MAHESH TUTORIALS I.C.S.E.

Model Answer Paper

Marks : 80

Time : 2½ hrs.

SUBJECT : MATHEMATICSICSE X

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

T17 I SP A003 Turn over

SECTION – A (40 Marks)

Attempt all questions from this Section.A.1(a) 2x – 5 < 5x + 4 < 11

2x – 5 < 5x + 4 and 5x + 4 < 11 2x – 5x < 4 + 5 and 5x < 11 – 4 –3x < 9 and 5x < 7

x > –3 and x <75

–3 < x <75

–3 < x < 1.4 –3 < x < 1, x I Solution set { –3, –2, –1, 0, 1 }

[3]

(b) L.H.S. =1 + sin A

cos A +

cos A1 + sin A

=2 2(1 + sin A) + cos A

cos A(1 + sin A)

=2 21 + 2sin A + sin A + cos A

cos A(1 + sin A)

=1 + 2sin A + 1

cos A(1 + sin A) [sin2A + cos2A = 1]

=2 + 2sin A

cos A(1 + sin A)

=2(1 + sin A)

cos A(1 + sin A) =2

cos A

= 2sec A = R.H.S.

1 + sinA

cosA +cosA

1 + sinA = 2secA [3]

Set B

–4 –3 –2 –1 0 1 2

Turn overT17 II SP B003

Set B

(c)x + + x -x + – x -

3 4 3 53 4 3 5

=91

By componendo and dividendo,

3 4 3 5 3 4 3 5

3 4 3 5 3 4 3 5

x x x x

x x x x

=9+19 –1

2 3 42 3 5

xx

=108

3 43 5

xx

=54

Squaring both sides,

3 43 5

xx

=2516

48x + 64 = 75x – 125 64 + 125 = 75x – 48x 27x = 189 x = 7 [4]

A.2.(a) Angles in the same segment are equal.

BAC = CDB = 49o

CDB = 49o

ADC = ABC = 43o

ABC = 43o

In ABC,ACB = 180º – (49º + 43º) [Remaining angle of ABC]ACB = 88º [3]

(b) 5x2 – 3x – 4 = 0Comparing with ax2 + bx + c = 0, we get,

By formula, a = 5, b = –3, c = –4

2–b± b – 4ac=

2ax

=2– (–3)± (–3) – 4(5) (– 4)

2(5)

=3±

109 + 80

=3± 89

10

89 = 9.434

... 2 ...

A

CD

B

43o

49o

Turn overT17 II SP B003

Set B

x =3±

109.434

x =3+

109.434

or x =3 –

109.434

=12.434

10or x =

–6.43410

x = 1.2434 or x = – 0.6434

x = 1.24 or x = – 0.643 [Correct to 3 significant figures] [3]

(c) Let the first term of the G.P. be ‘a’ and the common ratio be ‘r’t4 = 54

ar4 – 1 = 54 [ tn = arn – 1] ar3 = 54 ... (i)

t7 = 1458 ar7 – 1 = 1458 [ tn = arn – 1] ar6 = 1458 ... (ii)

Dividing equation (ii) by equation (i), we get,6

3

arar

=1458

54 r3 = 27 r3 = (3)3

r = 3 ... (iii) [Taking cube root on bothsides]

Substituting the value of ‘r’ in equation (i), we get,a(3)3 = 54

a × 27 = 54

a =5427

a = 2 ... (iv)The given G.P.= t1, t2, t3, t4, ............

= a, ar, ar2, ar3, ..........= 2, 2 × 3, 2 × 32, 2 × 33, .......[From (iii) and (iv)]= 2, 6, 2 × 9, 2 × 27, ..........= 2, 6, 18, 54, ...... [4]

A.3.(a) A represents the position of the kite and

AC is the length of the string.AB is the height of kite.ACB is the angle of elevation.ACB = 26032In ABC,

sin 26032=ABAC

... 3 ...

?100 m

A

BC26032

Turn overT17 II SP B003

Set B

0.4467 =AB100

AB = 0.4467 × 100 AB = 44.67 m.

Height of the kite from ground is 44.67 m. [3]

(b) Let number to be added to 3x3 – 5x2 + 6x be a Resulting polynomial = 3x3 – 5x2 + 6x + aPut (x – 3) = 0 x = 3 Remainder = f(3) = 3(3)3 – 5(3)2 + 6(3) + a 8 = 81 – 45 + 18 + a [ remainder = 8] 0 = 99 – 53 + a 0 = 46 + a a = – 46 The required number is –46 [3]

(c) Given : AB is diameter and a tangent at Cmeets AB produced at Q.

