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CBSEContinuous and Comprehensive Evaluation (CCE)

Sample

Question Papers

Term 1 (April to September 2013)

OSWAAL BOOKS

Class

Mathematics

Solutions



CONTENTS

l Solutions

q Sample Question Paper 6 1 - 12





ll



Solutions | 1

1. (C),243

2 5 73 2 3 =

3

3 3 3

7

2 5 7 = 3 3

1

2 51

Hence, above rational numbers have terminating decimal expansion.

2. (B), Given, p( x) = x2 – 5 x + 6

Sum of zeroes = + = –F

H

GI

K

J 5

1

= 5

Product of zeros = =6

1= 6

Now + – 3 = 5 – 3(6)

= 5 – 18 = – 13. 1

3. (A), XY || QR so

PQ

XQ =PR YR

7

3 =

63

YR

YR =3 × 63

7= 27 cm 1

4. (A), Given, sec + tan = 7

(sec tan )(sec tan )

(sec tan )

= 7

sec tan

sec tan

2 2

= 7 1

sec tan = 7 1

sec – tan =1

7

5. (B), The required LCM = p3 1

6. (C), 4 x –3 y = 92 x + ky = 11

For unique solutiona

a

1

2=

b

b

1

2

c

c

1

2

4

2=

3

k

9

11

2 =–3

k

k =3

2

1



2 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X

7. (B), Given 5 tan =12

tan =12

5

So

13

3

sin=

13

3

12

13

F

H G

I

K J = 4. 1

8. (B), In the distribution,

Median class = 160 – 165

Hence Upper limit of median class = 165

and Modal class = 150 – 155

So Lower limit of Modal class = 150

Hence, Sum of upper limit of Median class and lower limit of Modal class

= 165 + 150 = 315. 1

Question number 9 to 14 carry two marks each.

9. No.

15 does not divide 175.

LCM is exactly divisible by their HCF. 1

Two numbers cannot have their HCF as 15 and LCM as 175. 1

10. Given, p ( x) = 2 x2 – x – 6

= 2 x2 – 4 x + 3 x = 6

= 2 x ( x – 2) + 3 ( x – 2)

= (2 x + 3) ( x – 2)

So zeroes are 2 and 3

2 ½

Sum of zeroes (SOZ) = 2 +F

H GI K J

3

2

1

2½

Product of zeroes (POZ) = 2 ×

3

23

SOZ =coeff. of

oeff. of

x

c x2

1

2

1

2

( )

POZ =constant

coeff. of 2 x

6

23 ½

Thus relationship is verified. ½

11. According to question AO = 20 cm, BO = 12 cm, PB = 18 cm

In AQO and BPO

AOP = BOP (V.O.s)

A = B = 90º

AQO ~ BPO 1

AQ

BP=

QO

PO=

AO

BO

AQ

18=

20

12

AQ = 30 cm. 1

B

C

A

12

5

13

P

B

O A

Q



Solutions | 3

12. Given, 4cos = 11sin

cos =11

4sin

Now

11 7

11 7

cos sin

cos sin

=

1111

47

11 114

7

sin sin

sin sin

1

=

sin

sin

121

47

121

47

F H G

I K J

F H G

I K J

=121 28

121 28

93

149

–

1

13. p( x) = ax2 + bx + c

Let and1

be the zeroes of p( x), then

Product of zeroes, a ×

1

=

c

a 1

So, required condition is c = a. 1

14. Model class : 30 – 40. Here l = 30, f 1 = 45, f 2 = 12, f 0 = 30, h = 10 ½

Mode = l + f f

f f f 1 0

1 0 22

F

H GI

K J × h ½

= 30 +45 30

90 30 12

F H G

I K J × 10 ½

= 30 + 3125 = 33125 ½

Or

Class Interval Frequency

0-50 8

50-100 15

100-150 32

150-200 26

200-250 12

250-300 7

Total 100 2

15. In ABC and DEF AB

DE= 3 ;

AC

DF= 3; P1 = 3P2 BC = 3EF

A

B C

D

E F




AB

DE=

AC

DF=

BC

EF= 3

ABC ~ DEF 1

ar ABC

ar ( DEF)

( )

= AB

DE

F H G

I K J

2

= (3)2 = 9 2

16. 3 x3 + 4 x2 + 5 x – 13 = (3 x + 10) g( x) + (16 x – 43) ½

3 4 11 30

3 10

3 2 x x x

x

= g ( x) ½

3 10 3 4 11 30

10

6 11

6 20

9 30

9 30

0

2 33 2

2

2

2

2 x x x x

x x

x x

x x

x

x

x x

3 3

(–) (–)

–

– –

( ) ( )

(–)(–)

(

Hence, g( x) = x2 – 2 x + 3. 2

17. L.C.M. of 48 and 60

2 | 48, 60

2 | 24, 30

3 | 12, 154, 5 1

48 = 24 × 3

60 = 22 × 3 × 5 ½

L.C.M. = 24 × 3 × 5

= 16 × 15 = 240 1

240 books are required ½

Or

(i) Required number of minutes is the LCM of 15 and 12.18 = 2 × 32

and 12 = 22 × 3 1

LCM of 18 and 12 = 22 × 32 = 36Hence, Ravish and Priya will meet again at the starting point after 36 minutes. 1(ii) LCM of numbers. ½(iiii) Healthy competition is necesssary for personal development and progrees. ½

18.2 2

2 2

sin 63 sin 7

sec 20 cot 70

+ 2 sin 36° sin 42° sec 48° sec 54° 1

=2 2

2 2

sin (90 27 ) sin 7

sec (90 70 ) cot 70

+ 2sin (90° – 54°) sec 54° sin 42° sec (90° – 42°) 1

=2 2

2 2

cos 27 sin 7

cosec 70 cot 70

+ 2cos 54° sec 54° sin 42° cosec 42°

=11 + 2 (1) = 3 1



Solutions | 5

19. Given, p( x) = 3 x2 – 4 x – 7 and and are its zeroes.

sum of zeroes = + = –coeff. of

coeff. of 2

x

x= –

F H G

I K J

4

3

4

3

Product of zeroes =constant

coeff. of 2 x= –

7

3

7

3

F H G

I K J

1

for the new polynomial

Sum,1 1

=

4

37

3

4

7

Product,1 1

=

1 1

7

3

3

7

1

required quadratic polynomial = x2 – (sum of zeroes) x + product of zeroes

= x2 –F

H GI K J

F

H GI K J

4

7

3

7 x =

1

7(7 x2 + 4 x – 3)

= 7 x2 + 4 x – 3. 1

20. 7 x – 4 y = 49 ...(1)

on comparing with the equation a1 x + b1 y = 0

a1 = 7, b1 = –4, c1 = 49

Again 5 x – 6 y = 57 ...(2)

on comparing with the equation a2 x + b2 y = c2

a2 = 5, b2 = –6, c2 = 57

Sincea

a

1

2=

7

5and

b

b

1

2=

4

6

a

a

1

2

b

b

1

21

So, system has unique solution.

Multiply eqn. (1) by 5 and multiply eqn. (2) by 7 and subtract,

35 x – 20 y = 245

35 x – 42 y = 399 1

– + –

22 y = – 154

y = – 7

Put the value of y in eqn. (1)

5 x – 6(–7) = 57 5 x = 57 – 42 = 15

x = 3

Hence, x = 3 and y = –7. 1

Or

ax + by =a b

2

or 2ax + 2by = a + b ...(i)

and 3 x + 5 y = 4 ...(ii) 1

Multiplying (i) by 5 and (ii) by 2b, we get

10ax + 10 by = 5a + 5b




6bx + 10 by = 8b

– – –

On subtracting x(10a – 6b ) = 5a – 3b 1

x =5 3

2 5 3

1

2

a b

a b

( )

Putting x =1

2in (1), we get y =

1

2

Hence x =1

2and y =

1

2 1

21. Classes :125 115

2

=

10

2= 5

115 – 5 = 110 ; 115 + 5 = 120

C.I. f c.f.

110 – 120 6 6

120 – 130 25 31

130 – 140 48 79

140 – 150 72 151

150 – 160 116 267

160 – 170 63 330 1

n = 330,n

2= 165 ½

median class is 150 – 160,

So, Median = 150 +165 151

116

F H G

I K J × 10 1

= 150 +140

116= 151.21 ½

22. In s ADE and ABC 1

A is common

D = B [corr. s]

ADE ~ ABC [AA]

AD AB

=DEBC

½

AB AD

=BCDE

½

ADDB =

75 ½

DB AD

+ 1 =57

+ 1 ½

AB AD

=127

=BCDE

½

In s BFC and EFDBFC = EFD [V.O. s)FBC = EFD [alt. s]

BFC ~ EFD [AA]

arEFDarBFC

=2

2

DE

BC=

2712

½

= 49144 ½

A

F

ED

B C



Solutions | 7

Or

In TXM, We have

XM || BN

TBTX

=TNTM

...(i) 1

In TMC, we have

XM || CM

TX TC

=TNTM

...(ii) 1

From equation (i) and (ii), we get

TBTX

=TX TC

TX 2 = TB × TC 1

23. LHS. =sin cos

sin cos

sin cos

sin cos

=(sin cos ) (sin cos )

sin cos

2 2

2 2

=(sin cos ) – sin cos (sin cos ) sin cos

sin ( sin )

2 2 2 2

2 2

2 2

1

1

=1 1

12 2

sin sin 1

=2

2 12sin = RHS Proved 1

24.

C.I. f x u = i x ah fu

100-110 4 105 – 2 – 8

110-120 14 115 – 1 – 14

120-130 21 125 = a 0 0

130-140 8 135 1 8

140-150 3 145 2 6

Total 50 – 8 1½

x = a + fu

f

= 125 + 8

50

½

= 125 – 016

x = 12484 1

Question number 25 to 34 carry four marks each.

25. Let 3 – 5 is a rational number.

3 – 5 = p

q, q 0

5 = 3 – p

q

=3q p

q

T B X C

MN

A




3q p

q

is rational number. 1

But 5 is a irrational number. 1

Since irrational number cannot be equal to rational number.

