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CBSE Board Level X- S.A.- I Maths Chapterwise

Printable Worksheets with Solution

Session : 2014-15

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MATHEMATICS (Class X)

Index: S.A.-I

CBSE Chapter-wise Solved Test Papers

1. Real Numbers 01

2. Polynomials 12

3. Pair of Linear Equations in Two Variables 26

4. Triangles 50

5. Introduction to Trigonometry 84

6. Statistics 101

CBSE TEST PAPER-01

CLASS - X Mathematics (Real Number)

1. 7 11 13 15 15× × × + is a

(a) Composite number (b) Whole number

(c) Prime number (d) None of these

2. For what least value of ‘n’ a natural number, ( )24n

is divisible by 8?

(a) 0 (b) -1

(c) 1 (d) No value of ‘n’ is possible

3. The sum of a rational and an irrational is

(a) Rational (b) Irrational

(c) Both (a) & (c) (d) Either (a) or (b)

4. HCF of two numbers is 113, their LCM is 56952. It one number is 904. The other

number is:

(a) 7719 (b) 7119

(c) 7791 (d) 7911

5. Show that every positive even integer is of the from 2q and that every positive

odd integer is the four 2q+1 for some integer q.

6. Show that any number of the form 4n , nEN can never end with the digit 0.

7. Use Euclid’s division algorithm to find the HCF of 4052 and 12576

8. Given that HCF of two numbers is 23 and their LCM is 1449. If one of the

numbers is 161, find the other.

9. Find the greatest of 6 digits exactly divisible by 24, 15 and 36

10. Prove that the square of any positive integer is of the form 4q or 4q+1 for some

integer.

11. 144 cartoons of coke can and 90 cartoons if Pepsi can are to be stacked in a

canteen It each stack is of the same height and is to contain cartoons of the same

Drink. What would be the greaten number of cartoons each stack would have

12. Prove that Product of three consecutive positive integers is divisible by 6.

1

CBSE TEST PAPER-01

CLASS - IX Mathematics (Real Number)

Ans01. (a) and (b) both

Ans02. (c)

Ans03. (b)

Ans04. (b)

Ans05. Let a=bq+r : b=2

0 ≤ r<2 i.e. r=0, 1

a=2q+0, 2q+1,

If a=2q (which is even)

If a=2q+1 (which is odd)

So every positive even integer is of the form 2q odd integer is of the form 2q+1.

Ans06. 2 24 2 2nn n = =

If does not contains ‘5’.so 4 , 0.n n N can never end with the digit∈

Ans07. 12576 4052 3 420= × + 4052 420 9 272

420 272 1 148

272 148 1 124

148 124 1 24

124 24 5 4

24 4 6 0

= × += × += × += × += × += × +

HCF of 12576 and 4052 is ‘4’.

Ans08.

23 1449 161

23 1449

161207

207

HCF LCM a b

b

b

other number is

× = ×× = ×

×=

=∴

2

Ans09. The greater number of 6 digits is 999999.

LCM of 24, 15, and 36 is 360.

999999 360 2777 279

Re . 999999 279 999720quired No is

= × += − =

Ans10. Let a=4q+r when r=0, 1, 2 and 3

∴ Numbers are 4q, 4q+1, 4q+2 and 4q+3

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

2 2 22

2 2 2 2

2 2 2 2

2 2 2 2

4 16 4 4 4

4 1 16 8 1 4 4 2 1 4 1

4 2 16 16 4 4 4 4 1 4

4 3 16 24 9 4 4 6 2 1 4 1

int 4 4 1

a q q q m

a q q q q q m

a q q q q q m

a q q q q q m

square of anye ve ger is of the form q or q

= = = =

= + = + + = + + = +

= + = + + = + + =

= + = + + = + + + = +

∴ + +

Ans11. We find the HCF of 144 and 90

144 90 1 54

90 54 1 36

54 36 1 18

36 18 2 0

18

18.

HCF

so greatest number of cartoons is

= × += × += × += × +

∴ =

Ans12. Let three consecutive number be x(x+1) and (x+2)

Let x=6q+r ; 0 6r≤ <6 , 6 1, 6 2, 6 3, 6 4, 6 5x q q q q q q∴ = + + + + +

( 1)( 2) 6 (6 1)(6 2)

6 6

6 1

(6 1)(6 2)(6 3)

2(3 1).3(2 1)(6 1)

6(3 1).(2 1)(6 1)

6

product of x x x q q q

if x q then which is divisible by

if x q

q q q

q q q

q q q

which is divisible by

+ + = + +== +

= + + += + + += + + +

6 2

(6 2)(6 3)(6 4)

3(2 1).2(3 1)(6 4)

6(2 1).(3 1)(6 1)

6

if x q

q q q

q q q

q q q


= += + + += + + += + + +

3

6 3

(6 3)(6 4)(6 5)

6(2 1)(3 2)(6 5)

6

6 4

(6 4)(6 5)(6 6)

6(6 4)(6 5)( 1)

6

6 5

(6 5)(6 6)(6 7)

6(6

if x q

q q q

q q q


if x q

q q q

q q q


if x q

q q q

q

= += + + += + + +

= += + + += + + +

= += + + += 5)( 1)(6 7)

6

l n 6.

q q


product of any three natura umber is divisible by

+ + +

∴

4

CBSE TEST PAPER-02

Class X- Mathematics (Real Number)

1. A lemma is an axiom used for proving

(a) other statement (b) no statement

(c) Contradictory statement (d) none of these

[1]

2. If HCF of two numbers is 1, the Two number are called relatively ________ or________

(a) Prime, co-prime (b) Composite, prime

(c) Both (a) and (b) (d) None of these

[1]

3. 2.35 is

(a) a terminating decimal number (b) a rational number

(c) an irrational number (d) Both (a) and (b)

[1]

4. 2.13113111311113……is

(a) a rational number (b) a non terminating decimal number

(c) an irrational number (d) both (a) & (c)

[1]

5. Show that every positive odd integer is of the form (4q+1) or (4q+3) for same inter

q.

[2]

6. Show that any number of the form 6x, x∈N can never end with the digit 0 [2]

7. Find HCF and LCM of 18 and 24 by the prime factorization method. [2]

8. The HCF of two numbers is 23 and their LCM is 1449. If one of the number is 161,

find the other

[2]

9. Prove that (3- 5 ) is irrational. [3]

10. Prove that if x and y are odd positive integers then x2+y2 is even but not divisible

by 4

[3]

11. Show that one and only one out of n, (n+2) or (n+4) is divisible by 3, where n∈N [3]

12. Use Euclid’s division lemma to show that the square of any positive integer of the

from 3m or (3m+1) for some integer q

[3]

5

CBSE TEST PAPER-02

Class X - Mathematics (Real Number)

[ANSWERS]

Ans01. (a) Ans02. (a) Ans03. (b) Ans04. (c)

Ans05. Let a=4q+r : 0 4r≤ <4 2(2 ) an even integer

4 1 2(2 ) 1 an odd integer

4 2 2(2 1) an even integer

4 3 2(2 1) 1 an odd integer

This every positive odd integer is of the form

(4 1) (4 3)for some integ

a q q

a q q

a q q

a q q

q or q

∴ = == + = += + = += + = + +

+ + er

Ans6. ( )6 2 3 2 3nn n n= × = ×

5 6

0.

nis not a factor of

It never ends with

∴∴

Ans7. 218 2 3 3 2 3= × × = ×324 2 2 2 3 2 3

2 3 6HCF

= × × × = ×∴ = × =

Ans8. LCM HCF a b× = ×1449 23 161 b× = ×

1449 23

161207

b×=

=

Ans9. 3 5 =p

Letq

−

3- 5 = (where p & q are integer, coprime & 0)

3 5

35

Pq

q

p

q

q p

q

∴ ≠

− =

− =

6

( )

3(3 ) and are integer,so is a rational

number, but 5 is an irrational number. This contradiction

arises because of our wrong assumption. So 3 5 is an

irrational number.

q pq p q

q

−−

−

Ans10. Let x=2p+1 and y=2q+1

( ) ( )

( )( )

2 22 2

2 2

2 2

2 2

2 2

2 1 2 1

4 4 1 4 4 1

4 2

2 2 2 2 2 1

2 where m 2 2 2 2

x y p q

p p q q

p q p q

p q p q

m p q p q

∴ + = + + +

= + + + + +

= + + + +

= + + + +

= = + + + +( )2 2

1

is an even number but not divisible by 4.x y∴ +

Ans11. Let the number be (3q+r)

( ) ( )

( ) ( ) ( )( )

3 0 3

or 3 , 3 1, 3 2

If 3 then, numbers are 3 , 3 1 , 3 2

3 is divisible by 3.

If 3 1 then, numbers are 3 1 , 3 3 , 3 4

3 3 is divisible by 3.

n q r r

q q q

n q q q q

q

n q q q q

q

= + ≤ <+ +

= + +

= + + + +

+

( ) ( ) ( )( )

If 3 2 then, numbers are 3 2 , 3 4 , 3 6

3 6 is divisible by 3.

out of , ( 2) and ( 4) only one is divisible by 3.

n q q q q

q

n n n

= + + + +

+∴ + +

Ans12. 3 ; 0 3Let a q r r= + ≤ < 3 , 3 1 3 2 a q q and q∴ = + +

( ) ( ) ( )2 22 2 23 9 3 3 3 3a q q q m where m q= = = = =

( ) ( )( )

( ) ( )( )

22 2 2

2

22 2 2

2

3 1 9 6 1 3 3 2 1

3 1 3 2

3 2 9 12 4 3 3 4 1 1

3 3 4 1

a q q q q q

m where m q q

a q q q q q

m where m q q

= + = + + = + +

= + = +

= + = + + = + + +

= = + +

7

CBSE TEST PAPER-03

Class X - Mathematics (Real Numbers)

1. The smallest composite number is:-

(a) 1 (b) 2 (c) 3 (d) 4

[1]

2. 1.2348 is

(a) an integer (b) an irrational number

(c) a rational number (d) None of there,

[1]

3. π is (a) rational (b) irrational

(c) both (a) & (b) (d) neither rational nor irrational

[1]

4. (2+ 5 ) is (a) rational (b) irrational

(c) An integer (d) Not real

[1]

5. Prove that the square of any positive integer of the form 5g+1 is of the same form [2]

6. Use Euclid’s division algorithm to find the HCF of 4052 and 12576 [2]

7. Find the largest number which divides 245 and 1029 leaving remainder 5 in each

case

[2]

8. A shop keeper has 120 litres of petrol, 180 litres of diesel and 240 litres of

kerosene. He wants to sell oil by filling the three kinds of oils in tins of equal

capacity. What should be the greatest capacity of such a tin

[2]

9. Prove that in n is not a rational number, if n is not perfect square [3]

10. Prove that the difference and quotient of ( )3 2 3+ and ( )3 2 3− are irrational [3]

11. Show that (n2-1) is divisible by 8, if n is an odd positive integer [3]

12. Use Euclid division lemma to show that cube of any positive integer is either of the

form 9m. (9m+1) or 9m+8

[3]

8

CBSE TEST PAPER-03

CLASS - Mathematics (Real Numbers)

[ANSWERS]

Ans01. (c)

Ans02. (c)

Ans03. (b)

Ans04. (b)

Ans5. 5 1a q= +

( )( )

22 2

2

5 1 25 10 1

5 5 2 1

5 1

a q q q

q q

m

∴ = + = + +

= + +

= +

Ans6. 12576 4052 HCF of and

12576 4052 3 420

4052 420 9 272

420 272 1 148

272 148 1 124

148 124 1 24

124 24 5 4

24 4 6 0

4HCF

= × += × += × += × += × += × += × +

∴ =

Ans07. The required number is the HCF of (245-5) and (1029-5) i.e. 240 and 1024.

1024 240 4 64

240 64 3 48

= × += × +

64 48 1 16

48 16 3 0

16.number is

= × += × +

∴

9

Ans08. The required greatest capacity is the HCF of 120, 180, and 240.

240 180 1 60

180 60 3 0

60.

60, 120

120 60 2 0

120, 180 240 60.

the required capacity 60 litre.

HCF is

Now HCF of

HCF of and is

is

= × += × +

= × +∴∴

Ans9. Let be a rational number.n

2

2

2 2

2

2 2 2

2 2 2

2 2

2

divides

divides ( )

Let

divides

divides ( )

from

pn

q

pn

q

p nq

n p

n p i

p nm

p n m

n m nq

q nm

n q

n q ii

∴ =

⇒ =

=⇒

⇒ →=

⇒ =∴ =

=⇒

⇒ →(i) and (ii) n is a common factor of both and .

this contradicts the asssumption that and are co-prime.

So, our supposition is wrong.

p q

p q

Ans10. ( ) ( )Difference of 3 2 3 and 3 2 3+ −

( ) ( )3 2 3 3 2 3= + − −

3 2 3 3 2 3= + − +

4 3 which is irrational.

and quotient is

3 2 3 3 2 3

3 2 3 3 2 3

=

+ += ×− +

10

( )

9 12 12 3

9 12

21 12 3

3

7 4 3 7 4 3

+ +=−

+=−

= − − = − +

Ans11. Let n=4q+1 (an odd integer)

( )

( )

22

2

2

2

1 4 1 1

16 1 8 1

16 8

8 2 2

8

which is divisible by 8.

n q

q q

q q

q

m

∴ − = + −

= + + −= +

= +

=

Ans12. Let a=3q+r ; 0 3r≤ <

( )( )

3 3 3

3 3 2

3 2

3 2

3 ; then 27 9 ; where 3

when 3 1 ; then 27 27 9 1

9 3 3 1

9 8 where m 3 3

when

a q a q m m q

a q a q q q

q q q

m q q q

a

∴ = = = == + = + + +

= + + +

= + = + +

( )

( )( )

23

3 2

3 2

3 2

3 2 ; then 3 2

27 54 36 8

9 3 6 4 8

9 8 where m 3 6 4

Hence

q a q

q q q

q q q

m q q q

= + = +

= + + +

= + + +

= + = + +

( ) ( ) cubs of any positive integer is either of the form 9 , 9 1 or 9 8 .m m m+ +

11

CBSE TEST PAPER-01

Class X - Mathematics (Polynomials)

1. Which of the following is polynomial?

(a) 2 6 2x x− + (b)1

xx

+ (c)2

5

3 1x x− + (d) none of these

[1]

2. Polynomial 4 3 2 22 3 5 5 9 1x x x x x+ − − + + is a

(a) Linear polynomial (b) quadratic polynomial

(c) cubic polynomial (d) Biquadratic polynomial

[1]

3. If α and β are zero’s of 2 5 8x x+ + then the value of ( )α β+ is

(a) 5 (b) -5 (c) 8 (d) -8

[1]

4. The sum and product of the zeros of a quadratic polynomial are 2 and -15

respectively. The quadratic polynomial is

(a) 2 2 15x x− + (b) 2 2 15x x− − (c) 2 2 15x x+ − (d) 2 2 15x x+ +

[1]

5. Find the quadratic polynomial where sum and product of the zeros one a and

1

a.

[2]

6. If α and β are the zeroes of the quadratic polynomial ( ) 2 4,f x x x= − − find the

value of 1

α+

1 αββ

−

[2]

7. If the square of the difference of the zeroes of the quadratic polynomial

( ) 2 45f x x px= + + is equal to 144, find the value of p.

[2]

8. Divide ( )3 26 26 21x x x− − + by ( )7 3x− + [2]

9. Apply division algorithms to find the quotient q(x) and remainder r(x) an dividing

f(x) by g(x) where ( ) ( )3 2 26 11 6, 1f x x x x g x x x= − + − = + +[3]

10. If two zeroes of the polynomial 4 3 26 26 138 35x x x x− − + − are 2 3± , find the other

zeroes.

[3]

11. What must be subtracted from the polynomial ( ) 4 3 22 13 12 21f x x x x x= + − − + so

that the resulting polynomial is exactly divisible by ( ) 2 4 3g x x x= − +

[3]

12. What must be added to 5 4 3 26 5 11 3 5x x x x x+ + − + + so that it may be exactly

divisible by 23 2 4x x− +

[3]

12

CBSE TEST PAPER-01

CLASS - Mathematics (Polynomials)

[ANSWERS]

Ans01. (d) Ans02. (d)

Ans03. (b) Ans04. (b)

Ans05. Polynomial 2 2 21 19 i.e. 9 9 1

9 9x x x x − + − +

Ans06. 2( ) 4 i.e. f x x x= − −

( )

If and are the zeroes

1 + 1

14

. 41

,

1 1

14

41

44

15

4

So

α β

α β

α β

α βαβ αβα β αβ

∴ = =

−= = −

++ − = −

= − −−

= − +

=

Ans07.

