : 3

A(–1, 4)(x1, y1)

P(1, a)

B(4, –1)(x2, y2)

k 1


Set B... 6 ...

Now, a =1 2 2 1

1 2

++

m y m ym m

a =2(–1) + 3 (4)

2+ 3

a =–2 +12

5

a =105

a = 2 [3]

(b) P(a, b) is reflected in origin, M0(a, b) = (–a, –b)

Now (–a, –b) is reflected in y-axis to P My(–a, b) = (a, –b)

But P = (4, 6) (given) (a, –b) = (4, 6)

a = 4 and –b = 6i.e. b = –6Hence a = 4 and b = –6 [3]

(c) Let the two parts be ` x and ` (50,760 – x)For 1st part :-Sum invested = xRate of Dividend = 8%

Face Value = ` 100Discount = ` 8

Market Value = ` 100 – ` 8= ` 92

No. of shares =Sum investedM.V. of 1share

=92x

Income = No. of shares × Rate of Dividend × face value

=92x

×8

100 × 100 = `

223x

For 2nd part :-

Sum invested = 50, 760 – x

Rate of Dividend = 9%Face Value = ` 100

Premium = ` 8

Market Value = ` 100 + ` 8= ` 108


Set B... 7 ...

No. of shares =Sum investedM.V. of 1share

=50,760 –

108x

Income = No. of shares × Rate of Dividend × face value

=50760 –

108x

×9

100 × 100

=50760 –

12x

Income from both the investment are equal

223x

=50760 –

12x

2x × 12 = 23(50760 – x)

24x = 1167480 – 23x 24x + 23 x = 1167480 47x = 1167480

x =1167480

47 x = 24840

and (50760 – x) = 50760 – 24840= 25920

The two parts are ` 24,840 and ` 25,920 [4]

A.4.(a) Mean x = 62.8

Class f Mid-value x fx0 – 20 5 10 5020 – 40 f1 30 30 f1

40 – 60 10 50 50060 – 80 f2 70 70f2

80 – 100 7 90 630100 – 120 8 110 880

f = 30 fx = 30f1 + f1 + f2 +70f2 + 2060

f = 50 30 + f1 + f2 = 50 f1 + f2 = 20 ....(i)

x =fxf


Set B... 8 ...

62.8 = 1 130 + 70 + 206050

f f

3140 = 30f1 + 70f2 + 2060 30f1 + 70f2 = 3140 – 2060 30f1 + 70f2 = 1080 3f1 + 7f2 = 108 ....(ii)

Multiplying (i) by 3 and subtracting (ii) from3f1 + 3f2 = 603f1 + 7f2 = 108

– – ––4f2 = –48

f2 =–48–4

= 12

Put this value in (ii), f1 + 12 = 20

f1 = 20 – 12= 8

Hence f1 = 8, f2 = 12 [3]

(b) CONE :Radius (rc) = 5 cmHeight (hc) = 8 cmSPHERE :Radius (rs) = 0.5 cmLet ‘N’ spheres be formedVolume of Cone = N × Volume of Sphere

N =Volume of Cone

Volume of Sphere

N =

2

3

1 r34 r3

c c

s

h

N =5 5 8

4 0.5 0.5 0.5

N = 400 Number of spheres formed is 400 [3]

(c) Given : PQR in which XY || QRPX : XQ = 1 : 3 and QR = 9 cmand ar (PXY) = x cm2

To find : (i) Length XY(ii) ar(PQR)(iii) ar(trapezium XQRY)

Proof : (i) XY || QR PXY = PQR (Corresponding angles)Also, P = P (Common angle) PQR ~ PXY (AA postulate)

R

P

X

Q

Y


Set B... 9 ...

PQPX

=QRXY

.....(1) (Corresponding sides of similar

triangles are proportional)

NowPXXQ =

13

XQPX

=31

XQ + PX

PX=

3 +11

PQPX

=41

...... (2)

From (1) and (2)

41

=QRXY

41

=9

XY

XY =94

XY = 2.25 cm

(ii) Further

ar ΔPXY

ar ΔPQR =2

2

XYQR

ar ΔPQRx

=2

14

ar ΔPQRx

=1

16 ar (PQR) = 16x cm2

(ii) Area (trap. XQRY) = ar (PQR) – ar (PXY)= 16x – x= 15x cm2

Area (trap. XQRY) = 15x cm2 [4]

SECTION – II (40 Marks)Attempt any four questions from this Section.