Our assumption is wrong. 1

3 – 5 is an irrational number. 1

26. Let the cost of one pencil be ` x and the cost of one chocolate be ` y

According to question,

2 x + 3 y = 11 ...(1)

x + 2 y = 7 ...(2)

Now, 2 x + 3 y = 11 x =11 3

2

y

y 1 3 5

x 4 1 – 2 1

and x + 2 y = 7 x = 7 – 2 y

y 0 1 3

x 7 5 1 1

–4 –3 –2 –1 1 2 3 4 5 6 7

y'

–1

–2

–3

1

2

3

5

4

x' x

y

6

7

8

0–5–6

–4

–5

–6

–7

8

(1, 3)

(4, 1)(5, 1)

(7, 0)

2 + 3 = 11 x y

x y+ 2 = 7

(–2, 5)

Putting the above points and drawing the lines joining them, we get the graph of the aboveequation. Clearly two lines interesect at point (1, 3).

Solution of eqns. (1) and (2) is x = 1 and y = 2

Cost of one pencil = ` 1

and cost of one chocolate = ` 3. 1

27. LHS =cos

tan

sin

sin cos

2 3

1

=cos

sin

cos

sin

sin cos

2 3

1

½

1



Solutions | 9

=cos

cos sin–

sin

cos sin

3 3

1

=cos sin

cos sin

3 3

1

= (cos sin )(cos sin sin cos )(cos sin )

2 21

= 1 + sin cos = RHS Proved ½

28.

Classes f c.f.

5 – 10 2 2

10 – 15 12 14

15 – 20 2 16

20 – 25 4 20

25 – 30 3 23

30 – 35 4 27

35 – 40 3 30 2

Total f = 30 = N

SinceN

2= 15,

Median class is 15 – 20

Median = l +

N2

cf

f

× h 1

From table, l = 15, N = 30, c.f. = 14, f = 2, h = 5

Median = 15 +15 14

2

F H G

I K J × 5 = 15 + 25

= 175. 1

29. p( x) = 2 x2 + 5 x + k

Sum of zeroes = –coeff. of

coeff. of 2

x

x= + =

5

21


coeff. of 2 x= =

k

2

According to question, 2 + 2 + = 214

1

(2 – 2 + =21

4[ ()2 = 2 + 2 + 2]

F

H GI K J

5

2

2

–k

2=

21

4

25

4

21

4 =

k

21

1 =k

2k = 2

Hence, k = 2 1




Or

In cyclic quadrilateral ABCD, we have

A + C = 180º

and D = 180º 1

x + 7 + 3 y + 23 = 180

x + 3 y = 150 ..(1)1

and y + 8 + 4 x + 12 = 180

4 x + y = 160 ...(2)1

Solving equation (1) and (2), we get y = 40º, x = 30º

A = ( x + 7)º = 37º

B = ( y + 8)º = 48º

C = (3 y + 23)º = 143º

D = (4 x + 12)º = 132º 1

30. Given : ABC is a triangle in which DE || BC.

To prove :

AD

BD =

AE

CE

Constuction : Draw DN AE and EM AD, Join BE and CD. 1

Proof : In ADE, area ( ADE) =1

2× AE × DN ...(i)

In DEC,

area ( DCE) =1

2× CE × DN ...(ii)

By (i) / (ii)area ( ADE)

area ( DEC)

=

1

2

AE DN

1

2CE DN

area ( ADE)

area ( DEC)

= AE

CE...(iii) ½

Now in ADE, area ( ADE) =1

2× AD × EM ...(iv) ½

and in DEB, area (DEB) =1

2× EM × BD ...(v)

By (iv) / (v),area ( ADE)

area ( DEB)

=

1

2 AD EM

1

2BD EM

½

area ( ADE)

area ( DEB)

=

AD

BD ...(iv)½

DEB and DEC lies on the same base DE and between same parallel lines DE and BC.

area (DEB) = area (DEC)

From quation (iii),area ( ADE)

area ( DEB)

= AE

CE...(vii)

From equations (vi) and (vii), we get

AE

CE=

AD

BD Proved 1

Or

Let BD = DE = EC = x

A

M N

D E

B C



Solutions | 11

BE = 2 x

BC = 3 x

AE2 = AB2 + BE2 = AB2 + 4 x2 1

AC2 = AB2 + BC2 = AB2 + 9 x2

AD2 = AB2 + BD23 = AB2 + x2 1

Now, 8AE2 = 8AB2 + 32 x2

and 3AC2 + 5AD2 = 3(AB2 + 9 x2) + 5 (AB2 + x2) 1

= AB2 + 27 x2 + 5AB2 + 5 x2

= 8AB2 + 32 x2

3AC2 + 5AD2 = 8AE2. Proved 1

31. Given x = r sin A.cos C

y = r sin A.sin C

z = r cos A

then x2 = r2sin2 A cos2C

z2 = r2cos2 A 1

y2 = r2sin2 A sin2C 1 x2 + y2 + z2 = r2sin2 A cos2C + r2sin2 Asin2C + r2cos2 A 1

= r2sin2 A (cos2C + sin2C) + r2cos2 A

= r2sin2 A + r2 cos2 A

= r2 (sin2 A + cos2 A)

= r2 proved 1

32. FEC GBD

EC = BD ...(1)1

It is given that

1 = 2

AE = AD ...(2)1

From eqns. (1) and (2), AE

EC=

AD

BD

DE || BC [converse of BP.T.]

1 = 3 and 2 = 4 1

Thus in ADE and ABC, A = A

1 = 3

2 = 4

So by AAA criterion of similarity, we have

ADE ~ ABC. Proved 1

33. cos ºsin º6525 = cos ºsin( º º )

cos ºcos º

6590 65

6565 = 1 1

tan º

cot º

20

70=

tan( º–70º )

cot º

cot º

cot º

90

70

70

70 = 1

sin 90º = 1

tan 5º tan 35º tan 60º tan 55º tan 85º = tan (90º – 85º) tan (90º – 55º) tan 55º tan 60º tan 85°

1

= cot 85º. tan 85º. cot 55º tan 55º. 3

= 1 × 1 × 3 = 3 1

Given expression = 1 – 1 – 1 + 3 = 3 – 1. 1

A

BF C

E1 2

G

D

3 4




34.

Less than c.f. More than c.f.

10 8 0 35

20 15 10 2730 20 20 20

40 30 30 15

50 35 40 5 2

Scale

x-axis 10 units = 1 cm

y-axis 1 cm = 5 c.f.

35

30

25

20

15

10

5

10 20 30 40 50

CLASSES

C . F .

Scale

-axis 10 units = 1 cm-axis 1 cm = 5 cf x y y

Median =24

0

more than ogive

less than ogive

2



Solutions | 13

1. (D), Since 16 = ± 4, so it is a rational number. 1

2. (B), Given, a = – 15 and b = 15

So, quadratic polynomial = ( x – ) ( x – ) = ( x – 15 ) ( x + 15 )

= x2 – ( 15 )2 = x2 – 15. 1

3. (A), In the figure AB || ED then ABC and DEC are similar. 1

4. (C), (1 + tan2 ) cos2 =2

2

sin1cos

cos2

=2 2

2

(sin cos )

cos

× cos2

= 1 1

5. (C),81

2 3 53 2 2 = 3 2

9

2 51

Hence, this rational number has a terminating decimal expansion. 1

6. (D), For a unique solution

a

a

1

2

b

b1

2

i.e.,3

6

2

k

– 4 k

i.e., all real numbers except – 4. 1

7. (D), Given tan = 1

Then5sin 4 cos5sin 4cos

=

cos (5tan 4)

cos (5tan 4)

=51 451 4

= 9 1

8. (B), The median of the data is 425. 1

9. a =9009

3003= 3 ½

b =1001

143= 7 ½

Since 143 = 11 × 13, so c = 11 or 13 ½d = 13 or 11. ½




10. Given quadratic polynomial = 4 x2 + 7 x + 8

Sum of zeroes = –coeff. of

coeff. of 2

x

x= –

7

41

and Product of zeroes =constant

coeff. of 2

x

=8

4

= 2. 1

11. In 's ADE and ABC,

ADE = C [Given] ½

A = A [Common] ½

By AA criterion of similarity, ADE ~ ACB 1

12. x = 30°

sin 30° + cos y = 1

12

+ cos y = 1 ½

cos y =12

= cos 60° 1

y = 60° ½

13. Given, p( x) = 2 x2 – 9

2 x2 – 9 = 0

( 2 x)2 – (3)2 = 0

( 2 x + 3) ( 2 x – 3) = 0 1

x = –3

2and

3

2Hence, zeroes of the polynomials are

3

2and –

3

2. 1

14. Modal class : 20 – 30

Here, l = 20, f 1 = 40, f 0 = 24, f 2 = 36, h = 10 ½

Mode = l f f

f f f h

( )1 0

1 0 22½

= 20 +( – )

– –

40 24

80 24 36× 10 ½

= 20 +16 10

20

= 28 ½

Or

Classes 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80

Frequency ( f i) 3 4 2 2 1 f i = 12

xi (Class marks) 35 45 55 65 75

f i xi 105 180 110 130 75 f i xi = 600

Mean, x–

=

x f

f

i i

i

600

12 = 50 1



Solutions | 15

Mean weight of a worker = 50 kg. 1

Section - C

15. In ∆ ABC, given

A

BE D C

F

AB = AC ⇒ ∠ ABC = ∠ ACB 1

or ∠ ABD = ∠ECF

∠ ADB = ∠EFC, [Each 90º] 1

∴ ∆ ABD ~ ∆CEF. [AA Similarity] 1

16. If α, β are zeroes of the polynomial

p( y) = 6 y2 – 7 y + 2

α + β = –n

n 2

coeff of 76coeff of

y

y=

αβ =2

Constant term

Coefficient of y=

26

=13

Now,1 1

α β+ =

7613

α + β =αβ

=72

1

1

α×

1

β= 3 1

∴ Required polynomial = y2 –1 1 + α β y +

1 1·

α β

∴ = y2 –72

y + 3 = 21(2 7 6)

2y y− + 1

17. By Euclid’s division algorithm, 510 = 92 × 5 + 5092 = 50 × 1 + 42

50 = 42 × 1 + 8

42 = 8 × 5 + 2 18 = 2 × 4 + 0.