45

pα βαβ+ = −

=

( )

( )( )

2

2 2

2

2

2

144

2

4 144

4 45 144

144 180

18

p

p

p

α β

α β αβ

α β αβ

− =

+ −

+ − =

− − × =

= += ±

13

Ans08. 2

3 2

3 2

2

2

2 5 3

3 7 6 26 21

6 -14

15 26 21

15x 35

9 21

9 21

x x

x x x x

x x

x x

x

x

x

−

+ +

− + − −

− −−

−−

0

+

+-

- +

2 2 5 3quotient x x∴ = + +

Ans09. ( ) ( ) ( ) ( )f x g x q x r x= × ×

2 3 2

3 2

2

2

7

1 6 11 6

7 10 6

7 7 7

17 1

x

x x x x x

x x x

x x

x x

x

−

+ + − + −+ +− + −− − −

− ++ +

-

+

- -

( ) ( ) ( )3 2 2 6 11 6 2 1 7 17 1x x x x x x x∴ − + − = + + − + +

Ans10. Two zero’s are 2 3 ±

( )2 4 3 2

of zero's 4

product of the zero's 1

4 1 is the factor of 6 26 138 35

Sum

and

x x x x x x

∴ ==

∴ − + − − + −

2

2 4 3 2

4 3 2

3 2

3 2

2

2 35

4 1 6 26 138 35

4

2 27 138 35

2 8 2

35 140 35

x x

x x x x x x

x x x

x x x

x x x

x x

− −

− + − − + −− +

− − + −− + −

− + −235 140 35

0

x x− + −

+ +

- + -

-

- ++

14

Now,

( ) ( )( )( )

2

2

2 35

7 5 35

7 5 7

5 7

zeros are

7 and 5

other two zeros are 7 5

x x

x x x

x x x

x x

x x

and

− −+ −

− + −

− −∴

= = −∴ −

Ans11. 2

2 4 3 2

4 3 2

3 2

3 2

2

6 8

4 3 2 13 12 21

4 3

6 16 12 21

6 14 18

8

x x

x x x x x x

x x x

x x x

x x x

x

+ +

− + + − − +− +

− − +− +

2

30 21

8 32 24

2 3

x

x x

x

− +− +

−

+

+

- + -

- -

--We must be subtract (2x-3) to become a factor.

Ans12. 2

2 5 4 3 2

5 4 3

4 3 2

4 3 2

6 8

3 2 4 6 5 11 3 5

6 4 8

9 3 3 5

9 6 12

x x

x x x x x x x

x x x

x x x x

x x x

+ +

− + + + − + +− +

+ − + +− +

3 2

3 2

2

2

9 15 5

9 6 12

9 11 5

9 6 12

x x x

x x x

x x

x x

− + +− +− − +− + −

17 17x− +

+

+

- + -

- -

--

-+ +

So we must be added ( ) ( )23 2 4 17 17x x x− + − − +

2

2

3 2 4 17 17

3 15 13

x x x

x x

= − + + −= + −

15

CBSE TEST PAPER-02

Class X - Mathematics (polynomials)

1. If P(x)= 2x2-3x+5,3x+5,then P(-1) is equal to

(a) 7 (b) 8

(c) 9 (d) 10

[1]

2. Zeroes of P(x) = x2-2x-3 are

(a) 3 and 1 (b) 3 and -1

(c) -3 and -1 (d) 1 and -3

[1]

3. If α and β are the zeros of 2x2+5x-10, them the value of αβ is

(a) 5

2− (b) 5

(c)-5 (d) 2

5

[1]

4. A quadratic polynomial, the sum and product of whore zeros are 0 and

5 respectively is

(a) x2+ 5 (b) x2- 5

(c) x2-5 (d) None of these

[1]

5. Find the value of ‘k’ such that the quadratic polynomial x2-(k+6) x+2(2k+1) has

sum of the zeros is half of their product

[2]

6. If α and β are the zeroes of the quadratic polynomial f(x) = x2-p(x+1)-c, show that

(α +1) ( β +1)=1-c

[2]

7. If the sum of the zeroes of the quadratic polynomial f(t) = kt2+2t+3k is equal to

their product, find the value if ‘k’

[2]

8. Divide (x4-5x+6) by (2-x2) [2]

9. Find all the zeroes of the polynomial f(x) = 2x4-3x3-3x2+6x-2, if being given that

two of its zeroes are 2 and - 2

[3]

10. On dividing x3-3x2+x+2 by a polynomial g(x) the quotient and the remainder were

(x-2) and -2x+4 respectively find g(x)

[3]

11. Find all zeroes of f(x) = 2x3-7x2+3x+6 if its two zero one -

3

2and

3

2

[3]

12. Obtain all zeroes of the polynomial f(x)= 2x4+x3-14x2-19x-6, if two of its zeros are -

2 and -1

[3]

16

CBSE TEST PAPER-02

Class X - Mathematics (polynomials)

[ANSWERS]

Ans01. (d)

Ans02. (b)

Ans03. (c)

Ans04. (a)

Ans05. Sum of the zeros = 1

product of the zero's2

( ) ( )16 2 2 1

2 6 2 1

5

k k

k k

k

+ = +

+ = +⇒ =

Ans06.

( ) ( )( )

( )( ) ( ) ( )

2

2

1

1 1 1

1

1

f x x p x c

x px p c

p and p c

Now

p c p

c

α β αβα β αβ α β

= − + −

= − − +

∴ + = = − +

+ + = + + += − − + += −

Ans07.

( ) 2 2 3

Sum of the zeros Product of the zeros

2 3

f t kt t k

k

k k

= + +=

− =

2

3k⇒ = −

17

Ans08. 2

2 4

4 2

2

2

2

2 5 6

2

2 5 6

2 4

5 10

x

x x x

x x

x x

x

x

− −

− − +−

− +−

− ++

- +

-Quotient = 2 2x− − Remainder = -5x+10

Ans09. 2 2 are the zeros.and −

( )( )2 2 is the factor of the given polynomial.x x∴ − +

2

2 4 3 2

4 2

3 2

3

2

2 3 1

2 2 3 3 6 2

2 4

3 6 2

3 6

2

x x

x x x x x

x x

x x x

x x

x

− +

− − − + −−

− + + −− +

−2 2

0

x −

+-

+ -

- +

( )

( ) ( )( )( )

2

2

2 3 1

2 2 1

2 1 1 1

2 1 1

other two zero's are

1 1 and

2

q x x x

x x x

x x x

x x

x x

= − +

= − − += − − −

= − −∴

= =

Ans10. ( ) ( ) ( ) ( )p x q x g x r x= × +

( ) ( ) ( )( )

3 23 2 2 4

2

p x r xg x

q x

x x x x

x

−=

− + + + −=−

18

2

3 2

3 2

2

2

1

2 3 3 2

2

3 2

2

2

2

x x

x x x x

x x

x x

x x

x

x

− +

− − + −−

− + −− +

−−

0

+-

+ -

- +

( ) 2 1g x x x= − +

Ans11. ( ) 4 3 22 2 7 3 6f x x x x x= − − + +

( )

( ) ( )

2

2

3Two zero's are

2

3 3 1 2 3

2 2 2

2 3 is the factor of .

x x x

x f x

±

∴ + − = −

∴ −2

2 4 3 2

4 2

3 2

3

2

2

2 3 2 2 7 3 6

2 3

2 4 3 6

2 3

4 6

x x

x x x x x

x x

x x x

x x

x

− −

− − − + +−

− − + +− +

− +24 6

0

x− +

+-

+ -

-+

( )

( ) ( )( )( )

2

2

2

2 2

2 1 2

1 2


1 0 or 1

and 2 0 or 2


1 and 2

g x x x

x x x

x x x

x x

x x

x x

= − −

= − + −= − + −

= + −∴

+ = = −− = =

∴−

19

Ans12. ( ) 4 3 22 14 19 6, two zero's are 2 -1f x x x x x and= + − − − −

( ) ( ) ( )( )( ) 2

2 and 1 are the factors of .

2 1 3 2

x x f x

x x x x

∴ + +

∴ + + = + +

2

2 4 3 2

4 3 2

3 2

3 2

2 5 3

3 2 2 14 19 6

2 6x 4

5 18 19 6

5 15 10

x x

x x x x x x

x x

x x x

x x x

− −

+ + + − − −+ +− − − −− − −

2

2

3 9 6

3 9 6

0

x x

x x

− − −− − −

+

-

+ +

--

+++

( ) ( )( )( )

2

2

2 5 3

2 6 3

2 3 1 3

3 2 1

' are

3 0

3

Now x x

x x x

x x x

x x

zero s

x

x

− −= − + −= − + −

= − +∴

− ==

2 1 0

1

2 two zero's are

13 and

2

and x

x

other

+ =

= −

−

20

CBSE TEST PAPER-03


1. Degree of polynomial y3-2y2-

13

2y + is

(a) 1

2 (b) 2 (c) 3 (d)

3

2

[1]

2. Zeroes of P(x) = 2x2+9x-35 are

(a) 7 and 5

2 (b) -7 and

5

2 (c) 7 and 5 (d) 7 and 2

[1]

3. The quadratic polynomial whore zeros are 3 and -5 is

(a) x2+2x-15 (b) x2+3x-8 (c) x2-5x-15 (d) None of these

[1]

4. If α and β are the zeros of the quadratic polynomial P(x) = x2-px+q, then the value

of 2 2α β+ is equal to

(a) p2-2q (b) p

q (c) q2-2p (d) none of these

[1]

5. Find the zeros of the polynomial p(x) = 4 3 x2+5x -2 3 and verify the

relationship b/w the zeros and its coefficients

[2]

6. Find the value of ‘k’ so that the zeroes of the quadratic polynomial 3x2-kx+14 are in

the ratio 7:6

[2]

7. If one zero of the quadratic polynomial f(x) = 4x2-8kx-9 is negative of the other,

find the value of ‘k’.

[2]

8. Cheek whether the polynomial (t2-3) is a factor of the polynomial 2t4+3t3-2t2-9t-12

by Division method

[2]

9. Obtain all other zeroes of 3x4+6x3-2x2-10x-5. If two of its zeroes are

5

3and

5

3−

[3]

10. If the polynomial x4-6x3+16x2-25x+10 id divided by another polynomial x2-2x+k,

the remainder comes out to be (x+a), find ‘k’ and ‘a’

[3]

11. Find the value of ‘k’ for which the polynomial x4+10x3+25x2+15x+k is exactly

divisible by (x+7)

[3]

12. If α , and β are the zeros of the polynomial f(x) = x2+px+q form polynomial whore

zeros are (α + β )2 and (α - β )2

[3]

21

CBSE TEST PAPER-03


[ANSWERS]

Ans01. (c)

Ans02. (b)

Ans03. (a)

Ans04. (a)

Ans05. ( ) 24 3 5 2 3p x x x= + −

( ) ( )( ) ( )

2

2

4 3 8 3 2 3

4 3 2 3 3 2

4 3 3 2

zero's are 4 3 0 and 3 2 0

3 2

4 3Coefficient of

Sum of zero'sCoefficient of

x x x

x x x

x x

x x

x x

x

x

= + − −

= + − +

= − +

∴ − = + =

= = −

− −= =

( )

2

5

4 3

23

4 3

Constant term 2 3Product of zero's

Cofficient of 4 31

2

x

−= +

−= =

−=

3 2

4 3

− = ×

Ans06. Let the zero’s are 7p and 6p.

( )2 3 14

7 63 3

14 and 7 6

3

x k

k kp p

p p

− +− −

∴ + = =

× =

22

2 14 42

3 3

39

39 3

117

p

p

p k

k

k

⇒ =

=⇒ =∴ = ×∴ =

Ans07. 24 8 9, if one zero is then other is x kx α α− − −

Sum of the zero 0

80

40

k

k

∴ =

=

⇒ =

Ans08.

2

2 4 3 2

4 2

3 2

3

2

2t 3 4

3 2 3 2 9 12

2 6

3 4 9 12

3 9

4

t

t t t t t

t t

t t t

t t

t

+ +

− + − − −−+ − −

−

2

12

4 12

0

t

−−

- +

+

-

-

+

( )2Yes, 3 is the factor of given polynomial.t −

Ans09. 4 3 23 6 2 10 5x x x x+ − − −

5 zero's are

3±

( )2 25 5 5 1 is the factor given polynomial i.e. 3 5

3 3 3 3x x x x + − − −

23

2

2 4 3 2

4 2

3 2

3

2 1

3 5 3 6 2 10 5

3 5

6 3 10 5

6 10

3

x x

x x x x x

x x

x x x

x x

+ +

− + − − −−+ − −

−2

2

5

3 5

0

x

x

−−

- +

+-

- +

( )

( )

2

2

1

1

other zero's are

1 0

1 and 1 0

1

other two zero's -1 and -1

Now x x

x

x

x x

x

+ + +

⇒ +∴

+ == − + =

= −∴

Ans10.

( )

( )

2

2 4 3 2

4 3 2

3 2

3 2

4 8

2 6 16 25 10

2

4 16 25 10

4 8 4

x x k

x x k x x x x

x x kx

x k x x

x x kx

− + −

− + − + − +− +− + − − +

− + −

( ) ( )( ) ( ) ( )( ) ( )

2

2 2

2

8- 4 25 10

8 - 16 2 8

2 9 8 10

k x k x

k x k x k k

k x k k

+ − +

− − + −

− + − +

- +

+ -

- +

-

+

-

24

( )but remainder is

equating the cofficient of and constant term.

so 2 -8 10

25 40 10

5

5 and 5

x a

x

k k a

a

a

k a

+∴

+ =− + =

− =∴ = = −

Ans11. ( ) 4 3 210 25 15p x x x x x k= + + + +

( )( )

( ) ( ) ( ) ( )4 3 2

7 is the factor.

7 0

7 10 7 25 7 15 7 0

2401 3430 1225 105 0

91

x

p

or k

k

k

∴ +

∴ − =

− + − + − + − + =− + − + =

=

Ans12. ( ) 2 , if and are zero'sf x x px q α β= + +

( ) ( )( ) ( )

( )

2 2

2 2

2

-

If zero's are and

4

4

p and q

p q

α β αβ

α β α β

α β α β αβ

∴ + = =

+ −

− = + −

= − −

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2

2 2 2 2

2

2 2 2 2

4 2

4

Now sum of zero's

4

2 4

Pr oduct of zero's

4

4 4

required polynomial is

p q

p p q

p q

p p q

p p q

α β

α β α β

α β α β

− = − −

+ + − = − + −

= −

+ − = − + −

= −∴

( )( )

2

2 2 4 2

2 2 4 2

sum of zero's product of zero's

2 4 4 4

2 4 4

x x

x p q x p p q

x p x qx p p q

− +

− − + −

− − + −

25

CBSE TEST PAPER-01

Class X - Mathematics (Pair of Linear Equation)

1. A pair of Linear equation in two variables which has a common point i.e which has

only one solution is called a

(a) Consistent pair (b) Inconsistent pair

(c) Dependent pair (d) None of there.

[1]

2. If a pair of linear equation 1 1 1 0a x b y c+ + = and 2 2 2 0a x b y c+ + = represents

coincident lines, then

(a) 1 1

2 2

a b

a b≠ (b) 1 1 1

2 2 2

a b c

a b c= ≠

(c) 1 1 1

2 2 2

a b c

a b c= = (d) None of these

[1]

3. The value of ‘k’ for which the system of equation 2x+3y=5 and 4x+ky=10 has

infinite number of solutions is

(a) k=1 (b) k=3

(c) k= 6 (d) k=0

[1]

4. If the system of equation 2x+3y=7 and 29x+(a+b) y=28 has infinitely many

solution then

(a) a=2b (b) b=2a

(c) a+2b=0 (d) 2a+b =0

[1]

5. The cost of two kg of apples and 1kg of grapes on a day was found to be Rs 160.

After a month the cost of 4 kg apples and 2kg grapes is Rs 300. Represent the

[2]

26

situation algebraically and graphically.

6. Find the value of ‘k’ for which the system of equation kx+3y=k-3 and 12x+ky=k

will have no solution.

[2]

7. Can (x-2) be the remainder on division of a polynomial p(x) by (2x+3)? Justify your

answer.

[2]

8. ABCD is a rectangle find the values of x and y. [2]

9. Solve the following system of equation graphically. x+2y=1, x-2y=-7 also read the

paints from the graph where the lines meet the x-axis and y-axis.

[3]

10. Salve 23x-29y=98 and 29x-23y=110 [3]

11. A man has only 20 paisa coins and 25 paisa coins in his purse. If he has 50 coins in

all totaling Rs 11.25. How many coins of each kind does he have?

[3]

12. A says to B “my present age is Five times your that age when I was an old as you

are now. It the sum of their present ages is 48 years, find their present ages.

[3]

13. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it

can go 40km upstream and 55 km down stream. Determined the speed of the

stream and that of the boat in still water.

[5]

x+y

x-y

12

8

A

D

B

C

27

CBSE TEST PAPER-01


[ANSWERS]

Ans01. (a)

Ans02. (c)

Ans03. (c)

Ans04. (b)

Ans05. Let the cost of one Kg of apple is x and one Kg of grapes is y.

According to question

2x+y=160 and 4x+2y=300

2x+y=160

x 0 80 40

Y 160 0 80

4x+2y=300

x 0 75 40

Y 150 0 70

.

..

.

..

28

Ans06. 3 3kx y k+ = −

( )

1 1 1

2 2 2

2

2

2

12

The system has no solution.

If

3 3

12

36

6 (i)

3 3If

3 3

6 0

6 0

6

x ky k

a b c

a b c

k k

k k

k

k

k

k k

k k k

k k

k k

k i

+ =

= ≠

−= ≠

=⇒ = ±

−≠

≠ −

− ≠− ≠

≠ ( ) 6

i

k∴ = −

Ans07. ' 'No

( ) ( ) ( )( ) ( )

2 can't be remainder on dividing p by 2 3 .

Because the degree of 2 is equal to the degree of 2 3 .

x x x

x x

− +

− +

Ans08. 12x y+ =8

on adding 2 20

10

10 12

2

x y

x

x

y

y

− ===

∴ + ==

Ans09. ( )2 1 x y i+ = →

x 0 1 -1

y 1/2 0 1

( )2 7 x y ii− = − →

x 1 5 9

y 4 6 8

29

..