CAB = 34o

To Find: (i)CBA (ii) CQBSoln. Angle between tangent and chord is

equal to the angle in the alternate segment. BCQ = CAB BCQ = 34o

AB is diameter ACB = 90o

ACQ = ACB + BCQ [Angle addition property]= 90o + 34o

= 124o

CQB = 180o – (CAB + ACQ) [Remaining angle of ACQ]= 180o – (34o + 124o)= 180o – 158o

CQB = 22o

Again, CBA = 180o – (CAB + ACB) [Remaining angle of ABC]= 180o – (34o + 90o)= 180o – 124o

CBA = 56o [4]

A.4.(a) 2 3 – 3 = 3

3 3 – 2 3 = 3The difference between the consecutive terms is constant

The given sequence is an arithmetic progression.where first term, a = 3

... 4 ...

C

A 34ºB Q

Turn overT17 II SP B003

Set B

and common difference, d = 3tn = a + (n – 1) d

t100 = 3 + (100 – 1) 3= 3 + 99 3= 100 3

100th term is 100 3 [3]

(b) Let the number be x

Its reciprocal =1x

x +1x

= 4.25

2 425

100 + 1

xx

100x2 + 100 = 425x 100x2 – 425x + 100 = 0

Dividing throughout by 25 4x2 – 17x + 4 = 0 4x2 – 16x – 1x + 4 = 0 4x (x – 4) – 1(x – 4) = 0

(x – 4) (4x – 1) = 0

x = 4 or x =41

The number is 4 or 41

[3]

(c) sin A + cos A = msec A + cosec A = n

L.H.S. = n(m2 – 1)= (sec A + cosec A)[(sin A + cos A)2 – 1]

= (sec A + cosec A)(sin2A + 2sin A cos A + cos2A – 1) = (sec A + cosec A) (1 + 2sin A cos A – 1) [sin2A + cos2A = 1]

=

1cos A +

1sin A (2sin A cos A)

=

sin A + cos Acos Asin A (2sin A cos A)

= 2(sin A + cos A)= 2m [sin A + cos A = m]

n(m2 – 1) = 2m [4]

... 5 ...

Turn overT17 II SP B003

Set B

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

A.5.(a) In right angled PAR,

cos 30º =ARPR

0.8660 =AR6

AR = 0.8660 × 6= 5.196 m

Now in right angled QBR,

sin 47º =RBRQ

0.7314 =RB5

RB = 0.7314 × 5= 3.6570= 3.657 m

AB = AR + RB= 5.196 + 3.657= 8.853

AB = 8.85 m [3]

(b) G.P.: 1, –12

,14

, –18

, ...........

In the given G.P., a = 1, r =–12

< 1

Sn =(1 – )1 –

na rr

[ r < 1]

S9 =

9–11 1 –2–11 –2

=

9

9(–1)1–2112

=9

(–1)1–2

2 12

... 6 ...

P

A30º

6 m

R B

Q

47º

5 m

Turn overT17 II SP B003

Set B

=9

112

32

= 9

112

32

= 9

112

×23

S9 =23 9

11+2 [3]

(c) x4 – 25x² + 144 = 0Put x² = y

y² – 25y + 144 y² – 16y – 9y + 144 = 0 y(y – 16) –9 (y – 16) = 0 (y – 16) (y – 9) = 0 y – 16 = 0 or y – 9 = 0 y = 16 or y = 9When y = 16 When y = 9

x² = 16 x² = 9 x = ± 4 x = ±3Thus, S.S. = {–4, –3, 3, 4} [4]

A.6.

(a)2 2

2 2

cot 41º 2 sin 75º–tan 49º cos 15º

= 2

2

tan 90º – 41ºtan 49º –

2

2

2 cos 90º – 75ºcos 15º

cot tan(90º – )sin cos(90º – )

=2 2

2 2

tan 49º 2 cos 15º–tan 49º cos 15º

= 1 – 2(1)= 1 – 2= – 1 [3]

(b) Let the two numbers be x and y x : 12 = 12 : y

12x

=12y

x =144

y ............(i)

x : y = y : 96

... 7 ...