HCF (510, 92) = 2

92 = 22 × 23

510 = 2 × 3 × 5 × 17

LCM = 22 × 23 × 3 × 5 × 17 = 23460

HCF × LCM = 2 × 23460 = 46920 1

Product of 2 numbers = 510 × 92

⇒ HCF × LCM = Product two numbers. 1

Or

By Euclid’s division algorithm, for two positive integers a and b, we have

a = bq + r, 0 ≤ r < b




Let b = 6 r = 0, 1, 2, 3, 4, 5

So a = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 1Clearly, a = 6q, 6q + 2, 6q + 4 are even, as they are divisible by 2.

But 6q + 1, 6q + 3, 6q + 5 are odd, as they are not divisible by 2. 1

Any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5. 1

18.sec ( º ) cot

(sin º sin º)

cos º tan º tan º

(sec º– cot º )

2 2

2 2

2 2 2

2 2

90

2 25 65

2 60 28 62

3 43 47

=(cos cot )

(sin º cos º)

tan º cot º

[sec º– tan º ]

ec 2 2

2 2

2 2

2 22 25 25

21

2

1

228 28

3 43 43

1

= 1

2 1

1

228

1

283

22

( )

tan ºtan º 1

=1

2

1

6 =1

3 1

19. x x x x x

x x x

x x

x x

x

x2 3 2

3 2

2

2

2 6 11 12

2

5 9 12

5 5 10

4 2

5

–

– –

– –

–

Quotient x – 5, remainder = 4 x – 2 1

Dividend = (Divisor × Quotient) + Remainder 1

= ( x2 – x + 2) ( x – 5) + 4 x – 2

= x3 – x2 + 2 x – 5 x2 + 5 x – 10 + 4 x – 2

= x3 – 6 x2 + 11 x – 12 Proved 1

20. Let1

y= a, the given equations become

4 x + 6a = 15 ...(1)

6 x – 8a = 14 ...(2) ½

Multiply eqn (1) by 4 and eqn (2) by 3 and adding

16 x + 24a = 60

18 x – 24a = 42

On adding 34 x = 102

x =102

34= 3 ½

Put the value of x in eqn (1)

4(3) + 6a = 15

6a = 15 – 12 = 3

a =3

6

=1

2



Solutions | 17

a =1

y=

1

2 y = 2 1

Hence x = 3 and y = 2

Again y = px – 2 2 = p (3) –2 3 p = 4 p =4

3 1

Or

29 x + 41 y = 169 ...(1)

41 x + 29 y = 181 ...(2)

Adding equations (1) and (2), we get

70 x + 70 y = 350 1

x + y = 5 ...(3)

Subtracting eqn. (1) from eqn. (2), we get

–12 x + 12 y = –12

– x + y = –1 ...(4)

Adding equations (3) and (4), we get2 y = 4

y =4

2= 2 1

Putting this value of y in eqn (3), we get

x + 2 = 5 x = 3

Hence, x = 3 and y = 2 . 1

21. Let assumed mean, a = 35 and h = 10.

xi di= x a

hi

f i f idi

(Class Marks)5 – 3 5 – 15

15 – 2 13 – 26

25 – 1 20 – 20

35 0 15 0

45 1 7 7

55 2 5 10

Total f i = 65 f idi = – 44 1½

Mean, x–

= a +

f d

f

i i

i× h ½

= 35 +44

65× 10 = 35 – 676

x–

= 2824. 1

22. In ADB, AB2 = AD2 + BD2 [Pythagoras Theorem] ...(1)

A

B C

D3 x x




In ADC, AC2 = AD2 + CD2 [Pythagoras Theorem] ...(2) 1


AB2 – AC2 = BD2 –CD2 1

=3

4BC

1

4BC

2 2F H G

I K J

F H G

I K J =

BC

2

2

2 (AB2 – AC2) = BC2

2 (AB)2 = 2AC2 + BC2. 1

Or

According to question, BD = CD =BC

21

BC = 2BD

Using Pythagoras theorem in the right ABC, we have

AC2 = AB2 + BC2

= AB2 + 4BD2 1

= (AB2

+ BD2) + 3BD

2

AC2 = AD2 + 3CD2. 1

23. LHS = (coses – cot )2

=1

2

sin

cos

sin

F H G

I K J 1

=1

2F H G

I K J

cos

sin

=( cos )

sin

1 2

2

=( cos )( cos )

cos

1 1

1 2

1

=( cos )( cos )

( cos )( – cos )

1 1

1 1

=1

1

cos

cos

1

= R.H.S. Proved

24. (i) Here class intervals are not in inclusive form. So, we first convert them in inclusive formby subtracting 1/2 from the lower limit and adding 1/2 to the upper limit of each cases. whereh is the difference between the lower limit of a class and the upper limit of the preceding class.

The given frequency distribution in inclusive form is as follows : ½

Age (in years) No of cases

45 - 145 6

145 - 245 11

245 - 345 21

345 - 445 23

445 - 545 14

545 - 645 5

From table it is clear from table that the modal class is 345 - 445.

Here, l = 34.5, h = 10, f 1

= 23, f 0

= 21, f 2

= 14 ½

CB

A

D



Solutions | 19

Now, Mode = l f f

f f f h+

–

2 –0

0

1

1 2–

Mode =23–21345+ 10

46–21–14

= 345 +2

11 10

= 345 + 181 = 3631 1

(ii) Mode of grouped data. ½

(iii) If we parctise habit of cleanliness we will be able to put disease at on arm’s length. ½

25. Let 5 be a rational number.

5 =

a

b , [a, b are co-prime integers and b 0]

a = b 5

a2 = 5b2

5 is a factor of a2 1

5 is a factor of a

Let a = 5c, [c is some integer]

25c2 = 5b2

5c2 = b2 1

5 is a factor of b2

5 is a factor of b

5 is a common factor of a, bBut this contradicts the fact that a, b are co-primes.

5 is irrational

Let 2 – 5 be rational 1

2 – 5 = a

2 – a = 5

2 – a is rational, so 5 is rational.

But 5 is not rational contradiction

2 – 5 is irrational. 1

26. 2 x – y = 1 y = 2 x – 1

x 0 1 3

y – 1 1 5

and x + 2 y = 13 y =13

2

x

x 1 3 5

y 6 5 4




Putting the above points and drawing the lines joining them we get the graphs of above equations.

–4 –3 –2 –1 1 2 3 4 5 6 7

y'

–1

–2

–3

1

2

3

5

4

x' x

y

6

7

8

0–5–6

–4

–5

–6–7

8

2

–

= 1

x

y

(1, 6)

(3, 5)

(5, 4)

x y+ 2 = 13

(1, 1)

(0, –1)B

Clearly two lines intersect at point A (3, 5).

2

Hence, x = 3 and y = 5 is the solution of above equations. 1

ABC is the trianguler shaded region formed by the lines and the y-axis. 1

27. L.H.S. =tan sec

tan sec

1

1=

(tan sec ) (sec tan )

tan sec

2 2

1 [1 + tan2 = sec2 ] 1½

=(tan sec ) (sec tan )(sec tan )

tan sec

1

1

=(tan sec ) [ sec tan ]

tan sec

1

11

= tan + sec = RHS [Proved] ½

28.

CI f cf

0 – 10 5 5

10 – 20 x 5 + x

20 – 30 6 11 + x

30 – 40 y 11 + x + y

40 – 50 6 17 + x + y50 – 60 5 22 + x + y

Here from table, N = 22 + x + y = 40 1

x + y = 18 ...(1)

Since median = 31, median class is 30 – 40.

Median = l +

N c f

f

2F

H

GGG

I

K

J J J

. .

h

31 = 30 +20 11– ( )L

N

MO

Q

P x

y

× 10 1



Solutions | 21

1 =( – )9 10 x

y

y = 90 – 10 x

From (1), 10 x + y = 90 ...(2)1

(On subtraction) x + y = 18

– – –9 x = 72

x =72

9= 8

From (1), y = 18 – 8 = 10. 1

29. According to question, and are zeroes of

p( x) = 6 x2 – 5 x + k

So, sum of zeroes + = –F

H GI K J

5

6

5

6...(1)

Product of zeroes = =k

6

– =1

6 [Given] ...(2) 1

Adding equation (1) and (2), we get 2 = 1

=1

21

Put the value of in equation (2), we get1

2– =

1

61

2

1

6 =

2

6=

1

3 1

=k

612

13

k = 1. 1

Or

Let the fraction be x

y, then according to the question

x

y

2

2=

9

11

11 x –9 y = –4 ...(1) 1

and x

y

3

3=

5

6 6 x – 5 y = –3 ...(2) 1

Now (1) and (2) 11 x –9 y + 4 = 0

and 6 x – 5 y + 3 = 0

Solving these equations by cross-multiplication method, we get x

(–9)( ) – (–5)( )3 4=

y

( )( ) – ( )( )4 6 11 3=

1

11 6( )(–5) – ( )(–9)

x

–27 20=

y

24 33

1

54– –55

x y

7 9

1

11

Hence x = 7, y = 9

Fraction =7

9 1




30. Proof :

In quadrilateral ABCD, AO

BO=

CO

DO[Given]

AO

CO=

BO

DO...(1)

In ABD EO || AB (Construction)

AE

ED=

BO

DO[By BPT] ...(2)1

From eqns, (1) and (2), AE

ED=

AO

CO

In ADC, AE

ED=

AO

CO

EO || DC [Converse of BPT] 2

EO || AB [Constructions]

AB || DC

In quad. ABCD, AB || DC 1

ABCD is a trapezium. Proved

Or

(i) Let the initial position of the man be at O and his final position be B, Since the man goes to10 m due east and then 24 m due north. Therefore, AOB is a right triangle right angled at A suchthat OA = 10 m and AB = 24 m.

S

O

E

N B

A

W

24 m

10 m

By Pythagoras theorem, we have ½

OB2 = OA 2 + AB2

OB2 = (10)2 + (24)2

= 100 + 576 = 676

OB = 676 = 26 m 1

Hence, the man is at a distance of 26 m from the starting point.