.

...

( ) ( )

( ) ( )

3

2

1The straight line meet the axis at 0, and 1,0

2

7and straight line meet the axis at 0, and 7,0 .

2

x

y

i

ii

= −=

−

Ans10. ( )23 29 98 x y i− = →

( )( ) ( )

( )

29 23 110

on adding eq and

52 52 208

4

on subtracting

x y ii

i ii

x y

or x y iii

− = →

− =− = →

23 29 98

29 23 110

6 6 12

x y

x y

x y

− =− =

− − = −- + -

( )( ) ( )

2

on adding and we get

2 6 i.e. 3

3 2

1

3 and 1

x y iv

iii iv

x x

y

y

x y

+ = →

= =∴ + =

= −∴ = = −

30

Ans11. Let the number of coin of

( )

( )

20 paise be ' ' and

25 paise be ' '


50

and

20 25 1125

x

y

x y i

x y ii

or

+ = →

+ = →

( )4 5 225

4 4 200 from

25

x y

x y i

y

+ =

+ =

=- --

50

25 50

number of 20 paise coin 25

and number of 25 paise coin 25

x y

x

+ =+ =

∴ ==

Ans12. Let the percentage of A yearsx=

( )

and B years


48

y

x y i

=

+ = →

( )[ ]

5

5 2

10 5

x y x y

x y x

x y x

= − −

= −= −

3 5x y=

( )3 48 5

18 years

and 48-18 years

30 years

y y

y

x

x

− =⇒ =

==

Ans13. Let the speed of boat is km/h in still water x

and stream km/h

According to question,

30 4410

y

x y x y+ =

− +

31

( )( )

( ) ( )

40 55and 13

1 1 Let and

30 44 10

40 55 13

on solving eq and we get,

1 5

5

x y x y

u vx y x y

u v i

u v ii

i ii

u x y

+ =− +

= =− +

+ = →

+ = →

= ⇒ − = → ( )

( )( ) ( )

1 11

11 on solving eq and we get,

8 /

3 /

iii

v x y iv

iii iv

x km h

y km h

= ⇒ + = →

==

32

CBSE TEST PAPER-02


1. A system of simultaneous linear equations is said to be inconsistent, if it has:-

(a) One solution (b) Two solutions (c) Three solution (d) No solution

[1]

2. The system of equation 2x+3y-7=0 and 6x+5y-11=0 has

(a) unique solution (b) No solution (c) Infinitely many solution (d) None of these

[1]

3. The value of ‘k’ for which the system of equation x+2y-3=0 and 5x+ky+7=0 has no

solutions is

(a) k=10 (b) k=6 (c) k=3 (d) k=1

[1]

4. The equation axn+byn+c=0 represents a straight line if

(a) n ≥ 1 (b) n ≤ 1 (c) n=1 (d) None of these

[1]

5. The path of a train A is given by the equation x+2y-4=0 and the path of another

train B is given by the equation 2x+4y-12=0 represent this situation Graphically.

[2]

6. For what value of ‘α ’the system of linear equations α .x + 3y =α -3, 12x+α y=α

has no solution

[2]

7. Find the values of ‘a’ and ‘b’ for which the following system of linear equations has

infinite number of solution 2x+3y=7, (a+b+1) x +(a+2b+2)y = 4(a+b)+1

[2]

8. Solve for ‘x’ and ‘y’ where x+y =a-b, ax-by=a2+b2 [2]

9. Draw graphs of the equations on the same graph paper 2x+3y=12, x-y=1. Find the

Area and co-ordinate of the vertices of the triangle formed by the two straight lines

and the y-axis.

[3]

10. Solve

2 3 17

3 2 3 2 5x y x y+ =

+ − and

5 12

3 2 3 2x y x y+ =

+ −

[3]

11. The sum of a two digit number and the number obtained by reversing the order of

digits is 99. If the digits differ by 3, find the number.

[3]

12. In a cyclic quadrilateral ABCD, ( )2 4A x∠ = + ° , ( )3 ,B y∠ = + ° ( )2 10C y∠ = + ° and

( )4 5D x∠ = − ° Find the four angles

[3]

33

CBSE TEST PAPER-02

CLASS - X Mathematics (Pair of Linear Equation)

[ANSWERS]

Ans01. (d)

Ans02. (a)

Ans03. (a)

Ans04. (c)

Ans05. 2 4 0x y+ − =2 4 12 0x y+ − =

4 2when x y= −

x 4 2 y 0 1

12 4

2

ywhen x

−=

x 6 2 y 0 2

Ans06. 1 1 1

2 2 2

a b c

a b c= ≠

( )

2

2

2

3 3. .

123

36 12

or 6

3 33 3

or 6

or 0

i e

If

i

If

α αα α

α αα

αα α α α

α αα αα

−= ≠

= ⇒ =

= ± →−≠ ⇒ − ≠

≠= ( )

( ) ( )and 6

from eq and

The value of is 6.

ii

i ii

α

α

= →

∴

0

1

2

1 2 3 4 5 6x

y

34

Ans07. Is Infinite number of number of sol.

1 1 1

2 2 2

a b c

a b c= =

2 3 7or

1 2 2 4 4 12 3

If 1 2 2

1

3 7 and if

2 2 4 4 1 5 2 11

on solving we get,

3 and 2

a b a b a b

a b a ba b

a b a ba b

a b

= =+ + + + + +

=+ + + +

⇒ − =

=+ + + +

⇒ − =

= =

Ans08. x y a b+ = −2 2and + ax by a b− =

( ) ( )

2

2 2

bx by ab b

ax by a b

a b x a a b

+ = −− = ++ = ++

x a

x y a b

a y a b

y b

=+ = −+ = −

= −

Ans09. 2 3 12x y+ =1x y− =

12 3

2

ywhen x

−=

x 6 0 y 0 4

1when x y= +x 1 2 y 0 1

0

1

2

1 2 3 4 5 6x

y

3

4

35

1

21

5 225 square unit

A b h= ×

= × ×

=

Ans10. 1

Let 3 2

ux y

=+1

and 3 2

17 2 3

5and 5 2

1on solving we get

5and 1

vx y

u v

u v

u

v

=−

⇒ + =

+ =

=

= 3 2 5

3 2 1

on solving we get

1 and 1

x y

x y

x y

∴ + =− =

= =

Ans11. Let the digit at unit place be ' ' and tens place be ' '.x y

( ) ( )( )( )


10 10 99

or 9

and 3

or 3

y x x y

x y i

x y ii

y x

+ + + =

+ = →

− = →

− = ( )( ) ( )

( ) ( )

on solving eq and we get

6 and 3

then the original number is 36.


3 and 6

The number is '63'

iii

i ii

x y

i iii

x y

→

= =

= =∴

36

Ans12. In cyclic quadrilateral

( )

( )( ) ( )

180 and 180

2 4 2 10 180

83

and

180

3 4 5 180

4 182


33 an

A C B D

x y

x y i

B D

y x

x y ii

i ii

x

∠ + ∠ = ∠ + ∠ =+ + + =

+ = →

∠ + ∠ =+ + − =

+ = →

= d 50

angles are

2 4 2 33 4 70

3 50 3 53

2 10 2 50 10 110

4 5 4 33 5 127

o

o

o

o

y

A x

B y

C y

D x

=∴∠ = + = × + =∠ = + = + =∠ = + = × + =∠ = − = × − =

37

CBSE TEST PAPER-03

CLASS -X Mathematics (Pair of Linear Equation)

1. The value of ‘k’ for which the system of equation kx-y = 2and 6x-2y=3 has a unique

solution is

(a) k=3 (b) k ≠ 3

(c) k=0 (d) k ≠ 0

[1]

2. The value of ‘k’ for which the system of equations x+2y =5 and 3x+ky+15=0 has no

solutions if

(a) k=6 (b) k=-6

(c) –k=3

2 (d) None of these

[1]

3. In the equation a1x+b1y+c1=0 and a2x+b2y+c2=0 if 1 1

2 2

a b

a b≠ then the equation will

represents

(a) coincident liner (b) parallel lines

(c) intersecting liner (d) None of these

[1]

4. Solve Graphically 2x-3y+13=0 and 3x-2y+12=0 [1]

5. Find the values of α and β for which the following system of linear equation has

infinite number of solution, 2x+3y=7, 2α x+ ( )α β+ y = 28

[2]

6. Find the condition for which the system of equations

x yc

a b+ = and bx+ay = 4ab

(a,b ≠ 0) is inconsistent

[2]

38

7. Find the value of ' 'α so that the following linear equation have no solution

( ) ( ) ( )23 1 3 2 0, 1 2 5 0x y x yα α α+ + − = + + − − =

[2]

8. Solve for x and y ax+by=a-b and bx-ay=a+b [2]

9. Draw the graph of x+2y-7=0 and 2x-y-4=0

Shade the area bounded by there line and y-axis.

[3]

10. A two digit number is obtained by either multiplying the sum of the digits by 8 and

adding 1. Or by multiplying number How many such numbers are these?

[3]

11. A leading library has a fixed charge for the fist three days and an additional change

for each day thereafter sarika paid Rs 27 for a book kept for seven days

While Sury paid Rs 21 for the book she kept for five days find the fixed change and

the charge for each extra day.

[3]

12. If 2 is added to the numerator of a fraction, it reduces to

1

2 and if 1 is subtracted

from the denominator, it reduces to 1

3. Find the fraction.

[3]

13. Abdul travelled 300km by train and 200km by Taxi, it took him 5 hours 30

minutes. But if he travels 260 km by Train and 240 km by Taxi he takes 6 minute

longer, Find the speed of the train and that of the taxi.

[5]

39

CBSE TEST PAPER-03

CLASS - Mathematics (Pair of Linear Equation)

[ANSWERS]

Ans01. (b)

Ans02. (a)

Ans03. (c)

Ans04. 2 3 13 0x y− + =3 2 12 0x y− + =

13 3

2

ywhen x

+=

x 13/2 5 y 0 -1

3 12

2

xwhen y

+=

x 0 -3 y 6 3

Ans05. ( )1 1 1

2 2 2

Infinite solutiona b c

a b c= =

2 3 7

2 28

4, and 8

α α βα β

−= =+ −

⇒ = =

Ans06. Inconsistent

1 1 1

2 2 2

1/ 1/

41 1

i.e. 4

or 4

a b c

a b c

a b c

b a abc

ab ab abc

= ≠

= ≠

= ≠

≠

0

1

2

1 2 3 4 5 6

x

y

3

4

5

6

-1

-3

-2

-3 -2 -1

40

Ans07. No solution

1 1 1

2 2 2

2

3 1 3 2i.e.

1 2 5

a b c

a b c

αα α

= ≠

+ −= ≠+ − −

2 2 3 6 2 3 3

5 5

or 1

3 2or

2 519

2

α α α ααα

α

α

− + − = +− =

= −

≠−

⇒ ≠

Ans08. ]ax by a b a+ = − × ]bx ay a b b− = + ×

( )

2 2

2

2 2 2 2

a x aby a ab

bx aby ab b

a b x a b

+ = −− = +

+ = +

1

1

1

1

x

a by a b

by b

y

x

y

⇒ =∴ + = −

= −= −

∴ == −

Ans09. 2 7 0x y+ − =2 4 0x y− − =

7 2when x y= −

x 5 1 y 1 3

2 4when y x= −x 2 3 y 0 2

0

1

2

1 2 3 4 5 6

x

y

3

4

5

6

-1

-3

-2

-3 -2 -1

41

1

21

1 221 square unit

A b alt= ×

= × ×

=

Ans10. Let digit at unit place and tens place then x y

( )

( )( )

( )

original number 10


10 8 1

or 7 2 1 0

or 13 2 10

12 23 2 0

y x

y x x y

x y i

x y y x

x y

= +

+ = + +

− + = →

− + = +

− + = → ( )( ) ( )on solving eq and we get,

2

16

ii

i ii

yy

=

( )( )

( ) ( )

which is not possible

13 3 2 10

or 14 3 2


1, 4

original number 41

only one number

y y x

x y iii

i ii

x y

∴ − + = +

− = →

= =∴ =

exist.

Ans11. Let the fixed change be Rs and additional charge be Rs .x y

( )( )

( )( )


7 3 27

or 4 27

and 5 3 21

2 21

on solving eq

x y

x y i

x y

x y ii

+ − =

+ = →

+ − =

+ = →

( ) ( ) and we get

15, 3

i ii

x y= =

42

Ans12. Let the fraction be x

y

( )

( )( ) ( )


2 1

2

or 2 4

1and

1 3

or 3 1


x

y

x y i

x

y

x y ii

i ii

+ =

− = − →

=−− = − →

3, 10

3 fraction is

10

x y= =

∴

Ans13. Let the speed of the train be and taxi be km/hx y

ATQ

( )

( )

( )

300 200 15

2

300 200 11or

2

260 240 11 1and

2 10

260 240 56

10

1 1Let and

11 300 200

2

and

x y

ix y

x y

iix y

u vx y

u v iii

+ =

+ = →

+ = +

+ = →

= =

∴ + = →

( )56 260 240

10on solving eq (iii) and (iv) we get,

1 1000 /

1001

80 /80

u v iv

u x km h

v y km h

+ = →

= ⇒ =

= ⇒ =

43

CBSE TEST PAPER-04

CLASS – X Mathematics (Pair of Linear Equation)

1. If am ≠ bl, then the system of equation ax + by = c and lx + my = n

(a) Has a unique solution (b) Has no solution

(c) Infinitely many solution (d) May or may not have a solution.

[1]

2. The value of ‘k’ for which the system of equation. 3x+5y=0 and kx +10y = 0 has a

non – zero solution is

(a) k = 0 (b) k = 2

(c) k = 6 (d) k = 8

[1]

3. If a paired linear equation a1x+b1y+c1=0 and a2x+b2y+c2=0 represents parallel

liner then

(a) 1 1

2 2

a b

a b≠ (b) 1 1 1

2 2 2

a b c

a b c= ≠

(c) 1 1 1

2 2 2

a b c

a b c= = (d) None of these

[1]

4. The graphical representation of the linear equation y-5=0 is

(a) A line (b) A point

(c) A curve (d) None of these

[1]

5. Given the linear equation 2x+3y-8=0 write another linear equation in two variable

such that the geometrical representation of the pair so formed is

(a) intersecting lines

(b) Parallel lines

(c) Overlapping

[2]

6. Find the value of ‘k’ for which the system of equation has infinitely many solutions [2]

44

2x+(k-2)y=k and 6x+(2k-1)y=2k+5

7. Find the relation between a, b, c and d for which the equations ax+by=c and

cx+dy=a have a unique solution

[2]

8. Solve for ‘x’ and ‘y’

(a-b)x+(a+b)y = a2-b2-2ab

(a+b) (x+y)=a2+b2

[2]

9. Determine graphically the coordinates of the vertices of the triangle the equation

of whose sides are y=x, 3y=x, x+y=8.

[3]

10. Father’s age is three times the sum of ages of his two children. After 5 years his age

will be twice the sum of ages of two children. Find the age of father.

[3]

11. On selling a T.V. at 5% gain and a fridge at 10% gain shop keeper gains Rs 2000.

But if he sells the T.V at 10% gain and the Fridge at 5% loss, he gains Rs 1500 on

the transaction. Find the actual Price of TV and Fridges.

[3]

12. A taken 3 hours more than B to walk a distance of 30km. But if A doubles his

speed, he is ahead of B by 1

12

hours. Find their original speed.

[3]

13. If in a rectangle the length is increased and breadth is decreased by 2 units each,

The area is reduced by 28 square units, if the length is reduced by 1 unit and

breadth is increased by 2 units, the Area increased by 33 sq units. Find the

dimensions of the rectangle.

[5]

45

CBSE TEST PAPER-04

CLASS – X Mathematics (Pair of Linear Equation)

[ANSWERS]

Ans01. (a)

Ans02. (c)

Ans03. (b)

Ans04. (a)

Ans05. 2x+3y-8=0 another linear equation representing.

(i) Intersecting lines is 3 8

(ii) Parallel lines is 4 6 3

(iii) Overlapping lines is 6 9 24

x y

x y

x y

+ =+ =

+ =

Ans06. for infinitely many solution

1 1 1

2 2 2

2 2

2 2i.e.

6 2 1 2 51 2

if 3 2 1

2 1 3 6

5

1 or if

2 5 3or 3 2 5

5

2if

2 1 2 5

2 5 4 10 2

a b c

a b c

k k

k kk

kk k

k

k

kk k

k

k k

k k

k k k k k

= =

−= =− +

−=−

− = −=

=+

= +=

− =− +

⇒ + − − = − 2 10

5

k

k

==

Ans07. 1 1

2 2

i.e. a b a b

a b c d≠ ≠

or ad bc≠

46

Ans08.