A.5.(a) Let A (–4, –2) (x1, y1)

B (2, –3) (x2, y2)

Slope of AB =2 1

2 1

– –

y yx x

=–3 – (–2)2 – (–4)


Set B... 10 ...

=–3 + 22 + 4

=–16

Now,line AB is perpendicular to the line passing through (a, 5) and (2, –1)

Slope of line passing through (a, 5) and (2, –1) =–1

Slope of AB

=–1–16

= 6

6 =–1 – 52 –a

6(2 – a) = –6 12 – 6a = –6 – 6a = –18 6a = 18

a =186

a = 3 [3]

(b) (i) Face value of 1 share = ` 80Premium = 30% of ` 80

= `30

100 × 80 = ` 24

Market value of 1 share = ` 80 + ` 24 = ` 104 Market value of 150 shares = ` (150 × 104) = ` 15,600

Market value of 150 shares is ` 15,600(ii) Dividend on 1 share = 18% of ` 80

= `18100

× 80 = `1440100

Dividend on 150 shares = (14.40 × 150)= ` 2,160

Dividend on 150 shares is ` 2,160

(iii) % return =2,160

15,600 × 100

=18012

= 13.846 = 13.85%

% return = 13.85% [3]


Set B... 11 ...

(c) Construction :

Steps :For triangle :1. Draw AB = 7.0 cm.2. Construct MBA = 60º.3. Mark a point C on ray BM, such that BA = 8.0 cm.4. Draw AC.

ABC is the required triangle.

For point P :1. Draw perpendicular bisector of BC.2. Draw angle bisector of ABC.3. Mark point of intersection of perpendicular bisector and angle

bisector as P.4. P is the required point.

(i) P is equidistant from B and C as it lies on perpendicularbisector of BC.

(ii) P is equidistant from AB and BC as it lies on anglebisector of ABC.

(iii) PB = 4.6 cm [4]

A.6.(a) Given : In the figure,

(i) AB and DE are perpendiculars to BC.(ii) AB = 9 cm, DE = 3 cm, AC = 24 cm.

To find : ADSoln. In ABC and DEC,

ABC = DEC, (Each 90º)C = C (Common angle)

ABC ~ DEC (By AA postulate)

ABDE

=ACDC

(Sides of similar triangles are proportional)

BA

P

C

M

7cm

60º


Set B... 12 ...

93

=24DC

DC =24 3

9

DC = 8 cmAD = AC – DC

= 24 – 8= 16

AC = 16 cm [3]

(b) A (5 , x)B (–4 , 3)C (y , – 2)Let O be the centroid of ABC

The centroid is the origin O (0 , 0) By centroid formula,

0 =5 4

3 y

0 = 5 – 4 + y 0 = 1 + y y = –1

0 =3 23

x

0 = x + 1

x = – 1 [3]

(c)

8 – 2

1 4 . X =

12

10

Order of matrix

12

10 is 2 × 1

Order of

8 – 2

1 4 is 2 × 2

Let order of matix X be a × b

i.e. 2×2

8 – 2

1 4 × Xa×b = 2×1

12

10(1st matrix) (2nd matrix) (resulting matrix)Since, product of matrices is possiblel, only where the number ofcolumns in the first matrix is equal to the number of rows in thesecond.

a = 2Also, the number of columns of product (resulting) matrix is equal to

B CE

D

A

?

9 cm

3 cm

24 cm


Set B... 13 ...

the number of rows of 2nd matrix. b = 1 Order of matrix X = a × b = 2 × 1

Let X =

xy

8 – 2

1 4

xy =

12

10

8 – 2

4

x y x y =

12

10

8x – 2y = 12 ...(i) x + 4y = 10 ...(ii)

Multiplying (i) by 2 adding (ii) to it, 16x – 4y = 24 x + 4y = 10

17x = 34

x =3417

x = 2Putting this value in (i);

8x – 2y = 12 8 × 2 – 2y = 12 –2y = 12 – 16 –2y = –4

y =–4–2

= 2

X =

xy =

2

2

Hence, (i) 2 × 1 (ii)

22 [4]

A.7.(a) Area of square = 142

n(S) = 196Side of square=2 × diameter of 1 circle

14 = 2 × d

d =142

d = 7 cm

r =2d

r =72

cm


Set B... 14 ...