(ii) Right angled triangle. ½

(iii) Slow and steady wins the race. ½

31. tan = AB

BC

1

5

In ABC, AC2 = AB2 + BC2 = 1 + 5 = 6 1

AC = 6

(i)cos sec

cos sec

ec

ec

2 2

2 2

=

( ) –

( )

66

5

66

5

66

5

66

5

22

2

2

F

H GI

K J

F

H GI

K J

=

24

36

2

3 1½

A B

CD

OE

A

B C

16

5



Solutions | 23

(ii) sin2 +cos2 =1

6

5

6

2 2F H G

I K J

F

H GI

K J

=1

6

5

6 = 1. 1½

32. Proof BC = 2BD [ AD is the median]

A

B D C

P

Q M Rand QR 2 QM [PM is the median] 1

Given, AB

PQ=

AD

PM=

BC

QR

AB

PQ=

AD

PM=

2BD

2QM1

In triangles ABD and PQM, AB

PQ=

AD

PM=

BD

QM

ABD ~ PQM [SSS Similarity]

B = Q [By CPCT] 1

In ABC and PQR, AB

PQ=

BC

QR

and B = Q [ ABC ~ PQR] 1

33. LHS = (1 + cot A – cosec A) (1 + tan A + sec A)

= 11

11

F H G

I K J

F H G

I K J

cos

sin sin

sin

cos cos

A

A A

A

A A 1

=sin cos

sin

sin cos

cos

A A

A

A A

A

F H G

I K J

F H G

I K J

1 1

=(sin cos )

sin cos

A A

A A

2 211

=sin cos sin cos –

sin cos

2 2 2 1 A A A A

A A

1

= 1 2 sin cossin cos

A A – 1 A A

= 2 = RHS. 1

34.

Monthly Wages c.f.

More than 80 200

More than 100 180

More than 120 150

More than 140 130 2

More than 160 90




x

y

Scale

-axis 2 cm = 10 units


x

y

Lower Lts

200

190180

170

160

150

140

130

120

110

100

90

80

70

60

50

40

30

20

10

080 100 120 140 160

More than ogive

(80, 200)

(100, 180)

(120, 150)

(140, 130)

(160, 90)

2



Solutions | 25

Section - A

1. (C), The number of rational numbers lying between 2 and 3 is infinite. 1

2. (B), Given, Polynomial = x2 – 3

Sum of zeroes =coeff. of

coeff. of 2

x

x= 0


coeff. of 2 x= – 3

Hence required answer 0, – 3. 1

3. (D), For similar triangles,

area of triangle 1

area of triangle 2= ( )

235

=925

1

4. (C), tan x = sin 45º cos 45º + cos 60º

=1

2

1

2

1

2

F H G

I K J F H G

I K J

+

=1

2

1

2+ = 1 = tan 45º

⇒ x = 45º. 1

5. (C), It is clear that 0·102003000...........is non-terminating non-repeating decimal, so it cannot

be expressed in the form of p

q. 1

6. (B), 4 x –5 y = 5

kx + 3 y = 3

For the condition of inconsistent

a

a

1

2=

b

c

c

c

1

1

1

2

≠

i.e.,4

k=

−≠

5

3

5

3

⇒4

k=

−5

3

⇒ k = −125

1

7. (B), tan 3A = cot (A – 26)

⇒ cot (90º – 3A) = cot (A – 26º)

⇒ 90º – 3A = A – 26º

⇒ 4A = 90º + 26º = 116º

⇒ Α =116

4

º= 29º. 1

Sample Question Paper–8




8. (B), The modal class of the distribution is 50 – 60. 1

9. 6762

2 3381

3

7

7

1127

161

23

= 7 × 161

=161

7

2 × 3381 =

Composite number x = 6762. 1

10. Sum = – 2; product of zeroes = 1 1

p( x) = x2 – (sum of the zeroes) x + product of the zeroes

= x2 – (– 2) x + 1 ½

p( x) = x2 + 2 x + 1 ½

11.

AB || QR

PA

AQ PBBR

5125 – 5

6 x

½

x = 4 1

PR = 4 + 6 = 10 cm ½

12. LHS =1

1

cos

cos

A

A =

1

1

1

1

cos

cos

cos

cos

A

A

A

A 1

=( cos )

( cos )

( cos )

sin

1

1

12

2

2

2

A

A

A

A

=1 1

cos

sin sin

cos A

A A

A

sin A

= cosec A – cot A = RHS Proved. 1

1

P

A B

Q R



Solutions | 27

13. x2 – x – 72 = x2 – 9 x + 8 x – 72

= ( x – 9) ( x + 8)

Zeroes of the given polynomial are 9 and – 8 1

Sum of the zeroesb

a

=(–1)1

= 1 = 9 – 8 ½

Product of the zeroesca

=721

= – 72 = – 72 = 9 × (– 8) ½

Hence Verified.

14.

Classes Frequency cf

135 – 140 4 4

140 – 145 18 22

145 – 150 7 29

150 – 155 11 40155 – 160 6 46

160 – 165 5 51 1

From above,

Median class is 145 – 150 ½

and Modal class is 140 – 145 ½

Or

More than Frequency

More than 75 5

More than 70 20

More than 65 34 1

More than 60 42

More than 55 48

More than 50 50 1

Question number 15 to 24 carry three marks each.

15. We have, PQR ~ PAB [P is common andPA

PQ=

PB

PR] 1

area PQR

area PAB

=PQ

PA

F H G

I K J

2

32

area PAB=

4

3

2k

k

F H G

I K J 1

area PAB = 18 cm2

area of AQRB = area of PQR – area of PAB = 32 – 18 = 14 cm2 1

P

Q R

3k

A B

k

4k




16. p( x) = x2 + 7 x + 9

q( x) = x + 2

r( x) = – 1

g( x) = ?

p( x) = g( x).q( x) + r( x) ½

x2 + 7 x + 9 = g( x).( x + 2) – 1 ½

g( x) =x x

x

2 7 10

2

1

=( )( )

( )

x x

x

2 5

2½

g( x) = x + 5. ½

17. Let 3 be a rational number

3 =a

b. [a and b are integers and co-primes]

On squaring both the sides 3 = ab

2

2 1

3b2 = a2 a2 is divisible by 3 a is divisible by 3 ...(1)We can write a = 3c for some integer c a2 = 9c2

3b2 = 9c2

b2 = 3c2 1 b2 is divisible by 3 b is divisible by 3 ...(2)From (1) and (2), we get 3 is a factor of ‘a’ and ‘b’.Which is contradicting the fact that a and b are co-primes. Hence our assumption that 3 is

rational number is false. So 3 is irrational number. 1Or

154

+5

40=

15 25 5 254 25 40 25

1

=375100

+125

10001

= 375 + 0125 = 3875 1

18. cos + sin = 2 cos ½

L.H.S. sin = 2 cos – cos

= cos ( 2 – 1) × 2 12 1

=cos (2 1)

2 1

½

sin =cos

2 1

½

2 sin + sin = cos ½

cos – sin = 2 sin ½

= 2 sin

= R.H.S. Proved. ½



Solutions | 29

19.

2 + 2 x

6 + 2 – 4 + 3 x x x3 2

– + –6 – 4 + 2 x x x3 2

6 – 6 + 3 x x2

– + –6 – 4 + 3 x x2

– 2 + 1 x

3 – 2 + 1 x x2

Quotient = 2 x + 2, Remainder is – 2 x + 1 1½

p( x) = g( x).q( x) + r( x)

= (3 x2 – 2 x + 1) (2 x + 2) + (– 2 x + 1) ½

= 6 x3 – 4 x2 + 2 x + 6 x2 – 4 x + 2 – 2 x + 1

= 6 x3 + 2 x2 – 4 x + 3 ½= p( x) Verified. ½

20. Let the sum of the ages of the 2 children be x and the age of the father be y years.

y = 2 x i.e. 2 x – y = 0 ...(1) 1

and 20 + y = x + 40

x – y = – 20 ...(2)

Subtracting (2) from (1), we get x = 20

From (1), y = 2 x = 2 × 20 = 40

y = 40 1

Hence, the age of the father = 40 years. 1Or

(i) Let the fixed charges of taxi be Rs x and the running charges be Rs. y per km. According to the question

x + 10 y = 75 ...(1)

x + 15 y = 110 ...(2) ½

Subtracting equation (2) from equation (1), we get

– 5 y = – 35

y = 7

Putting y = 7 in equation (1), we get x = 5 ½

Total charges for travelling a distance of 25 km.

= x + 25 y= Rs. (5 + 25 × 7)

= Rs. (5 + 175)

= Rs. 180 1(ii) Pair of Linear equations in two variables. ½

(iii) We should always be justified in our dealings. ½

21. Mode = l + f f

f f f

1 0

1 0 22

× h 1

Here, modal class : 30 – 40, l = 30, f 1 = 25, f 0 = 20, f 2 = 12, h = 10 ½

Mode = 30 +25 20

50 20 12

–

– –×10 1

= 30 +5 10

18

= 30 + 277 = 3277 (Modal age). ½




22. Given : ABC, right angled at A.

B A

C

M

L

BL and CM are medians.

To prove : 4(BL2 + LM2) = 5BC2.

Proof : In ABL,

BL2 = AB2 + AL2 1

= AB2 + AC

2

F H G

I K J

2

[BL is median]

In ACM, CM2 = AC2 + AM2

= AC2 + AB

2

2F H G

I K J , [CM is median] 1

BL2 + CM2 = AB2 + AC2 + AC

4

AB

4

2 2

4(BL2 + CM2) = 5AB2 + 5AC2

= 5(AB2 + AC2)

= 5BC2. Proved. 1

Or

Given : In ABC, ADBC and AD2 = BD × CD.

To prove : ABC is a right triangle.

Proof : In right triangle ABD and ACD, Applying Pythagoras theorem

AB2 = AD2 + BD2

and AC2 = AD2 + CD2 1

AB2 + AC2 = 2AD2 + BD2 + CD2

= 2(BD × CD) + BD2 +CD2, given AD2 = BD × CD 1

= (BD + CD)2 = BC2

i.e. AB2 + AC2 = BC2

ABC is a right triangle. 1

23. tan + cot = 2

tan +1

tan = 2

2tan 1

tan

= 2 1

tan2 + 1 = 2tan tan2 – 2tan + 1 = 0 (tan – 1)2 = 0 tan – 1 = 0 tan = 1 = tan 45°

= 45° 1



Solutions | 31

Now, tan7 + cot7 = tan7 45° + cot7 45°

= (tan 45°)7 + (cot 45°)7

= (1)7 + (1)7

= 1 + 1

= 2 1

24.