( ) ( )( ) ( )

( )

2 2

2 2

2

2 2

a b x a b y a b ab

a b x a b y a b

bx b b a

− + + = − −

+ + + = +

− = − +---

( )( ) ( )( )

2 2

2 2 2 2

2

2

2

x a b

a b a b a b y a b ab

a b a b y a b ab

aby

a b

= +

∴ − + + + = − −

− + + = − −−=

+

Ans09. when y x=x 1 2 y 1 2

3when y x=x 6 3 y 2 1

8 or 8 -when x y y x+ = =x 4 5 y 4 3

Ans10. Let the presentage of father be years and sum of present age of two son's be years.x y

( )

( )( )

ATQ

after five years

5 2 5 5

5 2 20

2 15

and 3

3 - 2 15

or 15

age of fa

x y

x y

x y i

x y ii

y y

y

+ = + ++ = +− = →

= →∴ =

=∴ ther 3

3 15

45 years

x y== ×=

0

1

2

1 2 3 4 5 6

x

y

3

4

5

6

-1

-3

-2

-3 -2 -1

47

Ans11. Let the selling price of TV Rs x=

( )

and fridge Rs

ATQ

5% of 10% of 2000

or 200020 10 2 40000

and 10% of 5% of 1500

or

y

x y

x y

x y i

x y

=

+ =

+ =

+ = →+ =

( )

150010 20 2 30000

on solving eq (i) and (ii) we get,

20000 50000 Rs ; Rs

3 3

x y

x y ii

x y

+ =

+ = →

= =

Ans12. Let the original speed of A and B are km/h and km/h respectively.x y

( )

( )

ATQ

30 303

1 1 1

10

1

1030 30 3

2 2

1 1 1

2 20

1

2 20 on adding (i) and (ii) we get,

x y

orx y

or u v i

andy x

y x

uv ii

− =

− =

− = →

− =

− =

− = →

4 3 1 10

2 20 310

or 3

5

speed of A is 3.3 km/h

u

x

and y

= ⇒ =

=

=∴

B is 2 km/hand

48

Ans13. Let the lenght and breadth of a rectangle be and meter.x y

( )( )

( )( )( )

ATQ

2 2 28

2 2 24

12

and 1 2 33

2 33

Area xy

x y xy

or x y

or x y i

x y xy

x y

=+ − = −

− =− = →

− + = +

− = ( ) on subtracting eq (ii) (i) we get,

21

and

21 12

21 12

9

length 21

breadth 9

ii

x

y

so y

y

m

m

→−

=

− == −=

∴ ==

49

CBSE TEST PAPER-01

CLASS-X Mathematics (Triangles)

1. In the fig , .ABC EDC∆ ∆ if we have AB = 4cm, ED 3cm CE = 4.2 cm

and CD = 4.8cm, then the values of CA and CB are

(a) 6cm, 6.4 cm (b) 4.8cm, 6.4cm

(c) 5.4cm, 6.4cm (d) 5.6, 6.4cm

[1]

2. The areas of two similar triangles are respectively 29cm and

216cm . Then ratio of

the corresponding sides are

(a) 3:4 (b) 4:3 (c) 2:3 (d) 4:5

[1]

3. Two isosceles triangles have equal angles and their areas are in the ratio 16:25.

Then the ratio of their corresponding heights is

(a) 4

5(b)

5

4(c)

3

6(d)

5

7

[1]

4. If ABC DEF∆ ∆∼ and 5 ,AB cm= area ( ) 220ABC cm∆ = area ( ) 245 ,DEF cm∆ = then

DE =

(a) 4

5cm (b) 7.5cm (c) 8.5cm (d) 7.2cm

[1]

5. In the given Figures, , 125ODC OBA BOC∆ ∆ ∠ = °∼ and

70 .CDO∠ = ° Find

(i) DOC∠ (ii) DCO∠ (iii) OAB∠ (iv) AOB∠ (v) OBA∠

[2]

6. (a) ABC DEF∆ ∆∼ and their areas are respectively 64

cm2 and 121cm2. If EF = 15.4cm, find BC

[2]

7. ABC is an isosceles right triangle right angled at C. Prove that

2 22AB AC=

[2]

A B

D C

O125

0

700

A

B C

50

8. In the figure DE||AC and

BE BC

EC CP= prove that DC||AP

[2]

9. In the given figure,

QT QR

PR QS= and 1 2.∠ = ∠ Prove that

PQS TQR∆ ∆∼

[3]

10. In the given figure PA, QB and RC are each perpendicular to AC.

Prove that 1 1 1

2x y+ =

[3]

11. In the given figure DE||BC and AD:DB = 5:4 find

( )( )

area DFE

area CFB

∆∆

[3]

12. Determine the length of an altitude of an equilateral triangle of side

‘2a’ cm

[3]

13. Prove that if a line in drawn parallel to one side of a triangle to intersect the other

two sides in district points other two sides are divided in the same ratio. By using

this theorem prove that in ABC∆ if ||DE BC then AD AE

BD AC=

[5]

51

CBSE TEST PAPER-01

CLASS - Mathematics (Triangles)

[ANSWERS]

Ans1. (D)

Ans2. (D)

Ans3. (A)

Ans4. (B)

Ans5. (i) 180 125 55DOC∠ = ° − ° = °

(ii) ( )180 70 55 [ is a st. line and OC stands on it]DCO DOB∠ = ° − ° + ° ∵

180 125 55 [ sum of angles of a tringle = 180 ]= ° − ° = ° °∵

(iii) 55DAB DCO∠ = ∠ = ° (given)

, ,

ODC OBA

DOC AOB ODC OBA DCO OAB

∆ ∴∠ = ∠ ∠ = ∠ ∠ = ∠

∵ ∼

(iv) 55AOB DOC∠ = ∠ = °

(v) 70OBA ODC∠ = ∠ = °

Ans6. Since ( )( )

2

2

area

area

ABC BCABC DEF

DEF EF

∆∆ ∆ ∴ =

∆∼

[∵the ratio of the areas of two similar triangles is equal to the ratio of the

squares of the corresponding sides]

( )2

22

64 64 154 154 64 14 14

121 121 10 10 10015.4

BCBC

× × × ×⇒ = ⇒ = =

× ×

8 14

10BC

×⇒ =

=11.2cm

Ans7. In rt. , . ABC rt A at C∆ ∠2 2 2AB AC BC= + [By Pythagoras theorem]

( )2 2 22 [ ]AC AC AC BC AC given= + = =∵

2 22AB AC= =

52

Ans8. In ,ABC DE AC∆ �

.......( )BD BE

iDA EC

∴ = [By Thale’s Theorem]

Also ( ).........( )BE BC

given iiEC CP

=

∴from (i) and (ii) we get

BD BCDC AP

DA CP= ∴ � [By the converse of Thale’s Theorem]

Ans9. Since [ ]QT QR

GivenPR QS

=

.......( )QT PR

iQR QS

∴ =

Since 1 2 [Given]∠ = ∠

........( )PQ PR ii=

[In PQR∆ sides opposites to opposite angles are equal]

........( ) [Form( )and( )]QT PQ

iii i iiQR QS

∴ =

Now in PQS∆ and TQR

From (iii) i.e.PQ QT PQ QS

QS QR QT QR= =

And Q Q∠ = ∠ [Common]

PQS TQR∴∆ ∆∼ [By S.A.S. Rule of similarity]

Ans10. In PAC∆ and QBC∆

[Each 90 ]PAC QBC∠ = ∠ = ° [Common]PCA QCB∠ = ∠

PAC QBC∴∆ − ∆

i.e. ........( )x AC y BC

iy BC x AC

= =

Similarly i.e. .......( )

z AC y ABii

y AB z AC= =

Adding (i) and (ii), we get

1 1BC AB y yy

AC x z x z

+ ⇒ = + = +

53

1 1 1 11

1 1 1

ACy

AC x z x z

y x z

⇒ = + ⇒ = +

⇒ = +

Ans11. In ADE∆

and ABC∆

1 1∠ = ∠ [Common]

2 ACB∠ = ∠ [Corresponding s∠ ]

ADE ABC∴∆ ∆∼ [By A.A Rule]

........( )DE AD

iBC AB

∴ =

Again in DEF∆ and CFB∆3 6∠ = ∠ [Alternate s∠ ]

4 5∠ = ∠ [Vertically opposite s∠ ]

DFE CFB∴∆ ∆∼ [By A.A Rule]

( )( )

22

2

Area DFE DE AD

area CFB BC AB

∆ ∴ = = ∆ [From (i)]

25 5 5 5

9 4 5 4 9

AD AD AD

DB AD DB DB = = ⇒ = ⇒ = + +

∵

( )( )

25

81

area DFE

area CFB

∆∴ =

∆

Ans12. In right triangles ADB∆ and ADC∆AB AC

AD AD

==

( ) Each 90ADB ADC∴∠ = ∠ = °

( ) . .ADB ADC R H S∴∆ ≅ ∆

( ) . . . .BD DC c p c t∴ =

[ ] 2BD DC a BC a∴ = = =∵

In rt. 2 2 2,ADB AD BD AB∆ + = (By Pythagoras Theorem)

( )22 2

2 2 2 2

2

4 3

3

AD a a

AD a a a

AD acm

⇒ + =

⇒ = − =

⇒ =

54

Ans13. Given: In ABC∆ DE BC� intersect AB at D and AC at E.

To Prove: AD AE

DB EC=

Construction: Draw EF AB⊥ and DG AC⊥ and join DC and BE.

Proof: 1

2ar ADE AD EF∆ = ×

1

2ar DBE DB EF∆ = ×

12 .......( )12

AD EFar ADE ADi

ar DBE DBDB EF

×∆∴ = =∆ ×

Similarly :

12 .......( )12

AE DGar ADE AEii

ar DEC ECEC DG

×∆ = =∆ ×

Since DBE∆ and DEC∆ are on the some base and between the same parallels

( ) ( )ar DBE ar DEC∴ ∆ = ∆

( ) ( )1 1

ar DBE ar DEC⇒ =

∆ ∆ar ADE ar ADE

ar DBF ar DFCAD AB

DB EC

∆ ∆∴ =∆ ∆

⇒ =

II Part DE BC∵ �

AD AE

DB EC=

AD AE p r p r

AD DB AE EC q s p q r s

⇒ = = ⇒ = + + + +

∵

AD AE

AB AC⇒ =

A

BC

D

F

E

G

B C

DE

A

55

CBSE TEST PAPER-02


1. A man goes 15 m due west and then 8m due north. Find distance from the starting

point.

(A) 17m (B) 18m

(C) 16m (D) 7m

[1]

2. In a triangle ABC, if AB = 12cm BC = 16cm, CA = 20cm, then ABC∆ is

(A) Acute angled (b) Right angled

(C) Isosceles triangle (d) equilateral triangle

[1]

3. In an isosceles triangle ABC, AB=AC=25cm and BC = 14cm Then altitude from A on

BC =

(a) 20 cm (b) 24cm

(c) 12cm (d) None of these

[1]

4. The side of square who’s diagonal is 16cm

(a) 16cm (b) 8 2cm

(c) 5 2 (d) None of these

[1]

5. The hypotenuse of a right triangle is 6m more than the twice of the shortest side. If

the third side is 2m le3ss than the hypotenuse. Find the side of the triangle

[2]

6. PQR is a right triangle right angled at P and M is a point on QR such that PM ⊥ QR.

Show that 2 .PM QM MR=

[2]

56

7. In the gives Fig ||DE OQ and || ,DF OR Prove that ||EF OQ [2]

8. In Fig DE||BC, Find EC [2]

9. In the given Fig, if 1 2∠ = ∠ and .NSQ MTR∆ ≅ ∆ Then prove that PTS PRQ∆ ∆∼ [3]

10. In the given fig. the line segment XY||AC and XY divides

triangular region ABC into two points equal in area,

Determine AX

AB

[3]

11. BL and CM are medians of ABC∆ right angled at A. prove that

( )2 2 24 5BL CM BC+ =

[3]

12. ABC is a right triangle right angled at C. Let BC – a, CA = b, AB = C and let P be the

length of perpendicular from C on AB prove that

(i) cp = ab (ii) 2 2 2

1 1 1

P a b= +

[3]

13. Prove that the ratio of areas of two similar triangles are in the ratio of the squares

of the corresponding sides. By using the above theorem solve In two similar

triangles PQR and LMN, QR = 15cm and MN = 10 Find the ratio of areas of two

triangles.

[5]

57

CBSE TEST PAPER-02


[ANSWERS]

Ans1. (A)

Ans2. (B)

Ans3. (B)

Ans4. (B)

Ans5. Let shortest side be xm in length

Then hypotenuse ( )2 6x m= +

And third side = ( )2 4x m+

We have

( ) ( )2 22

2 2 2

2 6 2 4

4 24 36 4 16 16

2 8 20 0

10 or 2

x x x

x x x x x

x x

x x

+ = + +

⇒ + + = + + +⇒ − − =⇒ = = −

10x⇒ =Hence the sides of triangle are 10m, 26m and 24m

Ans6. PQR∵ is a right triangle right angle at P and PM QR⊥PMR PMQ

PR PM MR

PQ QM PM

PM MR

QM PM

∴∆ ∆

∴ = =

⇒ =

∼

i. e. 2 .PM QM MR=

Ans7. In , ||OQP DE OQ∆

.........( )PE PD

iEQ DO

=

In DF OROPR∆ �

........( )PD PF

iiDO FR

=

P Q

R

M

P

Q R

D

FO

E

58

From (i) and (ii) we get

PE PF

EQ FR=

∴From PQR∆EF QR�

Ans8 DE BC∵ �

AD AE

DB EC∴ =

1.5 1

3 EC⇒ =

2EC cm∴ =

Ans9 Since NSQ MTR∆ ≅ ∆SQN TRM∴∠ = ∠Q R⇒∠ = ∠ in PQR∆

190

2P= ° − ∠

Again 1 2∠ = ∠ [given in PST∆ ]

( )11 2 180

2P∴∠ = ∠ = ° − ∠

190

2P= ° − ∠

Thus in PTS∆ and PRQ∆1

1 902

Q Each P ∠ = ∠ = ° − ∠

2 , R P P∠ = ∠ ∠ = ∠ (Common)

PTS PRQ∆ ∆∼

Ans10 Since XY AC�

BXY BAC

BYX BCA

∴∠ = ∠∠ = ∠[Corresponding angles]

BXY BAC∴∆ ∆∼ [A.A. similar ]

( )( )

2

2

ar BXY BX

ar BAC BA

∆∴ =

∆

But ar ( ) ( )BXY ar XYCA∆ =

( ) ( ) ( )2 BXY ar BXY ar XYCA∴ ∆ = ∆ +

59

( )ar BAC= ∆

( )( )

1

2

ar BXY

ar BAC

∆∴ =

∆2

2

1

21

2

BX

BABX

BA

∴ =

=

2 1

2

2 1

2

2 2

2

BA BX

BA

AX

AB

− −∴ =

−=

−=

Ans11 BL and CM are medians of a ABC∆ in which 90A∠ = °From 2 2 2 ........( )ABC BC AB AC i∆ = +

From right angled 2AB∆2 2 2BL AL AB= +

i.e.

22 2

2

ACBL AB

= +

2 2 24 4 ..........( )BL AC AB ii⇒ = +From right angle CMA∆

2 2 2CM AC AM= +

i.e.

22 2

2

ABCM AC = +

[Mis mid point]

22 2

2 2 2

44 4 .........( )

ABCM AC

CM AC AB iii

⇒ = +

⇒ = +Adding (ii) and (iii) we get [From (i)]

i.e. ( )2 2 24 5BL CM BC+ =

Ans12(i) Draw CD AB⊥Then CD P=

Now ar of ( )1

2ABC BC CA∆ = ×

1

2ab=

60

Also area of 1

2ABC AB CD∆ = ×

1

2CP=

Then 1 1

2 2ab CP=

CP ab=

(ii) Since ABC∆ is a right triangle with 90C∠ = °2 2 2

2 2 2

22 2

AB BC AC

C a b

aba b

P

∴ = +⇒ = +

⇒ = +

CP ab

abc

P

∴ =

⇒ =

2 2

2 2 2

2 2 2

1

1 1 1

a b

P a b

P b a

+⇒ =

⇒ = +

Thus 2 2 2

1 1 1

P a b= +

Ans13 Given: Two triangles ABC and DEF

Such that ABC DEF∆ ∆∼

To Prove: ( )( )

2 2 2

2 2 2

ar ABC AB BC AC

ar DEF DE EF DF

∆= = =

∆

Construction: Draw AL BC⊥ and DM EF⊥

Poof: ( )( )

( )( )

( )( )

12

12

BC ALar ABC

ar DEF EF DM

∆=

∆

1 of

2ar b h ∆ = × ∵

( )( ) ........( )

Area ABC BC ALi

Area DEF EF DM

∆⇒ = ×

∆

Again, in ALB∆ and DME∆ we have

[ ]90ALB DME Each∠ = ∠ = °

61

ABC DEFABL DEM

B E

∆ ∆ ∠ = ∠ ∴∠ = ∠

∵ ∼

ALB DME∴∆ ∆∼ [By AA rule]

AB AL

DE DM∴ = [∵Corresponding sides of similar triangles are proportional]

Further, ABC DEF∆ ∆∼

........( )AB BC AC

iiiDE EF DF

∴ = =

From (ii) and (iii)

BC AL

EF DM=

Putting in (i) we get

( )( )

Area ABC Al AL

Area DEF DM DM

∆= ×

∆2 2

2 2

2

2

AL AB

DM DE

AC

DF

= =

=

Hence ( )( )

2 2 2

2 2 2

ar ABC AB BC AC

ar DEF DE EF DF

∆= = =

∆

II Part: Since PQR LMN∆ ∆∼

( )( )

( )( )

22

22

15

10

225 9

100 4

ar PQR QR

ar LMN MN

∆∴ = =

∆

= =

Hence required ratio 9:4

62

CBSE TEST PAPER-03


1. In an isosceles triangle ABC If AC = BC and 2 22AB AC= then C∠ =

(a) 45° (b) 60° (c) 90° (d) 30°

[1]

2. If ABC EDF∆ ∆∼ and ABC∆ is not similar to DEF∆ then which of the following is

not true?