Area of 4 circles = 4 × r2

= 4 ×227

×72

×72

= 154 cm2

Area of shaded region = Area of square – Area of 4 circles= 196 – 154= 42 cm2

Let A be the event that the die falls on the shaded region n(A) = 42

P(A) =n(A)n(S)

=42

196

P(A) =3

14 [3]

(b) Internal diameter = 21 cm

Internal radius, r =212

= 10.5 cm

External diameter = 28 cm

External radius, R =282

= 14 cm

Volume of the material of the vessel =12

×43(R3 – r3)

=23

×227

33 2114 –

2

=23

×227

92612744 –8

=23

×227

21952 – 92618

=23

×227

×12691

8

=11×1813

3 × 2

=19943

6 = 3323.83 cm3

Volume of the material of the vessel = 3323.83 cm3 [3]

(c) Sum deposited per month (P) = ` 400Rate of interest (r) = 8%Maturity value = ` 16,176


Set B... 15 ...

Calculate : Period i.e., no. of months (n)

Maturity value = (P × n) + P ×( +1)2×12

n n × 100

r

Maturity value = (400 × n) + P ×( 1)2 12

n n

× 100r

16,176 = 400n + 400 ×( 1)2 12

n n

×8

100

16,176 = 400n +4 ( 1)

3n n

16,176 =21200 4 4

3n n n

16,176 × 3 = 4 (n2 + 301n)

n2 + 301n =16,176 3

4

n2 + 301n = 12,132 n2 + 301n – 12,132 = 0 n2 + 337 n – 36n – 12,132 = 0 n(n + 337) – 36 (n + 337) = 0 (n – 36) (n + 337) = 0 n – 36 = 0 or n + 337 = 0 n = 36 or n = – 337

No. of months can not be negative, hencen = – 337 is not applicable value.

no. of months = 36 Period of R.D. = 3 years. [4]

A.8.

(a) A =

1 0

2 1 B =

2 3

–1 0 A2 + AB + B2 = A.A + A.B + B.B

=

1 0

2 1

1 0

2 1 +

1 0

2 1

2 3

–1 0 +

2 3

–1 0

2 3

–1 0

=

1 + 0 0 + 0

2 + 2 0 + 1 +

2 + 0 3 + 0

4 – 1 6 + 0 +4 – 3 6 + 0

–2 + 0 – 3 + 0

=

1 0

4 1 +

2 3

3 6 +

1 6

–2 – 3

=

1 + 2 + 1 0 + 3 + 6

4 + 3 – 2 1 + 6 – 3

A2 + AB + B2 =

4 95 4 [3]


Set B... 16 ...

(b) Sum Invested = ` 3072Rate of Dividend = 5%

Face Value = ` 10Market Value = ` 16

(i) His annual Income :

No. of shares =Sum Invested

Market Value of 1 share

=307216

= 192 No. of shares = 192

Annual Income = No. of Shares × Rate of Dividend× Face Value

= 192 ×5

100 × 10

= ` 96 Annual Income = ` 96(ii) His Percentage income on his investment :

Rate of dividend × F.V = Income % × Market Value 5 × 10 = Income % × 16

Income % =5 ×10

16Percentage income = 3.125% [3]

(c) P = (4 , 2)Q = (–1 , 5)O = (–3 , 2)

Let R = (h , k) andS = (x , y)

PQRS is a parallelogram(Given) Diagonals of PQRS bisect each other Point O is the midpoint of diagonals PR & QS.

For PR :By mid-point formula,

– 3 =4

2 h

– 6 = 4 + h – 6 –4 = h h = – 10

2 =2

2k

4 = 2 + k k = 4 – 2 = 2

P (4 , 2) Q (–1, 5)

S(x, y) R (h , k)

O (–3 , 2)


Set B... 17 ...

R = ( –10 , 2)For QS :By mid-point Formula,

–3 =12

x

–6 = –1 + x –6 +1 = x x = – 5

2 =5

2y

4 = y + 5 y = – 1 S = (–5 , –1) [4]

A.9.(a) Scale factor, m = 0.8

LMN ~ LMN

L MLM

=M NMN

=L NLN

= m ...(i)

(i)M NMN

= m (From (i))

M N

8

= 0.8.

M N = 0.8 × 8 M N = 6.4 cm

(ii)L MLM

= m (From (i))

5.4LM

= 0.8

LM =5.40.8

LM =548

LM = 6.75 cm [3]

(b) Sum deposited per month (P) = ` 300No.of months (n) = 24 monthsMaturity value = ` 7725Rate of interest (r) = ?

Maturity value = (P × n) + P ×( +1)2×12

n n × 100

r

Maturity value = ` (300 × 24) + P ×24(24 1)

2 12

× 100

r


Set B... 18 ...