Class f Class x u = x a

h

fu

0 – 10 7 5 – 2 – 14

10 – 20 12 15 – 1 – 12

20 – 30 13 25 0 0

30 – 40 10 35 1 10

40 – 50 8 45 2 16

Total f = 50 fu = 0 2

Let a = 25 and h = 10

Mean, x–

= a + fu

f × h ½

= 25 +0

50× 10 = 25 (Mean) ½

25. Let us assume that there is a positive integer n for which n n 1 1 is rational and equal

toP

Q, where P and Q are positive integers (Q 0) 1

n n 1 1 = PQ ...(i)

Q

P=

1

– 1 +1n n

=n n

n n n n

– 1 +1

– 1 +1 – 1 +1

( )( )

=n n

n n

n n– 1 +1 – 1 +1

( – ) – ( )1 1 21

But n n+1 – 1 =2Q

P...(ii)

Apply (i) + (ii), we get 2 n+1 =

P

Q

2Q

P

P 2Q

PQ

2 2

n +1 =P 2Q

2PQ

2 2...(iii)

Apply [(i) – (ii)], we get

n – 1 =P 2Q

2PQ

2 2...(iv)1

From (iii) and (iv) we can say n +1 and n – 1 both are rational because P and Q both are

rational. But it is possible only when n + 1 and n – 1 both are perfect squares. But they differ by 2and two perfect squares never differ by 2. So both n + 1 and n – 1 not be perfect squares, hence

there is no positive integer n for which n n– 1 +1 is rational. 1

26. Let the speed of the boat be x km/hr and speed of the stream be y km/hr.

According to the question,




32 36

x y x y

= 7

and40 48

x y x y

= 9

Let 1 x y = A, 1

x y = B, we get

32A + 36B = 7

and 40A + 48B = 9

Solving these equations, we get A =1

8, B =

1

21

Hence A =1

8=

1

x y

x – y = 8 ...(1)½

and B =1

12

1

x y

x + y = 12 ...(2)½

Adding equations (1) and (2), we get 2 x = 20

x = 10 1

Putting this value of x in eqn.(1), we get

y = x – 8 = 10 – 8 = 2

Hence, the speed of the boat = 10 km/hr and speed of the stream = 2 km/hr. 1

27. Given 5 cos = 4 tan =45

1

Now5sin 3cos

5sin 2cos

5sincos 3cos

5sincos 2cos

1

5tan 35tan 2

=

45 3545 25

=4 34 2

=

16

1

28.

Class c.f.

Less than 30 3Less than 40 7

Less than 50 17

Less than 60 25

Less than 70 34

Less than 80 42

Less than 90 54

Less than 100 60

1½ + 1½

10 20 30 40 50 60 70

10

20

30

40

50

60

70

80

80 90

Median - 65 (approx)

X

Y

O 100

Scale

x-axis 1 cm = 10 unitsy-axis 1 cm = 10 units



Solutions | 33

29. Since 3 x2 – 5 divides f ( x) completely

(3 x2 – 5) is a factor of f ( x)

3( x2 –5

3) is a factor of f ( x)

x x

F

H G

I

K J

F

H G

I

K J

5

3

5

3 is a factor of f ( x)

5

3and –

5

3are zeroes of f ( x) 1

3 x2 – 5 ) 3 x4 + 3 x3 – 11 x2 – 5 x + 10 ( x2 + x – 2

3 x4 – 5 x2

– +

3 x3 – 6 x2 – 5 x

3 x3 – 5 x 1

– +

– 6 x2 + 10

– 6 x2 + 10

+ –×

( x2 + x – 2) is a factor of p( x)

( x2 + 2 x – x – 2) is a factor of p( x)

( x + 2) ( x – 1) is a factor of p( x)

– 2 and 1 are zeroes of p( x) 1

all the zeroes of p( x) are5

3, –

5

3, – 2 and 1. 1

Or

x – y = 1 y = x – 1

x 2 3 – 1

y 1 2 – 2

2 x + y = 8 y = 8 – 2 x

x 2 4 0

y 4 0 8

Putting the above points and drawing a line joining them, we get the graph of the above equation

x – y =1 and x + y = 8.

–4 –3 –2 –1 1 2 3 4 5 6 7

y'

–1

–2

–3

1

2

3

5

4

x' x

y

(0, – 1)

6

7

8

A

(3, 2)

x

y –

=1

2 + = 8 x y

0

(0, 8)

(2, 4)

(2, 1)(4, 0)

(–1, –2)

2




Clearly, the two lines interesect at point A (3, 2). 1

Solution of given equations is x = 3, y = 2.

Again x – y = 1 intersects y-axis at (0, –1) 1

and 2 x + y = 8 intersects y-axis at (0, 8).

30. AB || CQ, BC is transversl 1 = 2 Also, A C || PQ, BC is transversal 3 = 4

ABC ~ QCP

ar (ABC)ar (QCP) =

2

2

BC

CP1

= 2

2

(BC)

2 BC3

1 2as BP = BC CP = BC3 3

1

=2

2

9BC

4BC The required ratio is 9 : 4. 1

Or

Let BC = 2 x and QR = x

AB =14

y and PQ = y 1

ar (ABC)ar (PQR) =

1 BC × AB21 QR × PQ2

1

=

124

x y

x y

=24 =

12 1

The required ratio is 1 : 2. 1

31. acot + bcosec = x2

bcot + acosec = y2

L.H.S. : x4 – y4 = ( x2 – y2)( x2 + y 2) ½

= (a cot + b cosec – b cot – a cosec )(a cot + b cosec + b cot + a cosec ) 1

= a2 (cot2 – cosec2– b2 (cot2 – cosec2 ) 1

= a2 (– 1) – b2 (– 1) ½

= – a2 + b2 ½

= b2 – a2 ½

32. In ABC, DE || AC [Given]

BD

DA =

BE

EC....(1) [BPT]1

In ABL, DC || AL [Given]

BD

DA =

BC

CL...(2) [BPT]1

From (1) and (2), we getBE

EC=

BC

CL

4

2=

6

CL

CL = 3 cm. ½

A

B LC

D

E4 2

B P C

A

O

4 2

31

A

B C2 x

y1

4

P

Q R x

y

1



Solutions | 35

33.cos A – sin A +1cos A +sin A – 1

=cot A – 1 + cosec A cot A + 1 – cosecA ½

[Dividing numerator and denominator by sin ]

=

(cot A + cosec A) – 1

cot A + 1 – cosecA ½

=2 2(cot A + cosec A) – (cosec A – cot A)

cot A + 1 – cosecA 1

=(cot A + cosec A) – [1 – cosecA + cotA]

cot A + 1 – cosec A 1

=cosA sinA +

1sinA =

1 cosA sinA

= R.H.S

Proved. 1

34.

Weight (in g) Cumulative Frequency

More than or equal to 0 120


More than or equal to 20 89More than or equal to 30 67


More than or equal to 50 18 2

Plotting the points

120

110

100

90

80

70

60

50

40

30

20

10

(0, 120)

(10, 106)

(20, 89)

(30, 67)

(40, 41)

(50, 18)

0 10 20 30 40 50 x

y

C u m u l a t i v e f r e q u e n c y

Scale



x

y

More than ogive

Lower Lts

60

2




1. (D) mn = 32 = 25

Hence n = 5k, and m = 2

Now nm + n = (5)5+2

= 57 1

2. (A), From the graph it is clear that the curve cuts the x-axis at four places, so number of zeroes

is 4. 1

3. (A),ar ( ABC)

ar ( PQR

AC

PR

2

2

)

= AC

PR

2F H G

I K J

81

169=

7.2

PR

F H G

I K J

2

7.2

PR

F H G

I K J

2

= 29

13

7.2PR

=9

13

PR = 13 7 29 .

= 10.4 cm. 1

4. (D), 2 tan22 1 sin 47ºcot68 2 cos43

= 2

tan22 1 sin47ºcot (90 22 ) 2 cos (90 47 )

= 2tan22 1 sin 47ºtan22 2 sin 47

= 2 +12

=52

1

5. (A), a = 5, b = 13 16. (C), 2 x + y = 7

6 x – py = 21For the condition of infinite number of solutions

a

a

1

2

=b

b

c

c

1

2

1

2

2

6=

1 7

21

p

1

3=

1

p

p = – 3. 1



Solutions | 37

7. (A),2 30

1 302

tan º

tan º=

21

3

11

3

2

3

11

3

2

34

3

2

F H G

I K J

F H G

I K J

=

2

3

3

4

=

3

2

= sin 60º. 1

8. (C), The upper limit of modal class is 175. 1

9. Any positive integer ending with the digit zero is divisible by 5. So prime factorization mustcontain 5.

(12)n = (22 × 3)n 1

By uniqueness of fundamental theoren 5 does not occur in the prime factorization of (12) n. ½

(12)n does not end with the digit zero. ½

10.

3 – 4 + 2 x x2

3 + 5 – 7 + 5 + 3 x x x x4 3 2 x x2 + 3 + 1

3 + 9 + 3 x x x4 3 2

– – –

– 4 – 10 + 5 + 3 x x x3 2

– 4 – 12 – 4 x x x3 2+ + +

2 + 9 + 3 x x2

2 + 6 + 2 x x2

– – –

3 + 1 = ( ) x r x

– (3 x + 1) is to be added so that the polynomial is exactly divisible by x2 + 3 x + 1. ½

11.

FED ~ STU

F = S ½E = T ½

D = U ½

EDTU

=FD EFSU TS

are proportional

andDE EFST TU

is wrong ½

These are not their corresponding sides. ½

12. 2 sin 45° + 3 tan 30° – tan 45° = 21

2

+ 31

3

– 1 1

= 1 + 1 – 1 = 1 ½

1½

E D T U

SF




13. p( x) = ax2 + bx + c ½

constant = 0 = c

p( x) = ax2 + bx ½

Product of zeroes =0

a= 0 ½

One of the zero must be zero, then only product is zero. ½

14. 2 mean = 3 median – mode

Mean =3 median – mode

21

=3 28 – 35

2

=84–35

2=

492

1

Mean = 245

Or

Class Fequency Less then cf

65 - 85 4 4

85 - 105 5 9

105 - 125 13 22

125 - 145 20 42 ½

145 - 165 14 56

165 - 185 7 63

185 - 205 5 68

½

Upper limit of median class = 145 ½

Lower limit of modalclass = 125 ½Difference = 20

15. Since XY || QR

PX

XQ=

PY

YR1

1

2=

PY

PR PY =

4

PR 4

RR – 4 = 8 PR = 12 cm

In right PQR, QR2 = PR2 – PQ2 1

= 122 – 62 = 144 –36 = 108

QR = 6 3 cm 1

16. Let and are zeroes

=

23

½

=23

+ = 15 ½

RQ

P

X Y

4 cm



Solutions | 39

23

= 15

53

= 15 ½

= 15 35 = 9

so =23

× 9 = 6 ½

Now, sum of zeroes = + = 6 + 9 = 15

product of zeroes = = 6 × 9 = 54 ½

Hence polynomial = x 2 – (sum of zeroes) x + product of zeroes

= x2 – 15 x + 54 ½

17. Euclid's algorithm

a = bq + r

0 r b

if b = 6

a = 6q + r

then r = 0, 1, 2, 3, 4, 5.