(a) . .BC EF AC FD= (b) . .AB EF AC DE=

(c) . .BC DE AB EF= (d) . .BC DE AB FD=

[1]

3. A certain right angled triangle has its area numerically equal to its perimeter. The

length of each side is an even integer what is the perimeter?

(a) 24 units (b) 36 units

(c) 32 units (d) 30 units

[1]

4. In the given fig. it AB ||CD, then x =

(a) 3

(b) 4

(c) 5

(d) 6

[1]

5. In the given fig, ABC and AMP are two right triangles, right angled at B and M

respectively prove that

( )

( )

i ABC AMP

CA BCii

PA MP

∆ ∆

=

∼

[2]

6. In the given fig OA.OB=OC.OD or ,

OA OD

OC OB= Prove

that A C∠ = ∠ and B D∠ = ∠

[2]

2x+1

63

7. In the given fig DE||BC and AD=1cm BD = 2cm what is the

ratio of the area of ABC∆ to the area of ?ADE∆

[2]

8. A right triangles has hypotenuse of length P cm and one side of length q cm. it p-q

=1. Find the length of third side of the triangle

[2]

9. In fig a triangle ABC is right angled at B. side BC is trisected at paints D and E prove

that 82 2 23 5AE AC AD= +

[3]

10. In fig DEFG is a square and 90BAC∠ = ° show that

2DE BD EC= ×

[3]

11. In a quadrilateral ABCD P,Q,R,S are the mid points of the sides AB, BC, CD and DA

respectively. Prove that PQRS is a parallelogram

[3]

12. Triangle ABC is right angled at C and CD is perpendicular to AB prove that

2 2BC AD AC BD× = ×

[3]

13. Prove that in a right triangle the square of the

hypotenuse is equal to the sum of the squares of the other

two sides use the above theorem in fig. to prove that

2 2 2 2 .PR PQ QR QM QR= + −

[5]

64

CBSE TEST PAPER-04


1. Length of an altitude of an equilateral triangle of side ‘2a’cm is

(a) 3a cm (b) 3 a cm

(c) 3

2

a cm (d) 2 3 a cm

[1]

2. If in two triangles ABC and PQR

AB BC CA

QR PR PQ= =

(a) PQR CAB∆ ∆∼ (b) PQR ABC∆ ∆∼

(c) CBA PQR∆ ∆∼ (d) BCA PQR∆ ∆∼

[1]

3. The area of two similar triangles are 281cm and 249cm respectively. It the altitude

of the bigger triangle is 4.5cm. then the corresponding altitude of the smaller

triangle is

(a) 2.5cm (b) 2.8cm

(c) 3.5cm (d) 3.7cm

[1]

4. In a right angled triangle if base and perpendicular are respectively 36015 cm and

48020cm then the hypotenuse is

(a) 69125 cm (b) 60025cm

(c) 391025cm (d) 60125 cm

[1]

5. The length of the diagonals of a rhombus are 24 cm and 10cm. find each side of

rhombus

[2]

6. In an isosceles right angled triangle prove that hypotenuse is 2 times the side of

a triangle

[2]

7. In fig express x in terms of a, b, c [2]

65

8. The perimeter of two similar triangle ABC and PQR are respectively 36cm and

24cm. if PQ=10cm Find AB

[2]

9. Triangle ABC is right angled at C and CD is perpendicular to AB. Prove that

2 2BC AD AC BD× = ×

[3]

10. In fig ABC and DBC are two triangles as the some base

BC. If AD intersect EC at O. prove that

( )( )

ar ABC AO

ar DBC DO

∆=

∆

[3]

11. In fig. ABC is a right triangle right angled at B. medians AD

and CE are of respective lengths 5 cm and 2 5cm find length

of AC

[3]

12. In the given fig

QR QT

QS PR= and 1 2∠ = ∠ show that

PQS TQR∆ ∆∼

[3]

13. Prove that the ratio of areas of two similar Triangles is equal to the square of their

corresponding sides using the above theorem do the following the area of two

similar ∆ are 281cm and

2144cm if the largest side of the smaller triangle is 27 cm.

then find the largest side of the largest triangle

[5]

A

BCD

E

66

CBSE TEST PAPER-04


[ANSWERS]

Ans1. (B)

Ans2. (A)

Ans3. (C)

Ans4. (B)

Ans5. 24 12AC AO cm= ∴ =

10 5BD cm OD cm= ∴ =

From right AOD∆

2 2 2

2 2

2

12 5

169

13

AD AO OD

AD

AD cm

= += +

==

Hence each side = 13cm

Ans6. Let hypotenuse of right angle unitsh∆ = and equal sides of unitsx∆

∴By Pythagoras theorem

2 2 2

2 22

2

h x x

h x

h x

= +

=

=

Ans7. AB OCD∆ ∆∼

( )

x x b

a cx ax ab

abx c a ab x

c a

+⇒ =

= +

− = ⇒ =−

67

Ans8. ABC PQR∆ ∆∼

AB BC AC

PQ QR PR∴ = =

perimetre of

perimetre of

36

10 2436 10

1524

AB BC AC ABC

PQ QR PR PQR

AB

AB cm

+ + ∆= =+ + ∆

=

×= =

Ans9. Given: ABC∆ right angle at C and CD AB⊥

TO Prove: 2 2BC AD AC BD× = ×

Proof: consider ACD∆ and DCB∆

Let A x∠ =

Then [ ]90 is right angledB x ACB∠ = − ∆∵

[ ] is right angledDCB x CDB⇒ ∠ = ∆∵

In and ADC CDB∆ ∆

[ ]90 eachADC CDB∠ = ∠ °

[ ][ ]

( )( )

2

2

from above

AA similarity

A DCB x

ACD CBD

ar ACD AC

ar VBD BC

∠ = ∠ =

∴∆ ∆

∆⇒ =

∆

∼

2

2

2

2

2 2

1212

.

AD CD AC

BCBD CD

AD AC

BD BC

BC AD AC BD

× ×⇒ =

× ×

⇒ =

⇒ × =

68

Ans10. Given ABC and DBC are two triangles on the same base BC but on the opposite

sides of BC, AD intersects BC at O

Construction : Draw AL BC⊥ and DM BC⊥

To Prove: ( )( )

ar ABC AO

ar DBC EO

∆=

∆

Proof: In and ALO DMO∆ ∆

[ ]90ALO DMO each∠ = ∠ °

[ ]Vertically opp. anglesAOL DOM∠ = ∠

[ ]

( )( )

By AA similarilyALO DMO

AL AO

DM DOar ABC AO

ar DBC DO

∴∆ ∆

⇒ =

∆∴ =

∆

∼

Ans11. Given: ABC∆ with 90B∠ = ° AD and CE are medians

To Find: Length of AC

Proof: In ABD∆ right angled at B

[ ]2 2 2

22

2 2

By pythagoras theorem

1 1=

2 2

1

4

AD AB BD

AB BC BD BC

AB BC

= +

= +

= +

∵

2 2 24 4 ........( )AD AB BC i= +

In BCE∆ right angle at B

2 2 2CE BE BC= +

( )

22

2 2 2

2 2 2

2 2 2 2 2 2

2 2 2

1

2

1

44 4 .......( )

4 4 5 5 5

4 4 5

AB BC

CE AB BC

CE AB BC ii

AD CE AB BC AB BC

AD CE AC

= +

= +

= +

+ = + = +

+ =

69

Given that AD = 5 and 2 5CE =

( ) ( )22 2

2

2

2

4 5 4 2 5 5

100 80 5

180

5

36 6

AC

AC

AC

AC AC cm

+ =

⇒ + =

⇒ =

= ⇒ =

Ans12. Given: and 1 2QR QT

QS PR= ∠ = ∠

Proof: As 1 2∠ = ∠

[ ]......( ) side oppasite to equal angles are equalPQ PR i=

Also ( ).......( )QR QT

given iiQS PR

=

From (i) and (ii)QR QT

QS PQ⇒ =

In PQS∆ and TQR, We have

From (ii)QR QT QS QR

QS QP QT QP= = ⇒

Also [ ]PQS TQR common∠ = ∠

[ ]SAS similarityPQS TQR∴∆ ∆∼

Ans13. Given: Two triangles ABC and DEF such that ABC DEF∆ ∆∼

To Prove: ( )( )

2 2 2

2 2 2

ar ABC AB BC AC

ar DEF DE EF DF

∆= = =

∆

Construction: Draw AL BC⊥ and DM EF⊥

Proof:- Since similar triangles are equiangular and their corresponding sides are

proportional

, ,

ABC DEF

A D B E C F

∴∆ ∆⇒ ∠ = ∠ ∠ = ∠ ∠ = ∠

∼

And .......( )AB BC AC

iDE EF DF

= =

70

In ALB∆ and DMB∆

1 2 and B E∠ = ∠ ∠ = ∠

ALB DME⇒ ∆ ∆∼ [By AA similarity]

.........( )AL AB

iiDM DE

⇒ =

From (i) and (ii) we get

.......( )AB BC AC AL

iiiDE EF DF DM

= = =

Now ( )( )

( )

( )

1212

BC ALarea ABC

area DEF BF DM

×∆=

∆ ×

( )( )( )( )

2

2

Area ABC BC AL

Area DEF EF DM

Area ABC BC BC BC

Area DEF EF EF EF

∆⇒ = ×

∆

∆⇒ = × =

∆

Hence 2 2 2

2 2 2

Area ABC AB BC AC

Area DEF DE EF DF

∆ = = =∆

II Part:- Let the largest side of the largest triangle be x cm

Using above theorem

2

2

144 12

27 81 27 9

x x= ⇒ =

36x cm⇒ =

71

CBSE TEST PAPER-05


1. In fig DE||BC and AD =1cm, BD = 2m. the ratio of the area of ABC∆ to the area of

?ADE∆

(a) 9:1 (b) 1:9

(c) 3:1 (d) none of these

[1]

2. In the given fig. ABC PQR∆ ∆∼ Then the value of

x and y

(a) ( ) ( ), 6, 20x y = (b) ( )20,60

(c) ( ) ( ), 3,10x y =

(d) none of these

[1]

3. In fig P and Q are points on the sides AB and AC respectively of

ABC∆ such that AP = 3.5cm, AQ = 3cm and QC = 6cm. If PQ =

4.5cm, then BC is

(a) 12.5cm (b) 5.5cm

(c) 13.5cm (d) none of these

[1]

4. D,E,F are the mid-points of the sides AB, BC, and CA respectively of ABC∆ then

( )( )

ar DEF

ar ABC

∆∆

is

(a) 1:4 (b) 4:1

(c) 1:2 (d) none of these

[1]

5. In the given fig DE||BC. If

, 2, 2, 1AD x DB x AE x EC x= = − = + = − find the value of x

[2]

6. The hypotenuse of a right angled triangle is P cm and one of sides is q cm. if P =

q+1, find the third side in terms of q.

[2]

A B

C

D

E

72

7. In the given fig.

1

2

AO BO

OC OD= = and AB = 5cm find the value of DC.

[2]

8. In ,ABC AB AC∆ = and D is a point on side AC, Such that 2BC AC CD= × . Prove

that BD = BC

[2]

9. Given a triangle ABC. O is any Point inside the triangle ABC, X,Y,Z are points on OA,

OB and OC, such that XY||AB and XZ||AC show that YZ||AC

[3]

10. PQR is a right triangle right angled at Q. If QS = SR, show that 2 2 24 3PR PS PQ= − [3]

11. A ladder reaches a window which is 12m above the ground on one side of the

street. Keeping its foot at the same point, the ladder is turned to the other side of

the street to reach a window 9 m high. Find the width of the street if the length of

the ladder is 15m.

[3]

12. In fig 3

XP XQ

PY QZ= = if the area of XYZ∆ is 232cm then find the

area of the quadrilateral PYZQ

[3]

13. In a triangle if the square of one side is equal to the sum of the squares on the

other two sides. Prove that the angle apposite to the first side is a right angle use

the above theorem to find the measure of PKR∠ in fig.

[5]

24cm

26cm

6cmK

Q

P

X

R

73

CBSE TEST PAPER-05


[ANSWERS]

Ans1. (A)

Ans2. (B)

Ans3. (C)

Ans4. (A)

Ans5. In the given fig

2 2

2

2 1

4 4

DE BC

AD AE

DB ECx x

x x

x x x x

∴ =

+⇒ =

− −− = − ⇒ =

�

Ans6. Let third side be x cm

2 2 2.........( )

Also 1.......( )

P q x i

P q ii

∴ = += +

From (i) and (ii), we get

( )2 2 2 21 2 1

2 1

q q x x q

x g cm

+ = + ⇒ = +

⇒ = +

Ans7. In AOB∆ and COD∆

AOB COD∠ = ∠ [Vertically opposite angles]

AO BO AO OC

OC OD OB OD= ⇒ = [Given]

AOB COD∴∆ ∆∼ [By SAS similarity]

AO BO AB

CO DO CD∴ = =

74

1 1

2 2

1 5

210

AB AO BOis given

DC OC OD

DCDC cm

= = =

=

=

Ans8. Given: A ABC∆ in which AB = AC, D is a point on BC such that

To Prove: BD = BC

Proof: 2BC AC CD= × [given]

BC DC

AC BC⇒ =

In ABC∆ and BDC∆

BC DC

CA CB⇒ = and [ ]C C Common∠ = ∠

ABC BDC∴∆ ∆∼ [SAS similarity]

[ ]AB AC AC ACAB AC

BD BC BD BCBD BC

⇒ = ⇒ = =

⇒ =

∵

Ans9. Given: A ,ABC O∆ is a point inside , , ABC X Y and Z∆ are points on OA, OB and OC

respectively such that XY||AB and XZ||AB and XZ||AC

To show: YZ||BC

Proof: In ,OAB XY AB∆ �

........( )[ B.P.T]OX OY

i ByAX BY

=

In ,OAC XZ AC∆ �

.........( )[ B.P.T]OX OZ

ii ByAX CZ

∴ =

From (i) and (ii) we get ..........( )OY OZ

iiiBY CZ

=

Now in ( )( )OY OZ

OBC from iiiBY CZ

∆ =

YZ BC⇒ � [Converse of B.P.T]

B C

D

A

A

B C

X

YZ

O

75

Ans10. Given: PQR is a right Triangle, right angled at Q.

Also QS = SR

To Prove:- 2 2 24 3PR PS PQ= −

Proof:- In right angled triangle PQR right angled at Q.

2 2 2PR PQ QR= + [By Pythagoras theorem]

Also [ ]1

2QS QR QS QR= =∵

In right angled triangle PQS, right angled at Q.

2 2 2PS PQ QS= +

[ ]2

12 2 ( )

2PS PQ QR From ii

⇒ = +

2 2 24 4 ........( )PS PQ QR iii⇒ = +

From (i) and (iii) we get

2 2 2 24 4PR PQ PS PQ= + −

2 2 24 3PR PS PQ⇒ = −

Ans11. Let AB be the width of the street and C be the foot of ladder.

Let D and E be the windows at heights 12m and 9m respectively from the ground.

In CAD∆ , right angled at A, we have

2 2 2

2 2 2

2

15 12

225 144 81

9

CD AC AD

AC

AC

AC m

= +

= +⇒ = − =⇒ =

In ,CBE∆ right angled at B, we have

2 2 2

2 2 2

2

2

15 9

225 81

144

12

CE BC BE

BC

BC

BC

BC m

= +⇒ = +⇒ = −⇒ =⇒ =

Hence width of the street AB=AC+BC=9+12=21m

15m

76

Ans12 Given XP XQ

givenPY QZ

=

......( )PQ YZ i⇒ � [By converse of B.P.T]

In XPQ∆ and ,XYZ∆ we have

[ XPQ Y∠ = ∠ [From (i) corresponding angles]

X X∠ = ∠ [common]

XPQ XYZ∴∆ ∆∼ [By AA sibilating]

( )( )

2

2.........( )

ar XYZ XYi

ar XPQ XP

∆∴ =

∆

We have 1 1 4

13 3 1 3

PY PY PY XP

XP XP XP

+= ⇒ + = ⇒ =+

4

3

XY

XP⇒ =

Substituting in (i) we get

( )( )

( )

( )

2

2

4 16

3 9

32 16

9

32 918

16

ar XYZ

ar XPQ

ar XPQ

ar XPQ cm

∆ = = ∆

⇒ =

×= =

Area of quad. 232 18 14PYZQ cm= − =

Ans13 I part Given: A ABC∆ such that

2 2 2AC AB BC= +

To Prove triangle ABC is right angled at B

Construction: construct a triangle DEF such that

,DE AB EF BC= = and 90E = °

Proof: DEF∆∵ is a right angled ∆ with right angle at E [construction]

∴By Pythagoras theorem we have

[ ]2 2 2

2 2 2 and

DF DE EF

DF AB BC DE AB EF BC

= += + = =∵

77

2 2 2 2 2DF AC AB BC AC ⇒ = + = ∵

2 2 2 2 2DF AC AB BC AC

DF AC

⇒ = + =

⇒ =

∵

Thus In ABC∆ and DEF∆ we have

, and AB DE BC EF AC DF= = = [By Construction and (i)]

90

ABC DEF

B E

∴∆ ≅ ∆⇒ ∠ = ∠ = °

Hence ABC∆ is a right triangle, right angled other part.