7725 = 7200 +300 24(25)

2 12

× 100

r

7725 – 7200 = 75 r525 = 75 r

r =52575

r = 7% Rate of interest = 7% [3]

(c)

Mode = 53 [4]

A.10.(a) x-intercept = 4 units

Corresponding point on x-axis = (4, 0)The other given point = (2, 3)

Slope m =3 – 02 – 4

=3–2

=3–2

Equation of line through (2, 3) and slope

–32 is

y – y1 = m (x – x1)

y – 3 =–32

(x – 2)

2y – 6 = –3x + 63x + 2y = 12 [3]

20

2

8

4

10

6

12

30 40 50 60 70 800

Mod

e

Class-intervals

Freq

uenc

y

BA

K

L

D

C

X

Y


Set B... 19 ...

(b) Cost Price of 15 identical articles = ` 840

Cost Price of 1 article =84015

= ` 56

Cost Price of 6 articles = 6 × Cost Price of 1 article Cost Price of 6 articles = 6 × 56 = ` 336

Selling Price of 1 article = ` 65 Selling Price of 6 articles = 6 × Selling Price of 1 article

Selling Price of 6 articles = 6 × 65 = ` 390Rate of Sales Tax = 8%VAT paid by shopkeeper against the sale of 6 articles = ?

C.P. Tax 8% S.P. Tax 8% VAT

Shopkeeper ` 336 ` 26.88 ` 390 ` 31.20 ` 4.32

Tax on Cost Price = Tax paidTax on Selling Price = Tax charged

Tax paid =8

100 × 336 = ` 26.88

Tax charged =8

100 × 390 = ` 31.20

VAT paid by Shopkeeper = Tax charged – Tax paid VAT paid by Shopkeeper = 31.20 – 26.88

VAT paid by Shopkeeper = ` 4.32 [3]

(c) Total Surface area of the cuboid = 2(lb + bh + lh)= 2(42 × 30 + 30 × 20 + 42 × 20)= 2(1260 + 600 + 840)= 5400 cm2

l2 = r2 + h2

=

2142

+ 242

= 72 + 242

= 49 + 576 l2 = 625 l = 25 cm

Curved surface area of the cone = rl

=227

×142

× 25

= 22 × 25= 550 cm2

Area of the circular base = r2

=227

×142

×142

= 11 × 14 = 154 cm2


Set B... 20 ...

Surface area of the remaining solid = Total surface area of the cuboid – Area of circular base + curved surface area of the cone= 5400 + 550 – 154 = 5796 cm2

Surface area of remaining solid is 5796 cm2 [4]

A.11.(a) (a) Line x = 0 is the equation of y-axis

My(–4, 3) = (4, 3)

Q is (4, 3)

(b) Q (4, 3) is reflected in liney = 0 i.e. x-axis get image R

Mx(x, y) = (x, –y)

R = (4, –3)

(c)

Plot the points P(–4, 3), Q(4, 3) and R(4, –3).Join PQ, QR and PR.

The figure PQR is right angled triangle.(d) PQ = 4 + 4 = 8 units

QR = 3 + 3 = 6 units

Area PQR =12

× PQ × QR

=12

× 8 × 6 = 24 sq. units [4]

2

3

1

-2

-3

-11 2 3 4-5 -4 -3 -2 -1 -0

-4

-5

4

5Y

XX 5

Y

R(4, –3)

Q(4, 3)P(–4, 3)


Set B... 21 ...

(b) Height (in cm) No. of Cumulativestudents frequency c.f.

135 – 140 0 0140 – 145 12 12145 – 150 20 32150 – 155 30 62155 – 160 38 100160 – 165 24 124165 – 170 16 140170 – 175 12 152175 – 180 8 160

20

40

60

80

100

160

1400 X

(140, 0)

120

140

145 150 155157.5

160 165 170 175 180

(175,152)

(180,160)

(145, 12)

(155,62)

Scale : On X axis : 2 cm = 5 cm height On Y axis : 2 cm = 20 students

(170,140)

(165,124)

(160,100)

(150, 32)No.

ofst

ude

nts

Height

B

P

Q

R


Set B... 22 ...

(i) Median :N = 160

N2

=160

2 = 80

N = (157.5, 0)

Median height = 157.5 cm

(ii)N4

=160

4 = 4

Q1 = 151 cm

3N4

=1603 ×

4= 3 × 40

= 120

Q3 = 163 cm

Inter quartile range = Q3 – Q1

= 163 – 151

= 12 cm

(iii) Number of students whose height is less than 172 cm = 146

Number of students whose height is above 172 cm = 160 – 146 = 14 [6]

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