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1111

PIONEER GUESS PAPER

+1 CBSE

MATHEMATICS

“Solutions” TIME: 3:00 HOURS MAX. MARKS: 100

General Instructions: (i) All questions are compulsory.

(ii) The question paper consists of 29 questions divided into three sections, A, B and C. Section A

comprises of 10 questions of one mark each, Section B comprises of 12 questions of four marks each

and Section C comprises of 7 questions of six marks each

(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact

requirement of the question.

(iv) There is no overall choice. However, an internal choice has been provided in 4 questions of four

marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all

such questions.

(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.

NAME OF THE CANDIDATE PHONE NUMBER

L.K. Gupta (Mathematics Classes)




2222

Section ‘A’

Question numbers 1 to 10 carry 1 mark each.

1. Find the domain of x 2

f(x)3 x

−=

−

Sol:

x 2

3 x

As 3 x 0 x 3

x 20

3 x

x 20

x 3

−

−

− ≠ ⇒ ≠

−⇒ >

−

−⇒ <

−

fD [2,3)∴ =

2. How many elements has P (A), if A = φ ?

Sol: 1

3. Let A = {1, 2, 3, 4} and R = { a , b} : a A, b A, a divides∈ ∈ b}. Write R explicitly.

Sol:

{ }R (1,1),(1,2),(1,3),(1,4),(2,2),(2,4),(3,3),(4,4)=

4. Find the locus of P if 2 2 2PA PB 2K ,+ = where A and B are the points (3, 4, 5) and

(–1, 3, – 7).

Sol:

2 2 2 2 2(x 3) (y 4) (z 5) (x 1) (y 3)− + − + − + + + − 2 2(z 7) 2k+ + =

2 2 2 2 2 2x y z x y z 9 6x 8y 16 25 10z 1 2x 9 6y 49 14z+ + + + + + − − + + − + + + − + + 22k=




3333

2 2 22(x y z )+ + 24x 14y 4z 109 2k 0− − + + − =

5. Differentiate nx tan x w.r.t. x.

Sol:

( )

n

n n

y x tan x

dy d dx tan x x tan x

dx dx dx

=

= +

( )n 1 n 2nx tan x x sec x−= +

( )n 1 2dyx n tan x xsec x

dx

−= +

6. Evaluate: π

x4

1 tan xlim

πx

4

→

−

−

Sol:

We have, π

x4

1 tan xlim

πx

4

→

−

−

h 0 h 0

πtan tanh

41π π

1 tan h 1 tan . tanh4 4lim lim

h h→ →

+

− − + −

= =

( )h 0 h 0

1 tan h1

1 tan h 1 tan h 1 tan hlim lim

h h 1 tan h→ →

+−

− − − − = =−

( ) ( )h 0 h 0

2 tan h tanh 1lim 2 lim 2 2.

h 1 tan h h 1 tanh 1→ →

−= = − = − × = −

− −

7. Check the validity of the statements given below by the method given against it.




4444

p : The sum of an irrational number and a rational number is irrational (by contradiction

method).

Sol:

Let a beirrationalnumberandbbea rational number. …(1)

b + p

a where p,q are co primeq

= −

∴ p

a bq

= − …. (2) L.H.S = a Anirrationalnumber=

R.H.S. = p

b A rationalnumberq

− =

It is a contradiction.

Therefore, the sum of a rational and irrational number is irrational.

8. Write the negative of the Statements.

Q : All cats scratch

Sol:

∼ q : All cats do not scratch

or we may say that there is at least one cat which does not scratch.

9. There are three colored dice of red, white and black color. These dice are placed in a

bag. One die is drawn at random from the bag and rolled its color and the number on its

uppermost face is noted. Describe the sample space for this experiment.

Sol:

Sample Space =

{(R,1),(R,2),(R,3),(R,4),(R,5),(R,6)

(B,1),(B,2),(B,3),(B,4),(B,5),(B,6),

(W,1) , (W,2) ,(W,3), (W,4),(W,5), (W,6)}




5555

10. The standard deviation of 6 observations is 4. If each observation is multiplied by 3, find

the new standard deviation of the resulting observations.

Sol: standard deviation =4

n=6

( ) ( )

( )

( ) ( )

( )

2

2

Var X S.D.

Var X 16

New Var X 3 16 144

New S.D. New Var X

12

=

∴ =

⇒ = × =

⇒ =

=

Section ‘B’

Questions numbers 11 to 22 carry 4 marks each.

11. tan o o o oθtan(θ 60 ) tanθtan(θ 60 ) tan(θ 60 )tan(θ 60 ) 3+ + − + + − = −

Sol.