In . 90QPR QPR∆ ∠ = °

2 2 224 26x⇒ + =

10 10x PR cm⇒ = ⇒ =

Now in 2 2 2 2 2 2, [ 10 8 6 ]PKR PR PK KR as∆ = + = +

PKR∴∆ is right angled at K

90PKR⇒ ∠ = °

78

CBSE TEST PAPER-01

CLASS – 10 Mathematics (Introduction to Trigonometry )

1. a sec2A – atan2A is equal to

(a) 9 (b) 1 (c) 2 (d) 3

[1]

2. 1+tan2A/1+cot2A is equal to

(a) Sec2A (b) tan2A (c) cot2A (d) tanA

[1]

3. If Sin2A = 2Sin A = 2Sin A than A is equal to

(a) 0 (b) 2 (c) 1 (d) 3

[1]

4. The value of Sec 90 ° is

(a) 0 (b) 3

2 (c) Not define (d) 2

[1]

5. Prove cos2θ +

2

11

1 cot θ=

+

[2]

6. Prove

( ) ( )( )

2cos 90 . 90sin

tan 90

A Sin AA

A

− −=

−

[2]

7. Find valve of Sin B and cos C and Cot B [2]

8. If tan A = 1 and tan B = 3 evaluate cos A. cos B – sin A. sin B [2]

9. Find the value of

cos 45

30 cos 30Sce ec

°° + °

[3]

10. Prove that

cos 1 1

sin cos 1 sec tan

Sinθ θθ θ θ θ

− + =+ − −

using the identity 2 2sec 1 tanθ θ= + [3]

A

B

C 12

135

79

CBSE TEST PAPER-01

CLASS - 10 Mathematics (Introduction to Trigonometry)

[ANSWERS]

Ans.1 (a)

Ans.2 (a)

Ans.3 (a)

Ans.4 (c)

Ans.5 L.H.S. = 2cos θ +2

1

cos ec θ2 2cos sin

1

θ θ= +=

Ans.6 L.H.S. =sin .cos

cot

A A

A2sin .cos

sincossin

A AA

A

A

= =

Ans.7 12 12

sin cos13 13

B and c= =

5cot

12B =

Ans.8

1cos

2A = 1

cos2

B =

1sin

2A = 3

2SinB =

1 1 1 3cos .cos sin

2 22 2A B B− = × − ×

1 3

2 2

−=

2k

A B

C

1k

1k AB

C

2k

1k

3k

80

Ans.9

1 1

2 22 2 2 2 3

13 3

=++

( )

1 3

2 2 2 3

1 3 2 2 3

2 2 2 3 2 2 3

1 2 3 6

4 122

1 2 3 6

82

2 3 6 2

8 2 2

2 6 6 2

16

2 3 2 6

16

3 2 6

8

= ×+

−= × ×+ −

−= −

−= −

−= ×−

−=−

− −=

−−=

Ans.10 Dividing numerator and denominator by cosθsin cos 1cos cos cossin cos 1cos cos cos

θ θθ θ θθ θθ θ θ

− +

+ −

( ) ( )( )

2 2

tan 1 sec

tan 1 sec

tan sec sec tan

tan 1 sec

θ θθ θθ θ θ θ

θ θ

− +=+ −+ − −

=+ −

( )( )( )

tan sec 1 sec tan

tan 1 sec

θ θ θ θθ θ

+ − +=

+ −

( )( )( )

sec tan sec tan

sec tan

θ θ θ θθ θ

+ −=

−1

sec tanθ θ=

−

81

CBSE TEST PAPER-02


1. The value of sinθ .sin ( ) ( )90 cos .cos 90θ θ θ− − − is

(a)1 (b) 0 (c) 2 (d) 3

[1]

2. If tanA = tanB, than A+B is

(a) 90° (b) 30 ° (c) 45 ° (d) 180 °

[1]

3. Value of tan 26 ° / cot 64 ° is equal to

(a) 1 (b) 2 (c) 0 (d) 5

[1]

4. The value of 2 21 tan 45 /1 tan 45− ° + is

(a) N.D (b) 2 (c) 0 (d) 3

[1]

5. Is it true sec A =

12

5for some value of angle A

[2]

6. Verify that

2

2 tan 30 360

1 tan 30 2Sin

°° = =+ °

[2]

7. If sec SA = ( )cos 36ec A − ° Find A [2]

8. If

5,

4Secα = evaluate

1 tan

1 tan

αα

−+

[2]

9. Evaluate

cos ( ) ( )2 2

2 2

cos 40 cos 5040 sin 50

sin 40 sin 50θ θ ° + °° − − ° + +

° + °

[3]

10. Prove

tan cot1 sec .cos

1 cot 1 tanec

θ θ θ θθ θ

+ = +− −

[3]

82

CBSE TEST PAPER-02


[ANSWERS]

Ans.1 (b)

Ans.2 (a)

Ans.3 (a)

Ans.4 (c)

Ans.5 yes, true since Sec A is always greater than 1.

Ans.6 L.H.S. =3

sin 602

=

R.H.S.= 2

1 22

3 311 11 33

×=

++

22 3 33

4 4 233

= = × =

Ans.7 ( )5 sec 36Sec A Co A= − °

( ) ( )cos 90 5 cos 36ec A ec A− = − °

90 5 36

6 36 90

6 126

21

A A

A

A

A

− = −− = − −− = −

= °

Ans.8 5

sec4

H

Bα = =

5k 3k

4k

α

A

BC

83

( ) ( )2 25 4

3

3 3tan

4 43

11 tan 1431 tan 714

AB k k

k

k

kα

αα

= −

=

= =

−− = =+ +

Ans.9 ( ) ( ) ( )( )

2 2

2 2

cos 40 cos 90 4090 40 50

40 sin 90 40Sin Sin

Sinθ θ

+ −− ° − − ° + + ° + −

( ) ( )2 2

2 2

cos 40 sin 4050 sin 50

40 401

0 11

SinSin cos

θ θ ° + °+ − ° + +° + °

= + =

Ans.10 L.H.S.

sin coscos sin

1 cos 1 sin1 sin 1 cos

θ θθ θ

θ θθ θ

= +− −

sin coscos sin

sin cos cos sinsin cos

θ θθ θ

θ θ θ θθ θ

= +− −

( ) ( )

( ) ( )

2 2

2 2

sin sin cos cos

cos sin cos sin cos sin

sin cos

cos sin cos sin cos sin

sin cos

cos sin cos sin sin cos

θ θ θ θθ θ θ θ θ θ

θ θθ θ θ θ θ θ

θ θθ θ θ θ θ θ

= × + ×− −

= +− −

= +− − −

2 2

3 3

1 sin cos

sin cos cos sin

1 sin cos

sin cos cos .sin

θ θθ θ θ θ

θ θθ θ θ θ

= − −

−= −

( )( )( )sin cos 1 sin .cos1

sin cos cos .sin

θ θ θ θθ θ θ θ

− + = −

1 sin .cossec .cos 1

cos .sin cos .sinec

θ θ θ θθ θ θ θ

= + = +

84

CBSE TEST PAPER-03


1. If ( )tan 2 cot 6θ θ= + the value of θ is

(a) 28 ° (b) 29 ° (c) 27 ° (d) 30 °

[1]

2. The value of sin60° . Cos30 ° + sin30 ° . Cos60°

(a) 1 (b) 2 (c) 4 (d) 3

[1]

3. If tan A=

4,

3than Sin A is equal to

(a) 4

3 (b)

4

5 (c)

5

4 (d)

1

4

[1]

4. The value of tan 48 ° . tan 23 ° . tan 42 ° . Tan67 ° is

(a) 1 (b) 2 (c) 3 (d) 4

[1]

5. Evaluate

2 22sin 47 cos 43

4cos 45cos 43 sin 47

° ° + − ° ° °

[2]

6. Prove ( )( )( )21 tan 1 sin 1 sin 1θ θ θ+ + − = [2]

7. Prove 4 2 4 2sec tan tanSec A A A A− = + [2]

8. Prove ( ) ( )2 2 2 2cos cos sec 7 tan cotSinA ecA A A A A+ + + = + + [2]

9. Evaluate

cos 70 cos55 .cos 35

sin 20 tan 5 . tan 25 .tan 45 .tan 65 .tan85

ec° ° °+° ° ° ° ° °

[3]

10. If

cos cos

cos sinm and n

α αβ β

= = show that ( )2 2 2 2cosm n nβ+ = [3]

85

CBSE TEST PAPER-03


[ANSWERS]

Ans.1 (a)

Ans.2 (a)

Ans.3 (b)

Ans.4 (a)

Ans.5 ( )

( )2cos 90 47sin 47 1

4cos 90 47 sin 47 2

− ° = + − − ° 2 2

sin 47 sin 47 14

sin 47 sin 47 2

° ° = + − ° °

1 1 2

0

= + −=

Ans.6 L.H.S. ( )( )2 2sec 1 sinθ θ= −

2 2sec .cos

1

θ θ=

Ans.7 L.H.S ( )2 2sec sec 1A A= −

( )( )2 2

2 4

1 tan tan

tan tan

A A

A A

= +

= +

Ans.8 L.H.S ( ) ( )2 2sin cos cos secA ecA A A= + + +

( )2 2 2 2

2 2 2 2

2 2

2 2

sin cos 2sin .cos cos sec 2cos .sec

sin cos cos sec 2sin .cos 2cos 3sec

1 11 1 cot 1 tan 2sin . 2cos .

sin cos

7 cot tan

A ec A A ecA A A A A

A A ec A A A ecA A A

A A A AA A

A A

= + + + + +

= + + + + +

= + + + + + +

= + +

86

Ans.9 ( ) ( )

( ) ( )cos 90 20 cos 90 35 .cos 35

sin 20 5 .tan 25 .1.tan 90 25 . tan 90 5

ec

tam

− − °+

° ° ° − − °

1sin 35 .sin 20 sin 35

sin 20 tan 5 .tan 25 .1.cot 25 .cot 5

°° °+° ° ° ° °

11

12

+

=

Ans.10 L.H.S ( )2 2 2cosm n β= +

( )

2

2

2 22

2

2 22 2

22 2

2 2

2

2

2

cos cos.cos

cos sin

cos cos.cos

cos sin 2

1 1cos .cos

cos sin

sin 2 coscos .cos

cos .sin

cos

sin

cos [

sinn n

α α ββ β

α α ββ β

α ββ β

β βα ββ β

αβ

αβ

= +

= +

= +

+=

=

= =∵

87

CBSE TEST PAPER-04


1. Value of θ when 2sin2θ = 3 is

(a) 0 ° (b) 45 ° (c) 30 ° (d) 90 °

[1]

2. Value of ( )2 2cos 1 tanθ θ+ is equal to

(a) 2 (b) -1 (c) 1 (d) 3

[1]

3. If sinθ 3

,5

= than value of ( )2tan secθ θ+ is

(a) 1 (b) 2 (c) 3 (d) 4

[1]

4. If sin(A-B) = . .SinA CosB CosA SinB− then value of Sin15 is

(a) 3 1

2 2

+ (b)

3 1

2 2

− (c) 3 (d) 2

[1]

5. If sin , tanx a y bθ θ= = prove

2 2

2 21

a b

x y− =

[2]

6. Prove

( ) ( ) 2sin 90 .cos 901 sin

tan

θ θθ

θ− −

= − [2]

7. If ( ) ( ) 3

sin 1,2

A B Cos A B+ = − = find the value of A and B[2]

8. Prove

1 1 1 1

cos cot sin sin cos cotecA A A A ecA A− = −

− +[2]

9. If

12cot ,

5B = prove 2 2 4 2tan sin sin .secB B B B− = [3]

10. Prove 2 2cos tan cotSec ecθ θ θ θ+ = + [3]

88

CBSE TEST PAPER-04


[ANSWERS]

Ans.1 (c)

Ans.2 (c)

Ans.3 (d)

Ans.4 (b)

Ans.5 ( ) ( )

2 2

2 2sin tan

a b

a bθ θ−

2 2

2 2 2 2

2 2

sin tan

cos cot

1

a b

a b

ec

θ θθ θ

= −

= −=

Ans.6 L.H.S. cos .sin

sincos

θ θθθ

=

2

2

cos

1 sin

θθ

== −

Ans.7 ( )sin 1 sin 90A B+ = =

( )

90..........(1)

3cos cos30

230 ..........(2)

A B

A B

A B

+ =

− = = °

− = °

On solving eq. (1) and (2)

60A = ° and 30B = °

89

Ans.8 L.H.S. =1 1

cos cot sinecA A A−

−

( )2 2

1 cos cot 1

cos cot cos cot sincos cot 1

cos cot sincos cot cos

cot

ecA A

ecA A ecA A AecA A

ec A A AecA A ecA

A

+= × −− ++

= −−

= + −=

R.H.S. = 1 1

sin cos cotA ecA A−

+

( )( )( )

2 2

cos cot1 1

sin cos cot cos cot

cos cotcos

cos cotcos cos cot

cot

. . . .

ecA A

A ecA A ecA A

ecA AecA

ec A AecA ecA A

A

L H S R H S

−= − ×

+ −−= −−

= − +=

=

Ans.9 ( ) ( )2 22 12 5AB K K= +

13AB K=

5 5 13tan ,sin ,sec

12 13 12B B B= = =

L.H.S. =

2 25 5

12 13 −

=

25 25

144 169169 144 25 25

25144 169 144 169

= −

− × = = × ×

R.H.S. =

4 25 13

13 12 = ×

25 25 13 13

13 13 13 13 12 1225 25

169 144

× ×= ×× × × ×

×=×

13K

5K

12K

A

B

C

90

Ans.10 L.H.S. 2 2

1 1

cos sinθ θ= +

2 2

2 2

2 22 2

sin cos

cos .sin

1sec .cos

cos .sinec

θ θθ θ

θ θθ θ

+=

= =

sec .cos ecθ θ=

R.H.S = tan cotθ θ+

2 2

sin cos

cos sin

sin cos

cos .sin1

cos .sinsec .cosec

θ θθ θθ θθ θ

θ θθ θ

= +

+=

=

=

91

CBSE TEST PAPER-05


1. Value of 2 2 22sin 30 3cos 45 tan 60° − ° + ° is

(a) 5 (b) 3 (c) 1 (d) 2

[1]

2. If

7cot

8θ = then value of 2tan θ is

(a) 64

49 (b)

49

64 (c)

77

8 (d)

8

7

[1]

3. If ( ) 1

sin2

A B− = and ( ) 1cos

2A B+ = value of A and B is

(a) 45 ° ,15 ° (b) 60 ° ,30 ° (c) 15 ° ,30° (d) 30 ° ,60 °

[1]

4. Value of 2

2

1cot

sinθ

θ− is

(a) 2 (b) -2 (c) 1 (d) -1

[1]

5. Given that ( )sin sin .cos cos .A B A B A SinB+ = + fin sin 75° [2]

6. If 3 tan 4,θ = find the value of

4cos sin

2cos sin

θ θθ θ

−+

[2]

7. Prove s 4 4 2 2sin cos 1 2sin .cosA A A A+ = − [2]

8. If

1sec

4x

xθ = + prove that sec tan 2xθ θ+ = or

1

2x

[2]

9. Prove ( )2 1 cos

cos cot1 cos

ecθθ θθ

−− =+

[3]

10. If ( )sin cos 2 sin 90θ θ θ+ = − determine cotθ [3]

92

CBSE TEST PAPER-05


[ANSWERS]

Ans.1 (d)

Ans.2 (a)

Ans.3 (a)

Ans.4 (d)

Ans.5 Put 45 , 30A B= ° = °

( )45 30 sin 45 .cos30 cos 45 .sin 30Sin + = ° ° + ° °

1 3 1 175

2 22 2Sin ° = × + ×

3 1

2 2

+=

Ans.6 Dividing by cosθ

4cos sincos cos

2cos sincos cos

4 tan 4 tan

2 tan 3

44 83

4 1023

4

5

θ θθ θθ θ

θ θθ θθ

−

+

− = +

−=

+

=

93

Ans.7 L.H.S. 4 4 2 2 2 2sin cos 2sin .cos sin .cosA A A A A A= + + −

( )2 2 2 2

2 2

sin cos 2sin .cos

1 2sin .cos

A A A A

A A

= + −

= −

Ans.8 1

1 4

xSec

xθ = +

2

2 2

22

4 1sec ............(1)

4

tan sec 1

4 11

4

x

x

x

x

θ

θ θ

+=

= −

+= −

4 2 2

2

4 2

2

222

2

16 1 8 16

16

16 1 8

16

4 1tan

4

4 1tan ...........(2)

4(1) (2)

x x x

x

x x

x

x

x

x

x

θ

θ

+ + −=

+ −=

−=

−= ±

+

( ) ( )

2 2

2 2

4 1 4 1tan

4 4

4 1 4 1

4

x xSec

x x

x x

x

θ θ + −+ = ±

+ ± −=

28 2

4 41

2 2

xor

x x

x orx

=

=

94

Ans.9 L.H.S. ( )2cos cotecθ θ= −

=

21 cos

sin sin

θθ θ

−

( ) ( )

( )( )( )

22

2 2 2

2

1 cos 1 cos1 cos

sin sin 1 cos

1 cos 1 cos

1 cos 1 cos 1 cos

θ θθθ θ θ

θ θθ θ θ

− −− = ⇒ ⇒ −

− −= =− + +

Ans.10 ( )cos 2 sin 90Sinθ θ θ+ = −

( )

sin cos 2 cos

sin 2 cos cos

sin cos 2 1

1cot

2 1

θ θ θ

θ θ θ

θ θ

θ

+ =

= −

= −

=−

1 2 1cot

2 1 2 1

2 1cot

2 1

2 1 cot

θ

θ

θ

+× =− ++ =

−+ =

95

CBSE TEST PAPER-01

CLASS-X Mathematics (Statistics)

1. 15, 3 36i i if f x P= = +∑ ∑ and mean of any distribution is 3, then p =

(a) 2 (b) 3 (c) 4 (d) 5

[1]

2. For what value of x the mode of the following data is 8:

4 5 6 8 5 4 8 5 6 x 8

(a) 5 (b) 6 (c) 8 (d) 4

[1]

3. The numbers are arranged in ascending order. If their median is 25 then x =

5 7 10 12 2x-8 2x+10 35 41 42 50

(a) 10 (b) 11 (c) 12 (d) 9

[1]

4. The median for the following frequency distribution is

X 6 7 5 2 10 9 3

F 9 12 8 13 11 14 7

(a) 6 (b) 5 (c) 4 (d) 7

[1]

5. The following data gives the number of boys of a particular age in a class of 40

students. Calculate the mean age of students:

Age (in years) 15 16 17 18 19 20

No. of student 3 8 10 10 5 4

[2]

6. For the following grouped frequency distribution find the mode.

Class 3-6 6-9 9-12 12-15 15-18 18-21 21-24

Frequency 2 5 10 23 21 12 3

[2]

96

7. Construct the cumulative frequency distribution of the following distribution:

Class 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 32.5-37.5

Frequency 2 22 19 14 13

[2]

8. The median and mode of a distribution are 21.2 and 21.4 respectively find its

mean.

