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-A-003

T18 I SP A003 Turn over

SECTION – I (40 Marks)

Attempt all questions from this Section. A.1

(a) 1 –12 =

12

32 – 1 =

12

The difference between the consecutive terms is constant The given sequence is an arithmetic progression.

where first term, a =12

and common difference, d =12

tn = a + (n – 1) d

t30 =12 + (30 – 1)

12

t30 =12 +

292

t30 =302

t30 = 15

30th term is 15 [3]

(b) f(x) = px3 + 4x2 – 3x + q f(x) is divisible by x2 – 1 f(x) is divisble by (x + 1) (x – 1) x + 1 = 0 x = –1Put x = –1 in f(x)

f(–1) = 0 p(–1)3 + 4(–1)2 – 3(–1) + q = 0 –p + 4 + 3 + q = 0 –p + q = –7 ... (i)Again x – 1 = 0 x = 1Put x = 1 in f(x)

f(1) = 0 p(1)3 + 4(1)2 – 3(1) + q = 0

Set A

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Set A

p + 4 – 3 + q = 0 p + q + 1 = 0 p + q = –1 ... (ii)

Adding (i) and (ii);–p + q = – 7

+ p + q = –12q = –8

q = –4Substituting in (i);

–p + (–4) = –7 –p = –7 + 4 –p = –3 p = 3Hence, p = 3 and q = –4 [4]

(c) Let ‘l’ be the slant height, ‘r’ be the base radius and ‘h’ be the height ofthe tent (cone)Diameter of cone = 48 m it’s radius (r) = 24 m

it’s height (h) = 7 m l2 = r2 + h2

l2 = (24)2 + (7)2

l2 = 576 + 49 l2 = 625 l = 25 m ...... [Taking square roots]

Curved surface area of the tent (cone)= rl

=227

× 24 × 25

=13200

7m2

Let the total area of the canvas be ‘A’Now, 10% of the canvas is used in folds and stitchings

Area of the canvas used in tent = A – 10% of A

= A –10100

A

=9A10

9A10

is the area of canvas required to make a tent

9A10

=13200

7Now, A = l × 1.5 (canvas is rectangular)

... 2 ...

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Set A... 3 ...

9 × × 1.5

10l

=13200

7

l =13200 109 1.5 7

l =13200 2 10

9 7 3

length of canvas =1396.825 cmLength of canvas required = 1396.83 cmCost of canvas at the rate of ` 24 per meter = 1396.83 × 24 = `33523.92

Cost of canvas at the rate of ` 24 per meter is ` 33523.92 [3]

A.2(a) AB = 4 cm

BX = 6 cmDX = 5 cm

Let,CD = x cm

AX = AB + BX [ A – B – X] AX = 4 + 6 AX = 10 cmand

CX = CD + DX [ C – D – X]CX = (x + 5) cm

Now,AX × BX = CX × DX [When two chords of a circle intersect

internally/externally, then the products ofthe lengths of segment are equal.]

10 × 6 = (x + 5) 5 60 = 5x + 25 60 – 25 = 5x 5x = 35

x =355

x = 7

CD = 7 cm [3]

(b) For the given G.P.a = 1

r =2

1

tt =

31

= 3tn = arn – 1

t7 = 1( 3 )7 – 1

= ( 3 )6

B

DA

X

C

4 cm

x

6 cm

5 cm

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Set A... 4 ...

= (123 )6

=1 623

[(am)n = amn]

= 33

t7 = 27 [3]

(c) Let the number to be added be x The nos. will be (16 + x), (7 + x), (79 + x) and (43 + x)

167

xx

=7943

xx

(16 + x)(43 + x) = (79 + x) (7 + x) 688 + 16x + 43x + x2 = 553 + 79x + 7x + x2

59x + 688 = 553 + 86x 86x – 59x = 688 – 553 27x = 135

x =13527

x = 5 [4]

A.3(a) Height (cm) No. of plants fx

x f

50 2 10055 4 22058 10 58060 f 60 f65 5 32570 4 28071 3 213

f = 28 + f fx = 1718 + 60f

Mean =fxf

60.95 =1718 + 60

28 +f

f 28 × 60.95 + 60.95 f = 1718 + 60f 1706.60 + 60.95 f = 1718 + 60f 60.95f – 60 f = 1718 – 1706.60 0.95 f = 11.40

f =1140

95 f = 12 [3]

Turn overT18 I SP A003

Set A... 5 ...