The odd positive integers n can be expressed as 6q + 1, 6q + 3, 6q + 5 ½

n = 6q + 1

n 2 = (6q + 1)2

= 36q2 + 12q + 1 ½

= 6 (6q 2 + 2q) + 1

= 6m + 1 where m = 6q2 + 2q

n = 6q + 3 ½n2 = (6q + 3)2

= 36q2 + 36q + 9

= 6(6q2 + 6q + 1) + 3

= 6m + 3where m = 6q2 + 6q + 1 ½

n = 6q + 5

n2 = (6q + 5)2

= 36q2 + 60q + 25 ½

= 6(6q2 + 10q + 4) + 1

= 6m + 1 where m = 6q2 + 10q + 4

Thus square of an odd positive integer is either of the form 6m + 1 (or) 6m + 3. ½

Or

(i) In order to arrage the books as required, we have to find the largest number that devides96, 240 and 336 exactly, clearly, such a number is their HCF.We have,

96 = 25 × 3240 = 24 × 3 × 5

and 336 = 24 × 3× 7 HCF of 96, 240, and 336 is 24 × 3 = 48So, there must be 48 books in each stack. 1

Number of stacks of English books =96

48= 2

Number of stacks of Hindi books = 24048

= 5




Number of stacks of sociology books =336

48= 7 1

(ii) HCF of numbers. ½(iii) Cleanlines has been discussed in this question, it is a good habit that leads to good health. ½

18. In ABC, A + B + C = 180°

2

= 90° – C2

1

tan A + B2 = cot

C2

L.H.S. = tan A + B2 . tan

C2

1

= cotC2

.tanC2

= 1 1

19. and are the zeroes of p ( x) p ( x) = 3 x2 – x – 4 ½

+ = – ba = –

13

= 13

=ca

= –43

½

43 + 34 = 33 ( + )= ()3 ( + ) ½

= 34 1

3 3

½

= –64 1

27 3

4

3

+ 3

4

= ––6481 1

20. Since BC || DE and BE || CD with BC CD, BCDE is a rectangle.

Opposite sides are equal.

i.e., BE = CD x + y = 5 ...(i)

and DE = BC = x – y 1

Since perimeter of ABCDE is 21

AB + BC + CD + DE + EA = 21

3 + x – y + x + y + x – y + 3 = 21

6 + 3 x – y = 21

3 x – y = 15 1

Adding (i) and (ii), we get 4 x = 20 ...(ii)

x = 5

On putting the value of x in (i), we get

y = 0

x = 5 and y = 0. 1

Or

Let the number of red balls be x and white balls be y.

According to the question,1

2 y =

1

3 x or 2 x – 3 y = 0 ...(1)

and 3( x + y) – 7 y = 6 1

or, 3 x – 4 y = 6 ...(2)



Solutions | 41

Mutliplying eqn. (1) by 3 and eqn. (2) by 2 and then subtracting, we get

6 x – 9 y = 0 1

6 x – 8 y = 12

– y = – 12

Subtracting from (1), y = 12

2 x – 36 = 0 x = 18 1 x = 18, y = 12

Hence, No. of red balls = 18 and No. of white balls = 12. 1

21. The chief motivational factor for .....ent.

Class Frequency

0 - 100 8

100 - 200 12

200 - 300 x

300 - 400 20

400 - 500 14

500 - 600 7

Since 340 is mode

300 - 400 is modal class ½

mode l +1 0

1 0 22

f f

f f f

× h ½

340 = 300 +20

40 14 x

x

× 100 1

40 = 2026

x x

× 100

52 – 2 x = 100 – 5 x ½

3 x = 48 x = 16 ½

22.

AM BCDN BC ½

In AMO and DNO

M = N = 90°

AOM = DON [Vertically opposite s] ½

AMO ~ DNO

AMDN

= AODO

...(1) ½

ar ( ABC)ar ( DBC)

=

1 × BC × AM21 × BC × DN2

½

= AMDN

Substituting from (1)

ar ( ABC)

ar ( DBC)

= AO

DO

Proved. 1

Or

M

O NCC

A

D




In PQR, CA || PR

PC

CQ=

RA

AQ[By BPT]

PC

15 =

20

12

PC =15 20

12

= 25

PC = 25 cm 1

In PQR, CB || QR

PC

CQ=

PB

BR

25

15=

15

BR1

BR =15 15

25

= 9 cm. 1

23. LHS = 2sec2 – sec4 – 2cosec2 + cosec4

= 2 sec2 – (sec2 )2 – 2 cosec2 + (cosec2 )2

= 2(1 + tan2 ) – (1 + tan2 )2 – 2 (1 +cot2 ) + (1 + cot2 )2 1

= 2 + 2tan2 – 1 – 2 tan2 – tan4 – 2 – 2cot2 + 1 + 2cot2 + cot4 1

= cot4 – tan4 RHS Proved 1

24.

C.I. f x v =i x a

h

fv

10 – 20 5 15 – 2 – 10

20 – 30 11 25 – 1 – 11

30 – 40 19 35 0 0

40 – 50 30 45 1 30

50 – 60 15 55 2 30

Total 80 fv = 39 2

x = a + hi i

i

f x

f

= 35 +3980

× 10

= 35 + 4875

= 39875 1

25. Let a = 4q + r, 0 r < 4

a = 4q, 4q + 1, 4q + 2 or 4q + 3 1

Case I : a2 = (4q)2 = 16q2 = 4(4q2) = 4m, m = 4q2 ½Case II : a2 = (4q + 1)2 = 16q2 + 8q + 1 = 4(4q2 + 2q) + 1

= 4m + 1, where m = 4q2 + 2q ½



Solutions | 43

Cases III : a2 = (4q + 2)2 = 16q2 + 16q + 4

= 4(4q2 + 4q + 1) = 4m, where m = 4q2 + 4q + 1 ½Cases IV : a2 = (4q + 3)2 = 16q2 + 24q + 9

= 4(4q2 + 6q + 2) + 1

= 4m +1, where m = 4q2 + 6q + 2 ½

From cases I, II, III and IV, we conclude that the square of any +ve integer is of the form 4 m or4m +1. 1

26. x + 3 y = 6 ...(1)

y =6

3

x

x 3 6 0

y 1 0 2 ½

2 x – 3 y = 12 ...(2)

y =2 12

3

x

x 0 6 3

y – 4 0 – 2 ½

Putting the above points and drawing a line joing them, we get the graphs of the equations x + 3 y

= 6 and 2 x – 3 y = 12.

–3 –2 –1 1 2 3 4 5 6

(6, 0)

(3, 1)(0, 2)

–1–2

–3

–4

y'

x x'

y

1

2

2–3

=1 2

x

y

x

y + 3

= 6

(0, –4)

A

(3, –2)

0B

Clearly, the two lines intersect at point B (6, 0).

Hence, x = 6 and y = 0 is the solution of the system. 1

Again OAB is the region bounded by the linee 2 x – 3 y = 12 and both the co-ordinate axes. ½

27. sin 62° = sin (90 – 28) = cos 28° ½

tan 73° = tan (90 – 17) = cot 17° ½sin 28° sec 62° = sin (90 – 62)° sec 62°

= cos 62° sec 62° = 1

= 7 [sec2 (90 – 58) – cot2 58] 1

= 7 [cosec2 58 – cot2 58] = 7 1

=5(1)cos 28 3 cot17 –

cos28 cot17 7(1) ½

1 + 3 –57

=(28– 5)

7

=

23

7 ½

1½




28.

Class Fequency Class mark x fx

0 - 20 5 10 50

20 - 40 8 30 240

40 - 60 f 50 50 f 60 - 80 12 70 840

80 - 100 7 90 630

100 - 120 8 110 880

Total 40 + f 2640 + 50 f 1

x = fx

f

½

628 =2640 50

40 f

2512 + 628 f = 2640 + 50 f ½

628 f – 50 f = 2640 – 2512128 f = 128

f = 10

Mode = l f f

f f f h+

–

2 –0

0

1

1 2– ½

= 60 +12 10

24 10 7

× 20 ½

= 60 +27

× 20 = 60 + 571 1

Mode = 6571

29. p( x) = x3 + 10 x2 + px + q

1 and – 2 are zeroes p (1) = 1 + 10 + p + q = 0

p + q = – 11 ...(1) ½

p (– 2) = (– 2)3 + 10 (–2)2 + p (– 2) + q = 0

– 8 + 40 – 2 p + q = 0

– 2 p + q = – 32 ...(2)

p + q = – 11

– 2 p + q = – 32 7 + q = – 11 1

+ – +

+ 3 p = + 21

p = 7 q = – 18 p ( x) = x3 + 10 x2 + 7 x – 18 ½

x + 9

x x x3 + 10 + 7 – 182 x x2 + – 2

– – +

9x x2 + 9 – 18

9 + – 18 x x2 2– – +

0

x x x3 + – 22

Third zero is – 9 2



Solutions | 45

Or

Let the speed of the car I from A = x km/hr.

Speed of the car II from B = y km/hr.