[2]

9. The following table shows the weekly wages drawn by number of workers in a

factory

Weekly wages (in Rs.) 0-100 100-200 200-300 300-400 400-500

No. of workers 40 39 34 30 45

[3]

10. Find the median for each of the following data:

Marks Frequency

Less than 10 0

Less than 30 10

Less than 50 25

Less than 70 43

Less than 90 65

Less than 110 87

Less than 130 96

Less than 150 100

[3]

11. Find the median of the following data.

Wages (in rupees) No. of workers

More than 150 Nil

More than 140 12

More than 130 27

More than 120 60

More than 110 105

More than 100 124

More than 90 141

More than 80 150

[3]

97

12. Draw a less than Ogive for the following frequency distribution.

Marks No. of students

0-4 4

4-8 6

8-12 10

12-16 8

16-20 4

[3]

13. In the following distribution locate the median mean and mode.

Monthly

consumption

of electricity

65-85 85-105 105-125 125-145 145-165 165-185 185-205

No. of

consumers

4 5 13 20 14 7 4

[5]

98

CBSE TEST PAPER-01

CLASS - Mathematics (Statistics)

[ANSWERS]

Ans1. (B)

Ans2. (C)

Ans3. (C)

Ans4. (A)

Ans5. We have

Age (in years)x No. of students (f) fx

15 3 45

16 8 128

17 10 170

18 10 180

19 5 95

20 4 80

40f =∑ 698fx =∑

Mean 698

17.4540

fxx

f= = =∑∑

years

Ans6. Since the maximum frequency = 23 and it corresponds to the class 12-15

∴modal class = 12-15

1 0 2

1 00

1 0 2

12, 3, 23, 10, 21

2

23 1012 3

2 23 10 21

l n f f f

f fM l h

f f f

= = = = =−= +

− −−= +

× − −

13 3912 3 12

46 31 15= + × = +

−

1312 12 2.6 14.6

5= + = + =

99

Ans7 The required cumulative frequency distribution of the given distribution is

given below:

Class Frequency Cumulative frequency

12.5-17.5 2 2

17.5-22.5 22 24

22.5-27.5 19 43

27.5-32.5 14 57

32.5-37.5 13 70

Ans8 We know that Mean = Mode + 3

2(Median-mode)

( )

( )

321.4 21.2 21.4

23

21.4 0.22

21.4 0.3 21.1

= + −

= + −

= − =

Ans9 We have

Weekly wages (in Rs.) No. of workers (f) C.F

0-100 49 40

100-200 39 79

200-300 34 113

300-400 30 143

400-500 45 188

188N f= =∑

Now 188

942 2

N = = and this is in 200-300 class.

∴Median class= 200-300

Here 1 200, 79, 100, 34, 942

Nl c h f= = = = =

We know that 12N

cMe l h

F

−= + ×

100

94 79200 100

341500

20034

750200 200 14.12

17244.12

−= + ×

= +

= + ⇒ +

=

Ans10 first of all we shall change cumulating series into simple series.

We have

X F C.F

0-10 0 0

10-30 10 10

30-50 15 25

50-70 18 43

70-90 22 65

90-110 22 87

110-130 9 96

130-150 4 100

100N f= =∑

Now 100

502 2

N = = which lies in 70-90 class

∴Median class = 70-90

Here 1 70, 43, 20, 22, 100l c h f N= = = = =

We know that Median, Me = 21

NC

l hf

−+ ×

( )2070 50 43

2220 7 70

70 7022 11

70 6.36

76.36

= + −

×= + = +

= +=

101

Ans11. Fist of all we shall find simple frequencies.

Wages (in Rupees) (X) No. of workers (F) C.F

80-90 9 9

90-100 17 26

100-110 19 45

110-120 45 90

120-130 33 123

130-140 15 138

140-150 2 150

150N f= =∑

Now 150

75,2 2



Here 1 110, , 45, 10, 45, 150l c h f N= = = =

We know that Me = 12M

Cl h

F

−+ ×

( )10110 75 45

4510 30 20

110 11045 3

110 6.67 116.67

= + −

×= + = +

= + =

Ans12. We have

Marks Frequency (F) C.F

0-4 4 4

4-8 6 10

8-12 10 20

12-16 8 28

16-20 4 32

32f =∑

102

Upper class

limits

4 8 12 16 20

Cumulative

frequency

4 10 20 28 32

Plot the

points

(4,4) (8,10) (12,20) (16,28) (20,32)

Join these points by a free hand curve. We get the required ogive which is as

follows:

Ans13.

Monthly

consumption

of electricity

No. of

consumers

C.F Class Mark

(X)

FX

65-85 4 4 75 300

85-105 5 9 95 475

105-125 13 22 115 1495

125-145 20 42 135 2700

145-165 14 56 155 2670

165-185 8 64 175 1400

185-205 4 68 195 780

63N f= =∑ ∑ƒx=9320

Now 68

342 2

N = = and this is in 125-145 class

103


Here 1 125, 22, 20, 20, 342

Nl c h f= = = = =

We know that 1

34 222 125 2020

Nc

Me l hf

− −= + × = + ×

125 12 137= + =

Hence Median = 137

Again Mean ( ) 9320137.05

68

fxx

f= = =∑∑

For mode, since the maximum frequency is 20 and this corresponds to the class

125-145

Here 1 0 2125, 20, 20, 13, 14l h f f f= = = = =

( )

1 00

1 0 22

20 13125 20

2 20 13 14

7125 20

13

140125

13125 10.76

135.76

f fM l h

f f f

−= +− −

−= + − −

= +

= +

= +=

Thus Median = 137

Mean = 137.05

Mode = 135.76

The three measures are approximately the same in the class.

104

CBSE TEST PAPER-02


1. In the formula ,

fiuix a h

fi

= +

∑∑

for finding the mean of grouped frequency

distribution, is =

(a) xi a

h

+ (b) ( )h xi a− (c)

xi a

h

− (d)

a xi

h

−

[1]

2. While computing mean of grouped data, we assume that the frequencies are

(a) Evenly distributed over all the class

(b) Centered at the class marks of the class

(c) Centre at the upper limits of the class

(d) Centre at the lower limits of the class

[1]

3. If 17, 4 63fi fixi P= = +∑ ∑ and mean = 7, then P=

(a) 12 (b) 13 (c) 14 (d) 15

[1]

4. If the value of mean and mode are respectively 30 and 15, then median =

(a) 22.5 (b) 24.5 (c) 25 (d) 26

[1]

5. The marks distribution of 30 students in a mathematics examination are given

below

Class Interval 10-25 25-40 40-55 55-70 70-85 85-100

No. of students 2 3 7 6 0 6

Find the mode of this data.

[2]

6. Construct the cumulative frequency distribution of following distribution:

Marks 39.5-49.5 49.5-59.5 59.5-69.5 69.5-79.5 79.5-89.3 89.5-99.5

students 5 10 20 30 20 15

[2]

7. If the values of mean and mode are respectively 30 and 15, then median =

(a) 22.5 (b) 24.5 (c) 25 (d) 26

[2]

105

8. If the mean of the following data is 18.75. find the value of P

xi 10 15 P 25 30

fi 5 10 7 8 2

[2]

9. Find the mean age in years form a frequency distribution given below:

Age(in yrs) 15-19 20-24 25-29 30-34 35-39 40-45 45-49 Total

Frequency 3 12 21 15 5 4 2 63

[3]

10. Find the median of the following frequency distribution:

Wages (in Rs.) 200-300 300-400 400-500 500-600 600-700

No. of Laborers 3 5 20 10 6

[3]

11. The following tables gives production yield per hectare of what of 100 farms of

village

Production yield (in hr.) 50-55 55-60 60-65 65-70 70-75 75-80

No. of farms 2 8 12 24 38 16

Change the distribution to a more than type distribution and draw its ogive

[3]

12. The A.M of the following distribution is 47. Determine the value of P

Classes 0-20 20-40 40-60 60-80 80-100

Frequency 8 15 20 P 5

[3]

13. Find the mean, mode and median for the following data:

Classes 0-10 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 5 8 15 20 14 8 5

[5]

106

CBSE TEST PAPER-02


[ANSWERS]

Ans1. (C)

Ans2. (B)

Ans3. (C)

Ans4. (C)

Ans5. Since the maximum frequency = 7 and it corresponds to the class 40-55.

The modal class= 40-55

Here 1 0 240, 15, 7, 3, 6l h f f f= = = = =

We know that mode Mo is given by

Mo = ( )

( )1 0 2

15 7 31 0 15 440 40 40 12 52

2 2 7 3 6 5

f fl h

f f f

−− ×+ ⇒ + ⇒ + ⇒ + =− − − −

Thus Mode marks = 52

Ans6. The required cumulative frequency distribution of the given distribution is

given below.

Marks No. of Students Cumulative Frequency

39.5-49.5 5 5

49.5-59.5 10 15

59.5-69.5 20 35

69.5-79.5 30 65

79.5-89.5 20 85

89.5-99.5 15 100

100N f= =∑

107

Ans7 Median = Mode2

3+ (Mean-mode)

( )215 30 15

32

15 153

15 10 25

= + −

= + ×

= + =

∴C holds

Ans.8 We have

xi fi xifi

10 5 50

15 10 150

P 7 7P

25 8 200

30 2 60

32N fi= =∑ 460 7fixi P= +∑

Now mean fixi

xfi

= ∑∑

= 18.75

406 718.75

32

P+=

32 1875460 7

100460 7 8 75 600

7 600 460

7 140

20

P

P

P

P

P

×⇒ + =

⇒ + = × =⇒ = −⇒ =⇒ =

108

Ans9 We have

Class-

Interval

Mid valve fi 32

5

xi a xiui

h

− −= = fiui

15-19 17 3 -3 -9

20-24 22 13 -2 -26

25-29 27 21 -1 -21

30-34 32 15 0 0

35-39 37 5 1 5

40-44 42 4 2 8

45-49 47 2 3 6

Total 63fi =∑ 37fiui = −∑

Let assumed mean ‘a’ = 32, Here h = 5

We know that Mean fiui

x a hfi

= + ×∑∑

37 532

63185

3263

32 2.94( )

29.06

nearly

years

− ×=

−=

= −=

Ans10 We have

Wages (in Rs.) No. of laborers (f) C.F

200-300 3 3

300-400 5 8

400-500 20 28

500-600 10 38

600-700 6 44

44N f= =∑

109

Now 44

222 2

N = = and this lies in 400-500 class.


Here 1 400, 8, 100, 20, 44l C h f N= = = = =

We know that 12N

CMe l h

F

−= + ×

22 8400 100

2014 100

40020

400 70

470

470Me

−= + ×

×= +

= +=

=

Ans11 More than type Ogive

Production yield (Kglha) C.F

More than or equal to 50 100






Now, draw the Ogive by plotting the points (50,100), (55,98), (60,90), (65,78),

(70,54), (75,16)

110

Ans12 We have

Class Interval Midvalue ( )xi Frequency ( )fi fixi

0-20 10 8 80

20-40 30 15 450

40-60 50 20 1000

60-80 70 P 70P

80-100 90 5 450

48fi P= +∑ 1980 70fixi P= +∑

Since Mean, 1980 70

4748

fixi Px

fi P

+= ⇒ =+

∑∑

2256 47 1980 70 70 47 2250 1980

27623 276 12

23

P P P P

P P

⇒ + = + ⇒ − = −

⇒ = ⇒ = =

Thus P = 12

Ans13 We have

Classes Midvalve

xi

Frequency

fi 10

x axi

−= fixi C.f

0-10 5 5 -3 -15 5

10-20 15 8 -2 -16 13

20-30 25 15 -1 -15 28

30-40 35 20 0 0 48

40-50 45 14 1 14 62

50-60 55 8 2 16 70

60-70 65 5 3 15 75

75fi =∑ 1fixi = −∑

Let assumed mean a = 35 h = length of class interval= 10

111

Mean 1

35 1075

fixix a h

fi= + × = − ×∑

∑

235

1535 0.13

34.87

= −

= −=

Since Maximum frequency = 20 ∴Modal class = 30-40

1 0 230, 20, 15, 14l f f f= = = =

Mode = 1 0

1 0 22

f fl h

f f f

−+− −

20 1530 10

40 50 14

5030

1130 4.55

34.55

− = + − −

= +

= +=

Hence mode = 34.55

Since 75

37.52 2

N = = Which lies in the class 30-40

i.e, Median class = 30-40

1 30, 37.5, 28, 20, 102

Nl C f h= = = = =

Median = 12N

C hl

f

− ×+

37.5 2830 10

209.5

302

30 4.75

34.75

−= + ×

= +

= +=

Hence Median = 34.75

112

CBSE TEST PAPER-03


1. The wickets taken by a bowler in 10 cricket matches as follows

2 6 4 5 0 2 1 3 2 3

Find the mode of the data

(a) 1 (b) 4

(c) 2 (d) 3

[1]

2. Mean of the data

Class Interval 50-60 60-70 70-80 80-90 90-100

Frequency 8 6 12 11 13

(a) 76 (b) 77

(c) 78 (d) 80

[1]

3. Construction of a cumulative frequency table is useful in determining the

(a) Mean (b) Median

(c) Mode (d) all these conditions

[1]

4. In the following distribution of the heights of 60 students of a class

Height (inch) 150-155 155-160 160-165 165-170 170-175 175-180

No. of

students

15 13 10 8 9 5

Then sum of the lower limit of the modal class and upper limit of the median class

is

(a) 310 (b) 315

(c) 320 (d) 330

[1]

5. Find the mean of the following data

Classes 10-20 20-30 30-40 40-50 50-60


[2]

6. The following data gives the information observed life times (in hours) of 225

electrical components. Determine the modal life times of the components.

[2]

113

Life time (in hours) 0-20 20-40 40-60 60-80 80-100 100-200

Frequency 10 35 52 61 38 29

7. Construct the cumulative frequency distribution of the following distribution

Class Interval 6.5-

7.5

7.5-

8.5

8.5-

9.5

9.5-

10.5

10.5-

11.5

11.5-

12.5

12.5-

13.5

Frequency 5 12 25 48 32 6 1

[2]

8. Calculate the median from

Marks 0-10 10-30 30-60 60-80 80-100

No. of students 5 15 30 8 2

[2]

9. Thirty women were examined in a hospital by a doctor and the number of heart

beats per minute were recorded and summarized as follows. Find the mean heart

beats per minute for these women choosing a suitable method.

Number of heart beats per minute No. of women

65-68 2

68-71 4

71-74 3

74-77 8

77-80 7

80-83 4

83-86 2

[3]

10. Following distribution shows the marks obtained by a class of 100 students

Marks 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 10 15 30 32 8 5

Change the distribution to less than type distribution and draw its ogive

[3]

11. Following table shows the daily pocket allowances given to the children of a

multistory building. The mean of the pocket allowances is Rs.18. Find out the

missing Frequency

[3]

114

Class Interval 11-13 13-15 15-17 17-19 19-21 21-23 23-25

Frequency 3 6 9 13 ? 5 4

12. A survey regarding the heights (in cm) of 51 girls of Class X of a school was

conducted and the following data was obtained. Find the median height.