LHS:- tanθ0 0

0 0

tanθ tan60 tanθ tan60tanθ

1 tanθtan60 1 tanθtan60

+ −+

− +

0 0

0 0

tanθ tan60 tanθ tan60

1 tanθtan60 1 tanθtan60

+ −+

− +

tanθ 3 tanθ 3tanθ tanθ

1 3 tanθ 1 3 tanθ

+ −= + − +

(tanθ 3)(tanθ 3)

(1 tan 3)(1 3 tanθ)

+ −+

− +

2 2 2

2

(tan θ 3 tanθ)(1 3 tanθ) (tan θ 3 tanθ)(1 3 tanθ) tan θ 3

1 3tan θ

+ + + − − + −⇒

−

2 3 2 2 3 2 2

2

tan θ 3 tan θ 3 tanθ 3tan θ tan θ 3 tan θ 3 tanθ 3tan θ tan θ 3

1 3tan θ

+ + + + − − + + −⇒

−

2 2

2 2

9tan θ 3 3(1 3tan θ)

1 3tan θ 1 3tan θ

− − −⇒ ⇒

− −

= –3

=RHS

Hence, Proved.




6666

12. Sketch the graph of y = cos2x and y = cos 2π

x4

−

on the same scale.

Sol:

OR

Solve 22sin x 3cos x 1 0+ + =

Sol:

22sin x 3cos x 1 0+ + =

2

2

2(1 cos x) 3 cosx 1 0

2 2cos x 3cosx 1 0

− + + =

− + + =




7777

2

2

2cos x 3cosx 3 0

3 ( 3) 4(2)( 3) 3 27cos x

2 2 4

− − =

+ + − − − + ±⇒ = =

×

3 27cos x

4

3 3 3cosx

4

±=

±⇒ =

4 3 2 3 3cos x , ;cos x

4 2 2

− −= =

5πx 2nπ

6= ± ;

5πcos x cos ,n Z.

6

= ∈

13. Prove by the principle of mathematical Induction:

2 3 n n 11.2 2.2 3.2 ....... n.2 (n 1)2 2++ + + + = − +

Sol:

2 3 n n 1P(n) 1.2 2.2 3.2 ...... n.2 (n 1)2 2+= + + + + = − +

2P(1) 2 (1 1)2 2 2= = − + =

Let P(k) be true then

2 2 k k 11.2 2.2 3.2 ...... k.2 (k 1)2 2++ + + + = − + ………(i)

We wish to show that P(k+1) is true. For this we have to show that

2 3 k k 1 k 21.2 2.2 3.2 ..... k(2) (k 1)2 k2 2+ ++ + + + + + = +

From -1

k 1 k 1LHS 2 (k 1)2 (k 1)2+ += + − + +

k 1 k 22 2 (k 1 k 1) k2 2+ += + − + + = + RHS=

So P(k) is true

P(k 1)∴ + is true

Hence by the principle of mathematical induction, the given result is true n N∀ ∈ .




8888

14. Find the coefficient of 10x in the expansion of

20

2 12x

x

−

Sol.

10x in the expansion of

20

2 12x

x

−

Suppose th(r 1)+ term contains 10x in the binomial expansion of

20

2 12x

x

−

Now ( )r

20 r20 2

r 1 r

1T C 2x

x

−

+

− =

20 20 r 40 2r r r

rc (2) (x) ( 1) (x)− − −= −

40 3r r 20 20 r

r(x) ( 1) c (2)− −= − …(1)

This term will contain 10x , if

40 3r 10− =

40 10 3r− =

30 3r=

r 10=

So ( )10 1 th+ i.e.11th term contain 10x .

Putting r 10= in (i) , we get

( ) ( )40 3(10) 10 20 20 10

10x 1 C (2)

− −−

10 20 10

10x c (2)×

∴ Coefficient of x10 is 20 10

10c (2)

15. The sum of three numbers in G.P. is 56. If we subtract 1 , 7 , 21 from these numbers in

that order, we obtain an A.P. Find the numbers.

Sol:




9999

A . T . Q.