Same dirrection :

Distance covered by car I = 150 + (distance covered by car II) 15 x = 150 + 15 y

15 x – 15 y = 150 x – y = 10 ...(1) 1

Opposite direction :

Distance covered by car I + distance covered by car II = 150 km

x + y = 150 ...(2)1

Adding eqns. (1) and (2), we get 2 x = 160 x = 80

Putting x = 80 in eqn. (1), we get y = 70 1

speed of the car I from A = 80 km/hr

speed of the car II from B = 70 km/hr. 1

30. ABC is a equilateral triangle so AB = AC = BC

Since BD =13

BC

DC =23

BC

BE = EC =23

BC

AD2 = AE2 + DE2

= AC2 – EC2 + (BE – BD)2 ½

AB2 – EC2 + BE2 + BD2 – 2BEBD ½

= AB2 + 2

1 BC3 – 2 12 BC 13 BC 1

= AB2 + BC2 1 1–9 3 1

= AB2 –29

BC2

= AD2 = AB2 –29

AB2

=27AB

9½

9AD2 = 7AB2 ½

Or

In ABC and APQ, BC || PQ [Given]

CBA = AQP, [Alternate angles]

BAC = PAQ, [Vertically opposite angles]

ABC ~ AQP [AA ] 1

AB

AQ=

BC

QP=

AC

AP1

65 AQ

=8

4=

AC28

1




AQ =652

= 3.25 cm, AC = 2 × 28 = 56 cm. 1

P

Q A

B

C

6.5 cm

4 cm

8 cm

2 . 8 c m

31. L.H.S. =3 3

2 2

tan cot

1 tan 1 cot

=3 3

2 2

tan cot

sec cosec

1

=

3 3

3 3

2 2

sin coscos sin

1 1cos sin

½

=3 3sin cos

cos sin

=

4 4sin coscos sin

½

=2 2 2 2 2(sin cos ) 2sin cos

cos sin

½

=2 21 2sin cos

cos sin

½

=sec .cosec 2sin .cos

= R.H.S 1

32. Let PQR be a right triangle right-angled at Q such that QR = b and A = Area of PQR.

Draw QN perpendicular to PR.

We have A = Area of PQR

=12

(QR × PQ)

=12

(b × PQ)

PQ =2A b

...(i)

Now, in 's PNQ and PQR, we have

PNQ = PQRQPN = QPR

So by AA – Criterion of similarity, we have

PNQ ~ PQQ

PQPR

=NQ

QR ...(ii)

Applyiong pythagoras theorem in PQR, we have

PQ2 + QR2 = PR2

2

22

4A b

b = PR2

PR =2 4

24A bb 2 4

4A bb 1

N

P

Q R

1



Solutions | 47

from (i) and (ii) we have

2A × PRb

=NQ

b

NQ =2A PR

= 2 4

2A

4A

b

b1

33. p = sec + tan

=1 sin 1 sin

cos cos cos

½

p2 =2

2

1 sin 2sin

cos

½

p2 + 1 =2 2

2

1 sin 2sin cos

cos

= 2

2 2sin

cos

1

p2 + 1 = 2 2

(1 sin )

cos

1

L.H.S.2 1

2

p

p

=

2

2(1 sin )

cos 12(1 sin ) cos

cos

1

= sec = R.H.S.

34.

Less than cf More than cf

40 3 38 35

42 5s 40 32

44 9 40 30

46 14 40 26

48 28 40 21

50 32 40 7

52 35 40 3 ½+½

2

40 42 44 46 48 50 52

5

10

15

20

25

30

35

40

Classes

X

Y

O

Median = 47

c.f.




1. (A),6243

2 53 4=

6243 2

2 54 4

=12486

104

Hence, decimal expansion of the rational number will terminate after 4 places of decimal. 1

2. (B), f ( x) = x2 – 7 x – 8

Let other zero be k, then – 1 + k = ––7

1

F H G

I K J

= 7

k = 8 1

3. (C), Given 2AB = DE and BC = 8 cm

ABC DEF

So, AB

BC=

DE

EF

AB

8=

2AB

EF

EF = 2 × 8 = 16 cm. 1

4. (D),11 11

2 2cot cos = 11

12

2 2

sin

cos cos

F

H G

I

K J

= 111

112

2

2

2

sin

cos

sin

cos

F

H G

I

K J

F

H G

I

K J

= – 11cos

cos

2

2

F

H G

I

K J = – 11. 1

5. (C), The smallest prime number = 2 and smallest composite number is 22.

Hence, required HCF (4, 2) = 2 1

6. (C), ad bc

ac

bd

Hence, the pair of linear equation has unique solution. 1

7. (A), cosec – cot =1

4

(cos cot )(cos cot )

(cos cot )

ec ec

ec

=

1

4

A

B C

D

E F



Solutions | 49

2 2cosec cotcosec cot

=1

4

2 21 cot cot

cosec cot

=

1

4

cosec + cot = 4. 1

8. (B), Median 1

9. If the number 4n, were to end with the digit zero, then it should be divisible by 5. ½

But 4n = 22n ½

Only prime in the factorization of 4n is 2.

So by fundamental theorem of Arithmetic, there are no other primes in the factorisation of 4n.½

4n can never end with the digit zero. ½

10. Let, f ( x) = 2 x2 – 5 x – 3

Let the zeroes of polynomial are and then

Sum of zeroes + =5

2, product of zeroes = –

3

2

According to question, zeros of x2 + px + q are 2 and 2

Sum of zeroes =coeff. of

coeff. of 2

x

x=

p

1

= 2 + 2 = 2 ( + ) = 2 ×5

2= 5 p = – 5 1

Product of zeros =constant

coeff. of 2

x=

q

1

= 2 × 2 = 4 = 4 F H G

I K J

3

2= – 6

Hence,

p = – 5 and q = – 6. 1

11. Since G is mid point of PQ. PG = GQPG

GQ= 1 1

According to question, GH || QR

PG

GQ =

PH

HR [BPT]

1 =PH

HR

PH = HR

Hence, H is mid points of PR.

12. 2 cosec2 30° + 3sin2 60° –34

tan2 30°. 1

= 2 × (2)2 + 3

2 23 3 1

2 4 3

1

= 8 +

9

4 –

1

4 = 10 ½+½

P

Q R

G H




13. Let p ( x) = 3 x2 + 11 x – 4

= 3 x2 + 12 x – x – 4

= 3 x ( x + 4) – 1 ( x + 4) ½

= (3 x – 1) ( x + 4)

So, zeroes are m = 12

and n = – 4. ½

Now,m

n

n

m =

1

3

4

4

1

3

F H G

I K J

F H G

I K J

=1

12– 12 ½

=145

12. ½

14.

Class Fequency Cumulative Frequency

( f ) (c.f.)

100 - 200 11 11

200 - 300 12 p

300 - 400 10 33

400 - 500 q 46

500 - 600 20 66

600 - 700 14 80

From above p + 10 = 33

p = 33 – 10 = 23 ½

33 + q = 46

q = 46 – 33 = 13 ½

Median Class = 400-500 ½

Model class = 500-600 ½

Or

C.I. f 1

100 - 120 12

120 - 140 14

140 - 160 8

160 - 180 6

180 - 200 1050

Modal Class : 120 – 140 ½

Mode = L +0 1

0 1 22

f f

f f f

× h ½

= 120 +14 12

28 12 8

× 20

= 120 + 5 = 125 1



Solutions | 51

15. AP

AB=

35105

=1

3

AQ AC

= 39

= 13

1

In ABC, AP

AB=

AQ

ACand A is common 1

APQ ~ ABC [SAS]

AP

AB=

PQ

BC1

3=

45BC

BC = 135 cm. 1

A

B

P

C

Q

3.5

7 6

3

4.5

16. Let and are the zeroes of the polynomial, then as per question

= 7

+ 7 = 8 = – F H G

I K J

8

3½

=

1

3 ½

and × 7 =2 1

3

k

72 =2 1

3

k

71

3

2F H G

I K J =

2 1

3

k 1

7 ×1

9=

2 1

3

k

7

3– 1 = 2k

2

3= k. 1

17. We have 117 = 65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0 1

Hence HCF = 13

65m – 117 = 13

65m = 117 + 13 = 130

m =130

65= 2 1

Now, 65 = 13 × 5

117 = 32 × 13 1

LCM = 13 × 5 × 32

= 585Or




(i) The number of rooms will be minimum if each room accomodates maximum number of participants. Since in each room the same number of participants are to be seated and all of themmust be of the same subject. Therefore, the number of paricipants in each room must be the HCFof 60, 84 and 108. The prime factorisations of 60, 84 and 108 are as under

60 = 22 × 3 × 5

84 = 2

2

× 3 × 7108 = 22 × 33

Hence, HCF = 22 × 3 = 12Therefore, in each room 12 participants can be seated. 1

Number of rooms required =Total number of participants

12

=60 84 108

12

=

252

12= 21. 1

(ii) HCF of numbers. ½(iii) Liberty and equality are the pay marks of democracy. ½

18.

cos ( º ) cos ( º– )

tan( º ) tan( º– )

2 245 45

60 30

+ cosec (75º + ) – sec (15º – )

=cos ( º ) sin ( º–45º )

tan( º ) cot( º–30º )

2 245 90

60 90

+ cosec (75º + ) – cosec (90º – 15º + ) 1

=cos ( º ) sin ( º )

tan( º ).cot( º )

2 245 45

60 60

+ cosec (75º + ) – cosec (75º + ) 1

=1

1= 1. 1

19.

4 3 2 8 14 2

8 6 4

8 2

8 6 4

4 4

4 3 2

7 2

2 2 12 4 3 2

4 3 2

3 2

3 2

2

2

2

x x x x x ax b

x x x

x x ax

x x x

x a x b

x x

a x b

x x

( )

( )

For exact division, remainder is zero, then(a + 7) x + b – 2 = 0 2

a + 7 = 0, b – 2 = 0

a = – 7, b = 2. 1

20. 141 x + 93 y = 189 ...(1)

93 x + 141 y = 45 ...(2)

Adding equations (1) and (2), we get

234 x + 234 y = 234

x + y = 1 ...(3)


48 x – 48 y = 144



Solutions | 53

x – y = 3 ...(4) 1

Adding eqns. (3) and (4), we get 2 x = 4

x = 2 1

Put the value of x in eqn. (3), we get 2 + y = 1

y = 1 – 2 = –1

Hence, x = 2 and y = –1. 1

Or

Let the speed of bus be x km/hr and the speed of the train be y km/hr.