Height (in cm) No. of girls

Less than 140 4

Less than 145 11

Less than 150 29

Less than 155 40

Less than160 46

Less than 165 51

[3]

13. Find the mean, mode and median for the following data:

Classes 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Frequency 4 8 10 12 10 4 2

[5]

115

CBSE TEST PAPER-03


[ANSWERS]

Ans1. (C)

Ans2. (C)

Ans3. (B)

Ans4. (B)

Ans5. We have

Classes Mid-value xi Frequency fi fixi

10-20 15 5 75

20-30 25 8 200

30-40 35 13 455

40-50 45 15 675

50-60 55 9 495

50fi =∑ 1900fixi =∑

Now mean

1900

50

38

fixix

fi= =

=

∑∑

Hence mean 38x =

Ans6 Since the maximum frequency = 61 and it corresponds to the class 60-80

∴Modal class = 60-80

Here 1 0 260, 20, 61, 52, 38l h f f f= = = = =


( )

1 0

1 0 22

61 5260 20

2 61 52 38

960 20

122 90

f fMo l h

f f f

−= +− −

−= +− −

= +−

116

20 960

3245

608

60 5.625

65.625hours

×= +

= +

= +=

Thus modal life times = 65.625 hours


given below

Class Interval Frequency Cumulative frequency

6.5-7.5 5 5

7.5-8.5 12 17

8.5-9.5 25 42

9.5-10.5 48 90

10.5-11.5 32 122

11.5-12.5 6 128

12.5-13.5 1 129

129N f= =∑

Ans8 We have

Marks No. of students (f) C.F

0-10 5 5

10-30 15 20

30-60 30 50

60-80 8 58

80-100 2 60

60N f= =∑

Since 302

N = which his in the class 30-60 ∴median class is 30-60

We know that median Me is given by

117

12N

CMe l h

f

−= + ×

Here 1 30, 30, 30, 20, 302

Nl h C F= = = = =

30 2030 30

30Me

−∴ = + ×

= 30 +10

= 40

Hence median = 40

Ans9 Let assumed mean ‘a’ = 75.5. we have

No. of heart

beats per

minute

No. of women

( fi )

Class Mark

i.e mid value

( )xi

xi axi

h

−= fixi

65-68 2 66.5 -3 -6

68-71 4 69.5 -2 -8

71-74 3 72.5 -1 -3

74-77 8 75.5=a 0 0

77-80 7 78.5 1 7

80-83 4 81.5 2 8

83-86 2 84.5 3 6

30fi =∑ 4fixi =∑

We know that

Mean fixi

x a hfi

= + ×∑∑

[By step Deviation Method]

475.5 3

3075.5 0.4

75.9

= + ×

= +=

118

Ans10 Less than type Ogive

Marks Marks Frequency Cumulative

Frequency

10-20 Less than 20 10 10

20-30 Less than 30 15 25

30-40 Less than 40 30 55

40-50 Less than50 32 87

50-60 Less than 60 8 95

60-70 Less than 70 5 100

Now, draw the ogive by plotting (20,10), (30,25), (40,55), (50,87), (60,95),

(70,100)

Ans11 Let the missing frequency = f we have

Class

interval

fi Mid-

value

18

2

xi a xixi

h

− −= =Fixi

11-13 3 12 -3 -9

13-15 6 14 -2 -12

15-17 9 16 -1 -9

17-19 13 18 0 0

19-21 F 20 1 F

21-32 5 22 2 10

23-25 4 24 3 12

40fi F= +∑ 8fixi f= −∑

119

Let assumed mean a = 18, Here h = 2

We know that mean fixi

x a hfi

= + ×∑∑

( )818 18 2

40

0 8

8

f

f

f

f

−⇒ = + ×

+⇒ = −⇒ =

Hence missing frequency = 8

Ans12 We have

Class Intervals Frequency (f) C.F

Below 140 4 4

140-145 7 11

145-150 18 29

150-155 11 40

155-160 6 46

160-165 5 51

51N f= =∑

Here 51

25.52 2

N = = which his in the class 145-150

Here 1 145, 5, 51, 11, 18l h N C F= = = = =

∴Median 12N

Cl h

f

−= + ×

25.5 11145 5

18

−= + ×

72.5145 149.03

18= + ⇒

∴Median height of the girls = 149.03

120

Ans13 We have

Classes Mid value xi fi xi aui

h

− = fiui c.f

10-20 15 4 -3 -12 4

20-30 25 8 -2 -16 12

30-40 35 10 -1 -10 22

40-50 45 12 0 0 34

50-60 55 10 1 10 44

60-70 65 4 2 8 48

70-80 75 2 3 6 50

50N fi= =∑ 14fixi =∑ -14

Let assumed mean a = 45, Here h = 10

We know that mean ( ) fixix a h

fi= + ×∑

∑

1445 10

5014

455

45 2.8

= − ×

= −

= −

Mean ( ) 42.2x =

Since max. frequency = 12 ∴modal class = 40-50

Hence 1 0 240, 12, 10, 10, 10l f f h f= = = = =

Now Mode 1

1 0 2

2

2

f fl h

f f f

−= + ×− −

12 1040 10

24 10 102

40 104

40 5

45

−= + ×− −

= + ×

= +=

∴Mode = 45

121

Now 50

252 2

N = = ∴Median class is 40-50

Now median 12N

cl h

f

−= + ×

Where 110, 22, 12, 10, 40N C F h l= = = = =

∴Median = 1

25 2210

12l

−+ ×

140 10

440 2.5

42.5

= + ×

= +=

Thus Median = 42.5

122

CBSE TEST PAPER-04


1. Choose the correct answer form the given four options in the formula

fixix a

fi= +∑

∑

For finding the mean of grouped data di’s are deviations from a of

(a) lower limits of the classes (b) Upper limits of the classes

(c) Mid points of the classes (d) Frequencies of the class marks

[1]

2. If mean of the distribution is 7.5

X 3 5 7 9 11 13

F 6 8 15 P 8 4

Then P:-

(a) 2 (b) 4 (c) 3 (d) 6

[1]

3. A shoe shop in Agra had sold hundred pairs of shoes of particular brand in a

certain day with the following distribution.

Size of the shoes 4 5 6 7 8 9 10

No. of pairs sold 1 4 3 20 45 25 2

Find mode of the destitution.

(a) 20 (b) 45 (c) 1 (d) 3

[1]

4. If the mode of a data is 45 and mean is 27, then median is

(a) 30 (b) 27 (c) 33 (d) None of these

[1]

5. Find the mean of the following data

Classes 0-10 10-20 20-30 30-40 40-50

Frequency 3 5 9 5 3

[2]

6. A survey conducted on 20 households in a locality by a group of students resulted

in the following frequency table for the number of family members in a household.

Find the mode.

[2]

123

Family size 1-3 3-5 5-7 7-9 9-11

No. of families 7 8 2 4 1


Class interval 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 5 3 10 6 4 2

[2]

8. If the values of mean and median are 26.4 and 27.2, what will be the value of

mode?

[2]

9. Consider the following distribution of daily wages 50 workers of factory

Daily wages (in kg) 100-120 120-140 140-160 160-180 180-200

No. of workers 12 14 8 6 10

Find the mean daily wages of the works of the factory by using an appropriate

method.

[3]

10. The distribution below given the weight of 30 students of a class. Find the median

weight of the students

Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75

No. of students 2 3 8 6 6 3 2

[3]

11. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs) 100-120 120-140 140-160 160-180 180-200

No. of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency

distribution and draw its ogive .

[3]

12. If the mean of the following distribution is 54, find the value of P.

Classes 0-20 20-40 40-60 60-80 80-100

Frequency 7 P 10 9 13

[3]

13. Find the mean, mode and median for the following data.

Classes 5-15 15-25 25-35 35-45 45-55 55-65 65-75

Frequency 2 3 5 7 4 2 2

[5]

124

CBSE TEST PAPER-04


[ANSWERS]

Ans1. (C) Ans2. (C) Ans3. (B) Ans4. (C)

Ans5 We have

Classes Mid-value ( )xi Frequency ( )fi xifi

0-10 5 3 15

10-20 15 5 75

20-30 25 9 225

30-40 35 5 175

40-50 45 3 135

25fi =∑ 625fixi =∑

Now Mean fixi

xfi

= ∑∑

62525

25= =

Ans6 Since the maximum frequency = 8 and it corresponds to the class 3-5

Modal class = 3-5

Here 1 0 23, 2, 8, 7, 2l h f f f= = = = =


( )( )

( )

1 0

1 0 22

8 73 2

2 8 7 2

13 2

7

f fMo l h

f f f

−= +− −

−= +

− −

= +

23

73 0.2857

3.286 nearly

+

= +=

125


given below:

Class Interval Frequency (f) Cumulative frequency

0-10 5 5

10-20 3 8

20-30 10 18

30-40 6 24

40-50 4 28

50-60 2 30

Total N= 30

Ans8 We know that

Mode = 3 median -2 mean

= 3(27.2) – 2(26.4)

= 81.6 – 52.8

Mode = 28.8

Ans9 Let assumed mean ‘a’ = 150, h = 120-100 = 20

We have

Daily wages No. of

workers ( )fi

Class mark

mid-value

( )xi

xi axi

h

−= fixi

100-120 12 110 -2 -24

120-140 14 130 -1 -14

140-160 8 150 = a 0 0

160-180 6 170 1 6

180-200 10 190 2 20

50fi =∑ 12fixi = −∑

We know that

126

Mean ( ) [ ]By step diviation methodfixi

x a hfi

= + ×∑∑

( )

12150 20

5024

150 150 4.85

Mean 145.20x

− ×

−= = −

=

Ans10 We have

Weight (in kg) No. of students (f) C.F

40-45 2 2

45-50 3 5

50-55 8 13

55-60 6 19

60-65 6 25

65-70 3 28

70-75 2 30

30N f= =∑

Here 30

152 2

N = = which his in 55-60 class.


Here 1 55, 13, 6, 5, 30l C f h N= = = = =

We know that the median Me is given by

21

15 1355 5

65

553

55 1.67 56.67

Nc

Me l hf

−= + ×

−= + ×

= +

= + =

Hence median weight = 56.67kg

127

Ans11 Less than type cumulative frequency distribution

Daily Income (in Rs.) Cumulative frequency

Less than 120 12

Less than 140 26

Less than 160 34

Less than 180 40

Less than 200 50

Let us draw the graph of the points (120,12), (140,26), (160,34), (180,40),

(200,50)

This is the required less than type ogive

Ans12 Here 39 , 2370 30fi P fixi P= + = +∑ ∑

Since Mean,

2370 3054

392106 54 2370 30

54 30 2370 2106

24 264

264

2411

fixix

fi

P

PP P

P P

P

P

P

=

+∴ =+

⇒ + = +⇒ − = −⇒ =

⇒ =

⇒ =

∑∑

5

10

15

20

25

30

35

40

45

50

100 120 140 160 180 200

.

...

(120,12)

(140,26)

(160,34)

(180,40)

(200,50).

128

Ans13 We have

classes Mid-value

( )xi

( )fi xi axi

h

−= fixi c.f

5-15 10 2 -3 -6 2

15-25 20 3 -2 -6 5

25-35 30 5 -1 -5 10

35-45 40 7 0 0 17

45-55 50 4 1 4 21

55-65 60 2 2 4 23

65-75 70 2 3 6 25

25fi =∑ 3fixi = −∑

Let assumed mean ‘a’ = 40, Here h=10

Mean fixi

x a hfi

= + ×∑∑

340 10

256

405

40 1.2

38.8

= − ×

= −

= −=

Since Max. Frequency = 7 ∴modal class = 35-45

1 0 235, 7, 5, 4l f f f= = = =

We know that

Mode 1 0

1 0 22

f fl h

f f f

−= + ×− −

7 535 10

14 5 42

35 105

35 4 39

−= + ×− −

= + ×

= + =

Since 25

12.52 2


129

Here 1 35, 7, 10l f c= = =

Median = 12N

cl h

f

−+ ×

12.5 1035 10

725

357

35 3.6 nearly

=38.6 nearly

−= + ×

= +

= +

130

CBSE TEST PAPER-05


1. If 'xi s are the mid points of the class intervals of grouped data, 'fi s are the

corresponding frequency and x is the mean, then ( )fixi x−∑ is equal to

(a) 0 (b) -1 (c) 1 (d)2

[1]

2. Mode of

Class Interval 0-20 20-40 40-60 60-80 80-100


(a) 65 (b) 66 (c) 75 (d) 70

[1]

3. Median OF

Class 0-500 500-1000 1000-1500 1500-2000 2000-2500


is

(a) 1000 (b) 1100 (c) 1200 (d) 1150

[1]

4. If the median of the distribution:

Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Total

Frequency 5 x 20 15 7 5 60

Is 28.5 find the value of x

(a) 8 (b) 10 (c) 4 (d) 9

[1]

5. The marks obtained by 30 students of class x of a certain school in a Mathematics

paper consisting of 100 marks are presented in table below. Find the mean of the

marks obtained by the students

Marks

obtained ( )xi

10 20 36 40 50 56 60 70 72 80 88 92 98

students ( )fi 1 1 3 4 3 2 4 4 1 1 2 3 1

[2]

131

6. A student noted the numbers of cars passing through a spot on a rod for 100

periods each of 3 minutes and summarized in the table given below. Find the mode

of the data.

No. of cars 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Frequency 7 14 13 12 20 11 15 8

[2]


consumption ( units) 65-85 85-105 105-125 125-145 145-165 165-185

Consumers ( )fi 4 5 12 20 14 8

[2]

8. If the values of mean and median are 53.6 and 55.81, what will be the value of

mode?

[2]

9. Calculate the mean for the following distribution

Class Interval 0-4 4-8 8-12 12-16 16-20 20-24 24-28 28-32

Frequency 2 5 8 16 14 10 8 3

[3]

10. The percentage of marks obtained by 100 students in an examination are given

below:

Marks 30-35 35-40 40-45 45-50 50-55 55-60 60-65

Frequency 14 16 18 23 18 8 3

Determine the median percentage of marks.

[3]

11. Draw a less than ogive for the following frequency distribution

Marks 0-4 4-8 8-12 12-16 16-20

No. of students 4 6 10 8 4

[3]

12. The A.M of the following frequency distribution is 53. Find the value of P

Classes 0-20 20-40 40-60 60-80 80-100

Frequency 12 15 32 P 13

[3]

13. From the following information, construct less than and more than ogive and find

out median form it

Wages (Rs.) 0-30 30-40 40-50 50-60 60-70 70-80

Mo. Of workers 10 15 30 32 8 5

[5]

132

CBSE TEST PAPER-05


[ANSWERS]

Ans1. (A)

Ans2. (D)

Ans3. (C)

Ans4. (A)

Ans5

Marks obtained ( )xi No. of students ( )fi fixi

10 1 10

20 1 20

36 3 108

40 4 160

50 3 150

56 2 112

60 4 240

70 4 280

72 1 72

80 1 80

88 2 176

92 3 276

95 1 95

30fi =∑ 1779fixi =∑

Mean x =1779

59.330

fixi

fi= =∑

∑

Thus mean x = 59.3

133

Ans6 Since the maximum frequency = 20

And it corresponds to the class 40-50

Modal class = 40-50

Here 1 0 240, 10, 20, 12, 11l h f f f= = = = =

We know that mode M0 is given by

( )

1 0

1 0 22

20 1240 10

2 20 12 11

8040

1740 4.705

44.705

44.7

f fMo l h

f f f

−− +− −

−= + − −

= +

= +==

Ans7 The required accumulative frequency distribution of the given distribution is

given below.

Monthly consumption (in

units)

No. of consumes ( )fi Cumulative frequency

( )cf

65-85 4 4

85-105 5 9

105-125 13 22

125-145 20 42

145-165 14 56

165-185 8 64

N = 64

Ans8 We know that

Mode = 3 Median – 2 mean

Mean = ( ) ( )3 55.81 2 53.6−

167.43 107.2 60.23= − =

134

Ans9 By Deviation Method

Let assumed mean a = 14

Class interval Mid-value ( )xi Frequency ( )fi Deviation

di xi a= −

Product fidi

0-4 2 2 -12 -24

4-8 6 5 -8 -40

8-12 10 8 -4 -32

12-16 14 16 0 0

16-20 18 14 4 56

20-24 22 10 8 80

24-28 26 8 12 96

28-32 30 3 16 48

Total 66fi =∑ 184fidi =∑

We know that Mean fidi

x afi

= +∑∑

18414

6614 2.866

16.866

= +

= +=

Ans10

Marks (class) No. of students (Frequency) Cumulative Frequency

30-35 14 14

35-40 16 30

40-45 18 48

45-50 23 71

50-55 18 89

55-60 8 97

60-65 3 100

135

Here n = 100

Therefore 502

n = which lies in the class 45-50

L1 (The lower limit of the median class) = 45

C (The cumulative frequency of the class preceding the median class) = 48

F (The frequency of the Median class)= 23

H (The class size) = 5

Median 12n

cl h

f

− = +

50 4845 5

23

1045 45.4

23

− = + ×

= + =

So, the median percentage of marks is 45.4

Ans11 We have

Marks Frequency ( )f C.F

0-4 4 4

4-8 6 10

8-12 10 20

12-16 8 28

16-20 4 32

32f =∑

Upper class

limits

4 8 12 16 20

Cumulative

Frequency

4 10 20 28 32

Plot the

points

(4,4) (8,10) (12,20) (16,28) (20,32)

136

Joint these points by a free hand curve, we get the required ogive which is as follows:

Ans12 We have

Class Interval Mid-value ( )xi Frequency ( )fi fixi

0-20 10 12 120

20-40 30 15 450

40-60 50 32 1600

60-80 70 P 70P

80-100 90 13 1170

72fi P= +∑ 3340 70fixi P= +∑

Since Mean fixi

xfi

=∑∑

3340 7053

723340 70 3816 53

17 3816 3340

476

1728

P

PP P

P

P

P

+⇒ =

+⇒ + = +⇒ = −

⇒ =

⇒ =

Thus P = 28

cum

ula

tive fre

quency

137

Ans13

Wages (Rs) Frequency

(less than)

Wages C.F (More

then)

Wages C.F

0-30 10 30 10 0 100

30-40 15 40 25 30 90

40-50 30 50 55 40 75

50-60 32 60 87 50 45

60-70 8 70 95 60 13

70-100 5 100 100 70 5

Median = 48.5

138

brilliant public school ,...

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