(b) 3x² + 12x + (m + 7) = 0a = 3, b = 12, c = m + 7D = b² – 4ac = (12)² – 4(3)(m+7)

= 144 – 12m – 84= 60 – 12m

Given equation has equal roots D = 0 60 – 12m = 0 12m = 60 m = 5Putting the value of m in the equation

3x² + 12x + (m + 7) = 0 3x² +12x + (5 + 7) = 0

3x² + 12x + 12 = 0 3(x² + 4x + 4) = 0 x² + 4x + 4 = 0 (x)² + 2(x) (2) + (2)² = 0 (x + 2)² = 0 x + 2 = 0 x = –2

Hence m = 5 and solution x = –2 [3]

(c) Let the three numbers in G.P. bea, ar, ar2

Sum =3910

a + ar + ar2 =3910

... (i)

Product = 1 a × ar × ar2 = 1 a3r3 = 1 ar = 1

a =1r

... (ii)

Substituting the value of ‘a’ in equation (i), we get,

1r

+1r

× r +1r

× r2 =3910

1r

+ 1 + r =3910

21+ r + r

r=

3910

10(r2 + r + 1) = 39 × r 10r2 + 10r + 10 = 39r 10r2 + 10r – 39r + 10 = 0 10r2 – 29r + 10 = 0

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Set A... 6 ...

10r2 – 25r – 4r + 10 = 0 5r(2r – 5) –2(2r – 5) = 0 (2r – 5) (5r – 2) = 0 2r – 5 = 0 or 5r – 2 = 0

r =52

or r =25

Case (i) : When r =52

,

a =1r

=152

=25

The numbers are : a, ar, ar2

=25

,25

×52

,25

×52

×52

=25

, 1,52

Case (ii) : When r =25

,

a =1r

=125

=52

The numbers are : a, ar, ar2

=52

,52

×25

,52

×25

×25

=52

, 1,25

The numbers in G.P. are25 , 1,

52 or

52 , 1,

25 [4]

A.4

(a)4

2

12

xx

=178

By Componendo Dividendo,4 2

4 2

1 21– 2

x xx x

=17 817 – 8

4 2

4 2

2 1– 2 1

x xx x

=

259

2 2 2 2

2 2 2 2

( ) 2. .1 (1)( ) – 2. .1 (1)x xx x

=

259

2 2

2 2

( 1)( –1)xx

=259

22

2

1–1

xx

=259

Taking square root on both sides, we get,

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Set A... 7 ...

2

2

1–1

xx

=259

2

2

1–1

xx

= 53

By Componendo – Dividendo, By Componendo – Dividendo,2 2

2 2

( 1) ( –1)( 1) – ( –1)x xx x =

5 35 – 3 2 2

2 2

( 1) ( –1)( 1) – ( –1)x xx x =

–5 3–5 – 3

2 2

2 2

1 –11– 1

x xx x

=82

22

2x

=–2–8

22

2x

= 4 x2 =14

x2 = 4 x = 14

x = 4 x = 12

x = 2 [3]

(b)

PB = 3.6 cm [4]

(c) Let son’s present age = xMan’s present age = x2

One year back,Son’s age = x – 1Father’s age = (x2 – 1)According to condition, (x2 – 1) = 8(x – 1) x2 – 1 = 8x – 8

X

M

A

l

B C

P

3.5

cm

6 cm60º

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Set A... 8 ...

x2 – 8x + 8 – 1 = 0 x2 – 8x + 7 = 0 x2 – 7x – x + 7 = 0 x (x – 7) –1(x – 7) = 0 (x – 7) (x – 1) = 0 x = 7 or x = 1 x2 = 49 or x2 = 1

But x2 = 1, is not possible x = 7 is the son present age, and x2 = 49 is the fathers present age Son’s present age = 7 years and father’s present age = 49 years. [3]

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

A.5

(a)3 – 2 3 – 8

=2 – 3 + 4

x xx x (3x – 2) (x + 4) = (3x – 8) (2x – 3) 3x2 + 12x – 2x – 8 = 6x2 – 9x – 16x + 24 3x2 – 6x2 + 10x + 25x – 8 – 24 = 0 –3x2 + 35x – 32 = 0 3x2 – 35x + 32 = 0 3x2 – 3x – 32x + 32 = 0 3x(x – 1) – 32 (x – 1) = 0 (x – 1) (3x – 32) = 0 (x – 1) = 0 or (3x – 32) = 0

x – 1 = 0 or 3x = 32

x = 1 or x =323

x = 1 or x = 10 32

[3]