2a ar ar 56+ + = ------------ (1)

2(a 1), (ar 7), (ar 21)⇒ − − −

22(ar 7) a 1 ar 21⇒ − = − + −

22 ar 14 a 22 ar⇒ − = − +

22 ar a ar 8⇒ = + −

From 1, we have

2 ar = 56 − ar − 8

3 ar = 48

ar = 16

a = 16

r

Put this value in 1

216 16 16r r 56

r r r

⇒ + + =

1616 16r 56

r⇒ + + =

216 16r 16r56

r

+ +⇒ =

216 16r 16r 56r⇒ + + =

216r 40r 16 0⇒ − + =

22r 5r 2 0⇒ − + =

22r 4r 1r 2 0⇒ − − + =

2r (r 2) 1 (r 2) 0⇒ − − − =

(2r 1) (r 2) 0⇒ − − =

(2r 1) 0 , r 2 0⇒ − = − =

r = 2 .




10101010

Hence a = 8 .

no. are 8, 16, 32.

OR

Sum the series to n terms: 2 + 10+ 30 + 68 + 130 + …

Sol:

n n 1

n n

n 1 n

n n 1

n n 1

S 2 10 30 68 130 ...............T

S 2 10 30 68 130 ...............T

____________________________________

0 2 8 20 38 62 ................T T

T 2 8 20 38 ..........T

T 2 8 20 38 ..........T

−

−

−

−

= + + + + +

= + + + + +

= + + + + + −

= + + + +

= + + + +

[ ]

( )( )

( )( )

( )( )

( )( ) ( )

n 2 n 1

n 1 n 2

n 1

n 1

n

2

2

n

____________________________________

0 2 6 12 18 24........T T

T 2 6 1 2 3 4............T

n 2 n 1T 2 6

2

T 2 3 n 2 n 1

T 2 3 n 1 n

3n 3n 2

S 3 n 3 n 2 1

n n 1 2n 1 n n 13 3 2n

6 2

− −

− −

−

−

= + + + + −

= + + + +

− −= +

= + − −

= + −

= − +

= − +

+ + += − +

∑ ∑ ∑

( )( )( )

( )

( )[ ]

( )

( )

2

3 2

n n 1 2n 1 3n n 1 2n

2 2

2n 1 3n n 1 2n

2 2

n n 1 n 1 2n

n n 1 2n

n n n n 1

+ += − + +

+ = + − +

= + − +

= − +

= + = +




11111111

16. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it . The

resulting mixture is to be more than 4% but less than 6% boric acid . if there are 640 litres of

the 8 % solution , how many litres of 2 % solution will have to be added?

Sol.

Let x litre of 80% acid solution be added to 640 litre of 2% solution of acid

Total quantity of mixture = (640 + x)

Total acid content in 640 + x litre of mixture

= 8 2

x 640100 100

× + ×

It is given that acid content in the resulting mixture must be between 4% to 6%

4% of (640 + x) < 8 2

x 640100 100

× + ×

< 6% of ( 640 + x)

( ) ( )

( ) ( )

( )

4 8x 2 640 6640 x 640 x

100 100 100 100

4 640 x 8x 2 640 6 640 x

640 4 4x 640 2 8x 6 640 6x 640 2

Taking 1st inequality

4x 640 2x

8

x 160 2x

2

x 320 2x

x 320 .....(1)

Taking 2nd inequality

8x 640 4 6x

640 4 6xx

× × + < + < +

+ < + × < +

× + − × < < × + − ×

+ ×⇒ <

+ ×⇒ <

⇒ + <

⇒ >

< +

× +⇒ <

8

3xx 320

4

or x 1280 ....(ii)

From (i) & (ii)

⇒ < +

<




12121212

More than 320 litres but less than 1280 litres of 2% solution will be added.

17. Solve 2(2 i)x (5 i)x 2(1 i) 0+ − − + − =

Sol:

2(2 i)x (5 i)x 2(1 i) 0+ − − + − =

25 i (5 i) 8(1 i)(2 i)x

4 2i

− ± − − − +=

+

x25 i 25 1 10i 8(2 2i i i )

4 2i

− ± − − − − + −=

+

5 i 24 10i 16 8i 8x

4 2i

− ± − − + −=

+

5 i 2i

4 2i

− ± −=

+

a ib 2i+ =

2 2a b 2iab 2i− + = −

2 2a b 0− =

2ab 2,

ab 1

= −

=

2 2 2 2 2 2 2 2(a b ) (a b ) 4a b+ = − +

2 2

0 4

4

a b 2

= +

=

+ = ±

a 1;b 1= = −

5 i 1 i

4 2i

− ± −⇒

+

5 i 1 i 5 i 1 ix ,x

4 2i 4 2i

− + − − − += =

+ +

6 2i 4x ,x

4 2i 4 2i

−= =

+ +




13131313

(6 2i)(4 2i) 4(4 2i)x ,x

16 4 16 4

− − −= =

+ +

24 12i 8i 4 16 8ix ,x

20 20

− − − −= =

4 2x 1 i ,x i

5 5= − = −

4 2x 1 i, i

5 5= − −

18. Express 1 – sinα + i cosα in the form r(cosθ isinθ):+

Sol:

1 sin icos− α + α

( ) ( )2 2 2

2

|z| 1 sin cos 2

cos 1 sin cos cos sintanθ

1 sin 1 sin 1 sin

= + − α + α =

α + α α + α α= × =

− α + α − α

( )

( ){ }( )

1

1 1

1tanθ tan

cos

tanθ sec tan

θ tan sec tan

r 2 cos tan sec tan isin{tan (sec tan }

−

− −

= + αα

= α + α

= α + α

= α + α + α + α

OR

Find the modulus and argument of 1 − i and hence express each of them in the

polar form:

Sol:

1 − i

Let z 1 i. then= −

2 2|z| 1 ( 1) 2= + − =




14141414

Im(z)let tan .then

Re(z)

1tan 1

1

π

4

α =

−α = =

⇒α =

Since the point (1, −1) lies in the fourth quadrant, therefore the argument of z is

given by θ = − α = − π

4

So, in the polar form of z = 1 – i is r (cos θ + i sinθ) = 2π π

cos isin4 4

− + −

= 2 π π

cos isin4 4

−

19. Show that the product of perpendiculars on the line x y

cosθ sinθ 1a b

+ = from the points

2 2 2( a b ,0) is b .± −

Sol:

Line given

x ycosθ sinθ 1

a b⇒ + =

xb cosθ y a sinθ ab⇒ + =

from the point ( )2 2a b , 0± −

2 2 2 2

2 2 2 2 2 2 2 2

a b b cosθ ab a b b cosθ ab

b cos θ a sin θ b cos θ a sin θ

− × − − − × −⇒ ×

+ +

( ) ( )2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2

a b b cos θ ab a b cosθ ab a b cosθ a b

b cos θ a sin θ

− − − − + − +⇒

+




15151515

2 2 2 4 2 2 2

2 2 2 2

a b cos θ b cos θ a b

b cos θ a sin θ

− + +⇒

+

( )2 2 2 2 2 2

2 2 2 2

cos θb a b a b

b cos θ a sin θ

− + + +⇒

+

( )( )( )

2 2 2 2 2

2 2 2 2

b cos θ a b a

b cos θ a 1 cos θ

− + +⇒

+ −

( )( )2 2 2 2 2

2 2 2 2 2

b cos θ a b a

b cos θ a a cos θ

− + +⇒

+ −

( )( )( )( )

2 2 2 2 2

2 2 2 2

b cos θ a b a

cos θ b a a

− + +⇒

− +

2b⇒

Hence Proved.

20. If y = x a

a x+ , prove that 2xy

dy x a

dx a x

= −

Sol:

1 1

2 2x ay

a x

= +

3

2

3

2

dy 1 1 1. a.

dx a 2 x2x

1 1 a

2 axx

⇒ = −

= −

3

2

A.T.Q

x a 1 1 a2.x. .

a x 2 axx

= + −




16161616

3

2

x a x x a

a x axx

x a x a

a x a x

= + −

= + −

2 2x a[by using (a b)(a b) a b ]

a x= − + − = −

Hence Proved

OR

Find the coordinates of the point of intersection of the axis and the directrix of the parabola

whose focus is (3,3) and directrix is 3x – 4y = 2. Find also its length of the latus – rectum.

Sol:

Directrix 3x 4y 2→ − =

⇒Slope = 3

3/ 4( 4)

−=

−

x 0 2/3

y –1/2 0

Axis is ⊥ to Directrix and passes through focus

neq of axis :




17171717

1y 3 (x 3)

3/ 4

3y 9 4x 12

−⇒ − = −

⇒ − = − +

⇒ 4x + 3y = 21

axis and directrix intersect at Q

3x – 4y = 2 (1) × 4

4x + 3y = 21 (2) × 3

12x – 16 y = 8

12 x + 9 y = 63

– – –

25y = 55

11

y5

= and 18

x5

=

18 11Q ,

5 5

→ Length of latus pectum = 2 (SQ)

=

2 218 11

2 . 3 35 5

− + −

2. 13

5⇒

21. Evaluate: π

x4

1 tan xlim

1 2sin x→

−

−

Sol.