Turn overT17 II SP B003

Set B

xy =

96y

y2 = 96x ............(ii)Substituting (i) in (ii)

y2 = 96144

y

y3 = 96 × 144 y3 = 12 × 12 × 12 × 8 y = 12 × 2 { Taking cuberoot on both sides } y = 24

x =144

y

x =14424

x = 6 The two numbers are 6 and 24. [3]

(c) The production of TV sets per year increases uniformly by a fixed numberevery year.Hence the production numbers form an A.P.Let first term = a and common difference = dProduction in 3rd year = 600 units t3 = 600 a + 2d = 600 ... (i)

Production in 7th year = 700 units t7 = 700 a + 6d = 700 ... (ii)

Subtracting (ii) from (i)a + 2d = 600a + 6d = 700

(–) (–) (–)–4d = –100

d = 25From (i)

a + 2d = 600 a + 2(25) = 600 a + 50 = 600 a = 550(i) Production in 1st year = a = 550 units(ii) Production in 10th year

= t10

= a + 9d= 550 + 9(25)= 550 + 225= 775 units [4]

... 8 ...

Turn overT17 II SP B003

Set B

A.7.(a) Amount of bill = ` 4800

Let the No. of persons staying be = x

Amount paid by each person = `4800

xAccording to given condition,No. of persons = x + 4

Amount paid by each = `4800

+ 4xAccording to condition,

–4800 4800+ 4x x

= 200

4800 –1 1 + 4

x x = 200

4800 + 4 – ( + 4)

x xx x = 200

2

44x x

=2004800

96 = x2 + 4x x2 + 4x – 96 = 0

x2 + 12x – 8x – 96 = 0 (x + 12) (x – 8) = 0 x = – 12 or x = 8 x = – 12 is not acceptable x = 8 Number of people staying overnight is 8 [3]

(b) PT = 12.5 cmPA = 10 cm

Let, AB = x cm PB = (x + 10) cm [ P – A – B] PA × PB = PT2

[if chord and tangent intersect externally, then product of lengths of segment is equal to the square of the length of tangent]

10(x + 10) = 12.52

10x + 102 = 12.52

10x = 12.52 – 102

= (12.5 – 10) (12.5 + 10)= 2.5 × 22.5

10x = 56.25

x =56.25

10

x =56251000

=458

... 9 ...

BP

T

12.5 cm

10 cm A

Turn overT17 II SP B003

Set B

x = 5.625

AB = 5.625 cm [3]

(c) f(x) = x3 + 6x2 + 11x + 6 ... (i)Put x = –1 in (i) f(–1) = (–1)3 + 6 (–1)2 + 11 (–1) + 6

= –1 + 6 – 11 + 6= –12 + 12= 0

(x + 1) is factor of f(x)Dividing (i) by (x + 1)

3 2 2

3 2

2

2

1 6 11 6 5 6

– –

5 11 65 5

– –

6 66 6

– –

0

x x x x x x

x x

x xx x

xx

f(x) = (x + 1) (x2 + 5x + 6)= (x + 1) (x2 + 3x + 2x + 6)= (x + 1) [x(x + 3) + 2 (x + 3)]= (x + 1) (x + 3) (x + 2) [4]

A.8.

(a) L.H.S. = 1 –2sin A

1 cos A

= 1 – 21 – cos A

1 cos A

= 1 –

1 – cos A 1 cos A

1 cos A

= 1 – (1 – cos A)= 1 – 1 + cos A= cos A

L.H.S. = R.H.S.Hence Proved [3]

... 10 ...

Turn overT17 II SP B003

Set B

(b) (m – 2)x2 – (5 + m) x + 16 = 0From the given equation,a = (m – 2), b = – (5 + m), c = 16Since the roots are equal,b2 – 4ac = 0

[ – (5 + m)]2 – 4 (m – 2) (16) = 0 (5)2 + 2(5) (m) + (m)2 – 64 (m – 2) = 0 25 + 10m + m2 – 64m + 128 = 0 m2 – 54m + 153 = 0 m2 – 51m – 3m + 153 = 0 m(m – 51) – 3 (m – 51) = 0 (m – 51) (m – 3) = 0 m = 51 or m = 3 [3]

(c)

[4]

A.9.(a) The natural numbers lying between 250 and 1000 which are divisible by

9 are : 252, 261, 270, ........... 999 This is an arithmetic progressiona = 252, d = 9l = 999

Let nth term be 999 tn = 999 a + (n – 1)d = 999 252 + (n – 1)9 = 999 (n – 1)9 = 747 n – 1 = 83 n = 84

Sn = 2n

(a + l)

S84 =842 (252 + 999)

= 42 × 1251= 52542 [3]

... 11 ...