(b) Let f(x) = 2x3 – 7x2 + ax – 6Put (x – 2) = 0 x = 2 Remainder = f(2) = 2(2)3 – 7(2)2 + a(2) – 6

f(2) = 16 – 28 + 2a – 6f(2) = – 12 + 2a – 6f(2) = 2a – 18 .......... (i)

Let g(x) = x3 – 8x2 + (2a + 1) x – 16Put (x – 2) = 0 x = 2 Remainder = g(2) = (2)3 – 8(2)2 + (2a + 1)(2) – 16

g(2) = 8 – 32 + 4a + 2 – 16 g(2) = 4a – 38 .... (ii)

f(2) = g(2) [ both polynomials leave same remainder]

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Set A... 9 ...

2a – 18 = 4a – 38 38 – 18 = 4a – 2a 20 = 2a a = 10 The value of a is 10 [3]

(c) First instalment = ` 3000Every other instalment is ` 100 less than the previous one.There are total 12 monthly instalments.Hence the instalments in ` are :

3000, 3000 – 100, 3000 – 2(100), 3000 – 3(100) ...... 3000 – 11(100)= 3000, 2900, 2800, 2700, ......., 1900

This is an A.P.Here a = 3000 and d = –100

(i) Amount of instalment paid in 9th month= t9

= a + (9 – 1)d= a + 8d= 3000 + 8(–100)= 3000 – 800= ` 2200

(ii) The total amount paid in the instalment scheme= Total of 12 instalments= Sum of 12 terms= S12

=122

[2a + (12 – 1)d]

= 6 [2(3000) + 11 × (–100)]= 6 [6000 – 1100]= 6 [4900]= ` 29,400 [4]

A.6(a) (i) SRT = 90º

(Angle in a semi circle)PRS + SRT + TRQ = 180º [Angles in a straight line]

PRS + 90º + 30º = 180º PRS + 120º = 180º PRS = 180º – 120º PRS = 60º

(ii) RST = TRQ (Angles in alternate segment) RST = 30º

ROT = 2 RST [Angle at the centre is twice theangle at the circumference]

= 2 × 300

ROT = 60º [3]

RP

SO

T30º

Q

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Set A... 10 ...

(b) (i) Let number of rows originally = x Number of seats in each row = x Total number of seats = x × x = x2

Number of rows in 2nd case = 2xNumber of seats in each row = x – 10

Total number of seats = 2x (x – 10) = 2x2 – 20xAccording to the given condition,

2x2 – 20x = x2 + 300 2x2 – 20x – x2 – 300 = 0 x2 – 20x – 300 = 0 x2 – 30x + 10x – 300 = 0 x(x – 30) + 10(x – 3) = 0 (x – 30) (x + 10) = 0Either x – 30 = 0 or x + 10 = 0 x = 30 or x = –10But x = –10 is not possible for rows. x = 30Number of rows in original arrangement = 30(ii) Number of seats after re-arrangement in a row

= x – 10 = 30 – 10 = 20Hence (i) 30 (ii) 20. [3]

(c) x =6ab

a b ....(i)

From (i),3xa =

2b

a bBy componendo dividendo,

3–3x a

x a =2 ( )2 – ( )b a bb a b

3

–3x a

x a =3

–b a

b a ....(ii)

From (i),3xb =

2a

a bBy componendo dividendo,

3– 3x b

x b =2 +( + )2 – ( )a a ba a b

3

–3x b

x b =3

–a + ba b ....(iii)

Adding equation (ii) and (iii), we get3

– 3x a

x a +3

– 3x bx b

=3

–b+ a

b a +3

–a + b

a b

=3

–b+ a

b a – 3

–a + b

b a

= 3 – 3

–b+ a a+ b

b a

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Set A... 11 ...