According to question,240 120

x y = 8

and120 240

x y = 7

Let1

x= a,

1

y= b, then 1

240a + 120b = 8 ...(1)

120a + 240b = 7 ...(2)

Apply [(1) × 2 – (2)], we get

480a + 240b = 16

120a + 240b = 7 ...(2)

– – –

On subtracting 360a = 9

a =9

360

1

40 1

Putting this value of a in eqn.(1), we get

b =1

60

b =1

60

1

y

y = 60

a =1

40

1

x x = 40 1

Hence, Speed of bus = 60km/hr and Speed of train = 40 km/hr.

21.

xi f i xi f i

3 10 30

9 p 9 p

15 4 60

21 7 147

27 q 27q

33 4 13239 1 39

Total f i = 26 + p + q xi f i = 408 + 9 p + 27q

Given, f i = 40,

26 + p + q = 40

p + q = 14 ...(i)½

Mean, x–

=

x f

f

i i

i½

147 =408 9 27

40

p q




588 = 408 + 9 p + 27q 1

180 = 9 p + 27q

p + 3q = 20 ...(2)

Subtracting eq. (1) from eq. (2), we get

2q = 6

q = 3 ½

Putting this value of q in eq. (1), we get

p = 14 – q = 14 – 3 = 11 ½

Or

(i) Let the assumed mean, A = 1400 and h = 400.

Calculation of Mean

Height (in m) x1 No. of Villlages f 1 D = xi –1400 ui = xi 1400

400 f i ui

200 142 –1200 – 3 –426

600 265 –800 – 2 –5301000 560 –400 – 1 –560

1400 271 0 0 0

1800 89 400 1 89

2200 16 800 2 32

Total N= f i= 1343 f iui = –1395 1

We have A = 1400, h = 400, f iui = –1395 and N = 1343.

Mean = A + h f ui i1

N RST

UVW

= 1400 + 400 ×

–1395

1343

F

H G

I

K J

= 1400 – 41549 = 98451 1

(ii) Mean by assumed mean method. ½

(iii) Villages are much necessary to keep a balance between the ecological problems. ½

22. Given, DBBC, DE AB and ACBC.

To prove :BE

DE=

AC

BC½

Proof : In ABC, 1 + 2 = 90º [C = 90º] 1

But 23 = 90º [Given]

1 = 3 ½

In ABC and BDE, 1 = 3 [Proved] ACB = DEB = 90º [Given]

ABC ~ BDE [AA] 1

AC

BC=

BE

DE

23. LHS =cos sin

cos sin

cos – sin

cos – sin

3 3 3 3

=(cos sin )(cos sin sin cos )

(cos sin )

(cos – sin )(cos sin sin cos )

(cos – sin )

2 2 2 2

1

= (1 – sin cos ) + (1 + sin cos ) 1

CB

A

D

E

1

23



Solutions | 55

= 2 – sin cos + sin cos 1

= 2 = RHS Proved

24.

CI f c.f.

50 - 60 8 8

60 - 70 10 18

70 - 80 12 30

80 - 90 16 46

90 - 100 18 64

100 - 110 14 78

110 - 120 12 90

120 - 130 10 100 1

Here N = 100 N2

= 50. So, median class is 90-100.

Median = l +

N . .2

c f h

f

½

= 90 + 50 4610

18 ½

= 90 +4018

= 922

Median weight = 922 gm. 1

25. n3 – n = n(n2 – 1) = n(n + 1)(n – 1) = (n – 1)n(n + 1)

= product of three consecutive positive integers.

Now, we have to show that the product of three consecutive positive integers is divisible by 6.

We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.

Let a, a + 1, a + 2 be any three consecutive integers. ½

Case I. If a = 3q.

a(a + 1) (a + 2) = 3q(3q + 1) (3q + 2)

= 3q (even number, say 2r) = 6qr,( Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer)

which is divisible by 6. 1

Case II. If a = 3q + 1.

a(a + 1) (a + 2) = (3q + 1) (3q + 2) (3q + 3)

= (even number say 2r) (3) (q + 1)

= 6r (q + 1), 1

which is divisible by 6.

Case III. If a = 3q + 2.

a(a + 1) (a + 2) = (3q + 2) (3q + 3) (3q + 4)

= multiple of 6 for every q

= 6r (say), 1




which is divisible by 6.

Hence, the product of three consecutive integers is divisible by 6. ½

26. 2 x + 3 y = 12 y =12 2

3

x

x 0 6 3

y 4 0 2

x – y = 1 y = x – 1

x 0 1 3

y – 1 0 2

Putting the above points and drawing a line joining them, we get the graph of the equations 2 x +

3 y = 12 and x – y = 1.

–5 –4 –3 –2 –1 1 2 3 4 5 6 7 8

y'

–1

–2

–3

–4

1

2

3

5

4

x' x

y

B

P(3, 2)

x – y = 1

(0, – 1)

2 3 = 12 x + y

A

(0, 4)

M

(1, 0) (6, 0)

Clearly, the two lines interesect at point P (3,2).

2

Hence x = 3 and y = 2 is the solution of the system 1

Area of shaded region = area of PAB

=1

2× base × height =

1

2× AB × PM

=1

2× 5 × 3

= 75 square unit. 1

27. sin A =1

5, cos A = 1 sin A 2

cos A = 11

5

2

F H G

I K J

= 11

5 =

2

51

sin B =1

10, cos B = 1 2 sin B

cos B = 11

10

2

F H G

I K J

= 11

10 =

3

101

Now cos ( A + B) = cos A cos B – sin A sin B



Solutions | 57

=2

5

3

10

1

5

1

10 –

=6

50

1

50 =

5

50=

5 1

5 2 2 1

cos (A + B) =1

2= cos 45º

A + B = 45º. 1

Or

2 52

38

70

20

sin º

cos º

tan º

cot º + cos2 63º + sin 63º cos 27º +

sin ºsec º

tan º.tan º.tan º

55 35

7 45 83

=2cos (90 – 52)º cot (90 – 70)º

cos38º cot20º + cos2 63º + sin 63º × sin (90º – 27º) +

sin55º cosec (90º –35º)

tan7ºtan45º cot(90º–83º)

1

=2cos38º cot20ºcos 38º cot 20º

+ cos2 63º + sin2 63º +sin55º cosec 55º

tan7º.tan45º.cot7º

1

= 2 –1 + 1 +

1sin55ºsin55º

1tan 7º 1tan7º

1

= 2 –1 + 1 +1

1= 3. 1

28. The mean of the following distribution is 628.

C.I. f

5 - 15 2

15 - 25 3

25 - 35 f

35 - 45 7

45 - 55 4

55 - 65 2

65 - 75 2

Modal class : 35-45 ½

Mode = l +1 0

1 0 22

f f h

f f f

½

39 = 35 +7

14 4

f

f

10 1

4 =70 10

10

f

f

½

40 – 4 f = 70 – 10 f ½

6 f = 30 f = 5 1

29. Since and are the zeroes of polynomial 3 x2 + 2 x + 1.

Hence + = 2

3

and =1

3

1




Now for the new polynomial

Sum of the zeroes =1

1

1

1

=

( ) ( )

( )( )

1 1

1 1

=2 2

1

22

3

1

2

3

1

3

Sum of zeroes =

4

32

3

= 2 1


1 1

F H G

I K J

F H G

I K J

1 –

=( )( )

( )( )

1 1

1 1

=1

1

1

1

( )

( )


2

3

1

3

12

3

1

3

6

32

3

3

1

Hence, required polynomial = [ x2 – (Sum of zeroes) x + Product of zeroes]

= x2 – 2 x + 3. 1

Or

Let two digit number is 10 x + y

According to question, 8( x + y) –5 = 10 x + y

2 x –7 y + 5 = 0 ...(1)

and 16( x – y) + 3 = 10 x + y

6 x –17 y + 3 = 0 ...(2) 1

Solving eqns. (1) and (2) by cross-multiplication method, we get x

(–7)( ) – (–17)( )3 5=

y

( )( ) – ( )( )5 6 2 3=

1

2 6( )(–7) – ( )(–7)1

x

21 85=

y

30 6

1

34 42

x

64=

y

24

1

8

x

8=

y

3= 1 1

Hence x = 8, y = 3

So required number = 10 × 8 + 3 = 83. 1

30. Proof : In POQ, AB || PQ, [given]

OA

AP=

OB

BQ...(i) [BPT] 1

In OPR, AC || PR, (given)

OA

AP=

OC

CR...(ii) [BPT ] 1

From (i) and (ii), we getOB

BQ=

OC

CR1

BC || QR, [By converse of BPT] 1



Solutions | 59

Or

31. LHS. =1

1

1

1

sin

sin

sin

sin

=( sin )

( sin )

( sin )

( sin )

( sin )

( sin )

( sin )

( sin )

1

1

1

1

1

1

1

1

1

=( sin )

sin

( – sin )

sin

1

1

1

1

2

2

2

2

1

=( sin )

cos

( sin )

cos

1 12

2

2

2

1

=1 1

sin

cos

sin

cos

=2

cos= 2 sec = RHS Proved 1

32. In ABC, DE || AC, [Given]

BD

DA =

BE

EC[BPT] ...(1)1

In ABE, DF || AC, (Given)

BD

DA =

BF

FE[BPT] ...(2)1

From (1) and (2), we haveBF

FE=

BE

EC Proved 1

33. sin =3

4 cos =

7

4and tan =

3

7½+½

2 2

2

cosec cot 72cot cos

sec 1 x

1

27

3

7 7

42tan

x1

1 2 7

3

7 7

4tan

x

7

3

2 7

3

7

4

7

x1

4 7 7

4

7

x

3 74

7 x x =4

3 1

1




34. Draw the ‘less then’ ogive and ‘more than’ ogive sumultaneously.

Less than c.f. More than c.f.

30 10 20 10040 18 30 90

50 30 40 8260 54 50 7070 60 60 4680 85 70 40

90 100 80 15 2

100

90

80

70

60

50

40

30

20

10

0 10 20 30 40 50 x

y

C u m u l a t i v e f r e q u e n c y

From the graph median = 58.3

60 70 80 90

(20, 100)

(30, 90)

(40, 82)

(50, 70)

(60, 46)

(70, 40)

(80, 15)

(90, 10)

(80, 85)

(70, 60)

(60, 54)

(50, 30)

(40, 18)

(30, 10)

More than ogive

Less than ogive

2

math class xth

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