C

I

DA B YX

Turn overT17 II SP B003

Set B

(b) Let f(x) = (3k + 2) x3 + (k – 1)Put (2x + 1) = 0

x =12

Remainder = f1

2

= (3k + 2)3

12

+ (k – 1)

0 = (3k + 2)1

8

+ k – 1

18

(3k + 2) = k – 1

3k + 2 = 8 (k – 1) 3k + 2 = 8k – 8 2 + 8 = 8k – 3k 10 = 5k k = 2 The value of k is 2 [3]

(c) Let speed of boat = x m/sAB = 150 m

Angles of elevation of A at C and D are;ACB = 60º

and ADB = 45º Time taken from C to D = 2 minutes

= 2 × 60= 120 sec.

Distance CD = Speed × Time= x × 120= 120 x m

In right angled ABC,

ABBC

= tan 60º

150BC

= 3

BC =150

3 BD = BC + CD

=150 120

3x

m ... (i)

Also from right angled ABD,ABBD

= tan 45º

150BD

= 1

BD = 150 ... (ii)

... 12 ...

DB C45º60º

A

150 m

Turn overT17 II SP B003

Set B

From (i) and (ii);150

3 + 120x = 150

120x = 150 –150

3

120x = 15011–3

x =150120

3 –13

x =54

1.732 –13

×33

x =54

×0.732

3 × 3

x =54

× 0.244 × 1.732

x = 5 × 0.061 × 1.732 x = 0.52826 x = 0.528 Speed of boat = 0.528 m/s [4]

A.10.(a) Angles in the same segment are equal

DBC = DAC DBC = 27ºABC = ABD + DBC [Angle addition property]

ABC = 50º + 27º = 77o

Opposite angles of a cyclicquadrilateral are supplementary

ADC = 180º – 77º = 103º CDB = ADC – ADB

= 103º – 33º = 70o

CAB = CDB [Angles in the same segment] CAB = 70o

DCB = 180º – DAB [Opposite angles of a cyclic quadrilateralare supplementary]

= 180º – (DAC + BAC)= 180º – (27º + 70º)= 180º – 97º

DCB = 83o [3]

(b) G.P. : 3, 6, 12, ..........., 1536 a = 3, l = 1536, r = 2 > 1

Sn =––1

lr ar [ r > 1]

... 13 ...

B

CD

A27o

33º

50o

Turn overT17 II SP B003

Set B

=1536 2 – 3

2 –1

=3072 – 3

1 Sn = 3069 [3]

(c) Distance travelled = 400km.Speed of areoplane during onward journey = x km/hr.Speed of aeroplane during return journey = (x + 40) km/hr.

Time taken by aeroplane for onward journey =400 hrs.

x

Time taken by aeroplane for return journey =400 hrs.+ 40x

According to the condition,

400+ 40x

=400

x –

3060

400

x –

400+ 40x

=12

4001 1-

+ 40 x x =

12

400 2

+ 40 -+ 40

x xx x =

12

400 × 40 × 2 = x2 + 40x x2 + 40x – 32000 = 0 x2 + 200x – 160x – 32000 = 0 x (x + 200) – 160 (x + 200) = 0 (x + 200) (x – 160) = 0 x = 160 orx = –200 (Not possible) Speed of aeroplane onward journey = 160 km / hr. [4]

A.11.

(a) Let, a cb d = k

a = bk, c = dk

L.H.S. =

a bc d

=

bk bdk d

= 11

b kd k

=bd

... 14 ...

Turn overT17 II SP B003

Set B

R.H.S. =

2 2

2 2

a bc d

=

2 2 2

2 2 2

b k bd k d

=

2 2

2 2

1

1

b k

d k

=bd [3]

(b) –2 <12

–23x 1

56

, x N

–2 <12

–23x

116

Taking

–2 –12

–23x

–4 –1

2 –23x

–52

–23x

52

23x

5 32 2 x

154 x

3.75 x ... (i)Taking

12

–23x

116

–23x

116

–12

–23x

22 – 612

–23x

1612

–23x

43

... 15 ...

Turn overT17 II SP B003

Set B... 16 ...

–x 43

×32

–x 2 x –2 ... (iii)

From equations (i) and (ii),{x N : –2 x 3.75} = {1, 2, 3}The graph on the number line is

[3]

(c)

ABCDEF is the required regular hexagon. [4]

–3 –2 –1 0 1 2 3 4

O

BA

CF

X

E D