=3 – 3

–b + a a – b

b a

=2 – 2

–b ab a

= 2 –

–b a

b a

+3– 3

x ax a +

+ 3– 3

x bx b = 2 [4]

A.7(a) x2 – 8x + 5 = 0

a =1, b = – 8, c = 5By formula,

2– ± –4=

2b b ac

xa

= 12– – 8 ± (– 8) – 4 ( ) (5)

2 (1)

=8± 64 – 20

2

=8 ± 44

2

=8 ±2 11

2

= 2 4± 11

2 = 4 11

= 4 3.317

x = 4 + 3.317 or x = 4 – 3.317

x = 7.317 or x = 0.683

x = 7.32 or x = 0.68 [Correct to 2 decimal places] [3]

(b) Given Arithmetic Progression is :1 + 4 + 7 + 10 + ....First term, a = 1Common difference, d = 4 – 1 = 3Let nth term be 52 tn = 52 a + (n – 1) d = 52 1 + (n – 1) 3 = 52 3(n – 1) = 51

n – 1 =513

n – 1 = 17

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Set A... 12 ...

n = 17 + 1 n = 18 18th term is 52 [3]

(c) Let f(x) = x3 + ax2 + bx – 12Put (x – 2) = 0 x = 2 Remainder = f(2) = 23 + a(2)2 + b(2) – 12 0= 8 + 4a + 2b – 12 [ x – 2 is a factor of f(x) f(2) =0] 0= 4a + 2b – 4

4a + 2b = 4Dividing throughout by 2

2a + b = 2 ..... (i)Put (x + 3) = 0 x = –3 Remainder = f(–3) = (–3)3 + a(–3)2 + b(– 3) – 12 0 = – 27 + 9a – 3b – 12 [ x + 3 is a factor of f(x)

f(-3) =0] 0 = 9a – 3b – 39 9a – 3b = 39

Dividing throughout by 3, 3a – b = 13 ..... (ii)

By adding (i) and (ii) we get,

+

2 2

3 13

5 = 15

a ba ba

a =155

a = 3Substituting a = 3 in equation (i) we get,

2a + b = 2 2(3) + b = 2

6 + b = 2 6 – 2 = – b

4 = – b

b = – 4 The value for a = 3 and b = –4 [4]

A.8(a) Let number of articles be x.

Cost price = ` 1200Profit = ` 60Selling price = ` (1200 + 60) = ` 1260Number of articles damaged = 10Remaining articles = x – 10

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Set A... 13 ...

Cost price per article = `1200

x and selling price per article = `

1260– 10x

As per condition,

1260– 10x

=1200

x+ 2

1260

– 10x–

1200x

= 2

1260 – 1200 + 12000

( – 10)x x

x x = 2

2

60 + 12000 – 10

xx x

= 2

2x2 – 20x = 60x + 12000 2x2 – 80x – 12000 = 0 x2 – 40x – 6000 = 0 [Dividing throughout by 2] x2 – 100x + 60x – 6000 = 0 x (x – 100) + 60 (x – 100) = 0 (x – 100) (x + 60) = 0

x = 100 or x = – 60But x = – 60 is not possible

No. of articles is 100 [4]

(b)

(i) Median

Median =N2

=802

= 40th term = 40

Median = 40

10

20

30

40

50

60

70

80

10X

Q1 = 18 Q3 = 66Md = 40

R

Y

20 30 40 50 60 70 80 90 100

(10,

5)

(20, 24)

(40, 40)

(50, 42) (60, 48)

(70, 70)

(80, 77)(90, 79)

(100, 80)

N2 = 40

N4 = 20

3N4 = 60

Marks (Less than)

c.f. (30, 37)

(0, 0)

Scale : On X axis : 2 cm = 10 units Y axis : 2 cm = 10 unitsOn

Turn overT18 I SP A003

Set A... 14 ...

(ii) Lower quartile, (Q1)

Q1 =N4

=804

= 20th term = 18

Q1 = 18

(iii) Upper quartile, (Q3)

Q3 =3N4

=3 × 80

4 = 60th term = 66

Q3 = 66Hence (i) 40 (ii) 18 (iii) 66 [6]

A.9(a) 5x – 3 < 5 + 3x < 4x + 2

5x – 3 < 5 + 3x 5 + 3x < 4x + 2 5x – 3x < 5 + 3 3x – 4x < 2 – 5 2x < 8 –x < –3

x <82 x > 3

x < 4 3 < x < 4Comparing with a < x < b

a = 3 and b = 4 [3]

(b) (i)4 34 3m nm n

=74

By componendo dividendo4 3 4 34 3 4 3

m n m nm n m n =

7 47 4

i.e.86mn =

113

mn =

11 63 8

mn =

114

(ii)mn =

114

2

2mn

=12116

(Squaring both sides)

2

2211

mn

=2 12111 16

(Multiplying both sides by211

)

2

2211

mn

=118

By componendo dividendo,2 2

2 22 112 –11m nm n

=11 811– 8

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Set A... 15 ...

2 2

2 22 112 –11m nm n

=193

2 2

2 22 -112 +11

m nm n =

319 [By invertendo] [3]

(c)

Mode = 23 [4]

A.10(a)

[3]

(b) Odd natural numbers less than 50 are :1, 3, 5, 7, ......................... 49This is an arithmetic progressiona = 1, d = 2, last term l = 49Let nth term be 49 tn = 49 a + (n – 1)d = 49 1 + (n – 1)2 = 49 2(n – 1) = 48 n – 1 = 24

P

C

A B

X

Y9 cm

6 cm

9 cm

4 cm

P

1

0

2

3

4

5

7

6

8

9

10

10

X

C

AY

20 30 40 50 60 70 80Class Interval

FRE

QU

EN

CY

MODE

D

B

L

K

Scale :X-axis : 2 cm = 10 unitsY-axis : 2 cm = 1 unit

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Set A

n = 25There are 25 terms

Sn = 2n

(a + l )

S25 =252 (1 + 49)

=252 × 50

= 25 × 25 S25 = 625 [3]

(c) (i) AOB = 2APB [ Angle at the centre is twice the angle at the circumference]

AOB = 2(75º)

AOB = 150º

(ii) OACB is a cyclic quadrilateral.

AOB + ACB = 180º [Opposite angles of a cyclic quadrilateral are supplementary]

150º + ACB = 180º

ACB = 30º(iii) ABCD is a cyclic quadrilateral.

ABD + ACD = 180º [Opposite angles of a cyclic quadrilateral are supplementary]

ABD + 30º + 40º = 180º [ ACB = 30º]

ABD = 180º – 70º

ABD = 110º(iv) ADB = ACB = 30º [Angles in the same segment]

ADB = 30º [4]

A.11(a) G.P. : 1, 4, 16, 64, ...............

In the given G.P., a = 1, |r| = 4 > 1Sn = 5461

( – 1)

– 1

na rr

= 5461 [ |r| > 1]

1(4 –1)

4 –1

n

= 5461

... 16 ...

B

P

CA

O

D

400

75o

Turn overT18 I SP A003

Set A

4 –14 –1

n

= 5461

4n – 1 = 5461 × 3 4n – 1 = 16383 4n = 16383 + 1 4n = 16384 4n = 47

n = 7 [ am = an m = n] 7 terms must be added [3]

(b) Diameter of base of cylindrical container = 42 cm

Radius (r) =422

= 21 cmLet rise in level of water = h cm

Size of rectangular solid = 22 cm × 14 cm × 10.5 cm Volume of solid = 3234 cm Now, Volume of rise in level of water = volume of solid r2h = 22 × 14 × 10.5

227

× 21 × 21 × h = 22 × 14 × 10.5

h =22 14 10.5 7

22 21 21

h =73

cm

h = 2.33 cm Rise in level of water is 2.33 cm. [3]

(c) 9 22

1+xx

– 91

+xx

– 52 = 0

Let x +1x

= y

x2 + 2

1x

+ 2 = y2 [Squaring both sides]

x2 + 2

1x

= y2 – 2

Given equation reduces to :9(y2 – 2) – 9 y – 52 = 0

9y2 – 18 – 9y – 52 = 0 9y2 – 9y – 70 = 0 9y2 – 30y + 21y – 70 = 0 3y(3y – 10) + 7(3y – 10) = 0 (3y – 10)(3y + 7) = 0

... 17 ...

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Set A... 18 ...

3y – 10 = 0 or 37 + 7 = 0

y =103

or y = –73

When y =103

,

x +1x

=103

2 1xx

=103

3x2 + 3 = 10x 3x2 – 10x + 3 = 0 3x2 – 9x – x + 3 = 0 3x(x – 3) –1(x – 3) = 0 (x – 3)(3x – 1) = 0

x = 3 or x =13

Similarly when y = –73

,

x +1x

= –73

,

2 1xx

= –73

3x2 + 3 = – 7

3x² + 7x + 3 = 0

a = 3, b = 7, c = 3

b² – 4ac = 49 – 4 × 3 × 3= 49 – 36= 13

Using formula x =2– ± – 4

2b b ac

a

By solving, we get x =13– 7 ±

6

The solution is 3,13 ,

– 7±6

13[4]