18181818

( )( )

( )( )( )

( )

πx

4

h 0

h 0

h 0

h 0

1 tan xlim

1 2sin x

πPut x h

4

π1 tan h

4lim

π1 2sin h

4

1 tanh1

1 tanhlim

1 11 2 cosh sinh

2 2

2tanhlim

1 tanh 1 cosh sinh

2tanhlim

1 cosh 1 cosh1 tanh sinh

1 cosh

2l

→

→

→

→

→

−

−

= +

− +

⇒

− +

+ −

− ⇒

− +

−⇒

− − −

−⇒

− +− − +

⇒ −

( )( )

( )( )

( )( )

( )( )

2h 0

2h 0

h 0

tanhim

sin h1 tanh sinh

1 cosh

sinh2lim

sin hcosh sinh sinh

1 cosh

12lim

sin hcosh sinh 1

1 cosh

12

1 0 0 1

2

→

→

→

− − +

⇒ −

− − +

⇒ −

− − +

⇒ −− −

⇒

22. What is the probability that a leap year has 53 Sundays and 53 Mondays?

Sol:




19191919

In a leap year, there are 366 days

We have,

366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays and 52 Mondays.

The remaining two days can be:

1) Sunday and Monday 2) Monday and Tuesday

3) Tuesday and Wednesday 4) Wednesday and Thursday

5) Thursday and Friday 6) Friday and Saturday

7) Saturday and Sunday

Let A be the event that a leap year has 53 Sundays and 53 Mondays.

∴ A will happen in case (1) only.

∴ Required probability = 1

7

Section ‘C’

Question numbers 23 to 29 carry 6 marks each.

23. Prove that

(i) cos 3π 3π

x cos x 2sin x4 4

+ − − = −

(ii) π π

cos x cos x 2cos x4 4

+ + − =

Sol:

(i) cos 3π 3π

x cos x 2sin x4 4

+ − − = −

LHS: cos 3π 3π

x cos x4 4

+ − −

= cos (135 o + x) – cos (135 o – x)

= – 2 sin o o o o135 x 135 x 135 x 135 x

sin2 2

+ + − + − +




20202020

= – 2 sin o270 2x

sin2 2

= – 2 sin 135 o sin x

0 02sin(90 45 )sin x= − +

= – 2(sin 90 o cos 45 o + cos 90 o sin 45 o ) sin x

= –21 1

(1) 0 sin x2 2

+

= 2sin x RHS− =

Hence Proved.

(ii) π π

cos x cos x 2cos x4 4

+ + − =

LHS:π π

x cos x cos(45 x) cos(45 x)4 4

cos

+ + − = + + −

= 2 cos 45 x 45 x 45 x 45 x

cos2 2

+ + − + − +

02cos45 cos x=

2cosx 2cosx

2= =

RHS=

Hence proved.

24. In a survey of 60 people, it was found that 25 people read newspaper H, 26

read newspaper T, 26 read newspapers I, 9 read both H and I, 11 read both H and

T, 8 read both T and I, 3 read all three newspapers. Find:

(i) the number of people who read at least one of the newspapers.

(ii) the number of people who read exactly one newspaper.




21212121

Sol:

n(H) 25

n(T) 26

n(I) 26

n(H I) 9

n(H I) 11

n(T I) 8

=

=

=

∩ =

∩ =

∩ =

n(H T S) 3∩ ∩ =

[ ]

(i)n(T' H' T) ?

n T H' ?

N n(T I H) ?

n(T I H) n(T) n(I) n(H) n(T I) n(I H) n(T H) n(F T H)

25 26 26 9 11 8 3

80 28

52

∩ ∩ =

∪ Ι ∪ =

− ∪ ∪ =

∪ ∪ = + + − ∩ − ∩ − ∩ + ∩ ∩

= + + − − − +

= −

=

(ii) n(T' H' I) n(T H' I') n(T' H I') ?

n(T' H' I) n(T H)' I)

n(I) n(T H) I)

n(I) n((T I) (H I))

n(I) [n(T I) n(H I) n(T H I)

26 [8 9 3]

26 [14]

12

∩ ∩ + ∩ ∩ + ∩ ∩ =

→ ∩ ∩ = ∪ ∩

= − ∪ ∩

= − ∩ ∪ ∩

= − ∩ + ∩ − ∩ ∩

= − + −

= −

=




22222222

n(T H' I') n(T (H I)')

n(T) n(H I) t)

n(t) (n(H T) n(I I)

→ ∩ ∩ = ∩ ∪

= − ∩ ∩

= − ∩ ∪ ∩

n(T) {n(H T) n(I T) n(H I I)}

26 {11 8 3}

26 16

10

= − ∩ + ∩ − ∩ ∩

= − + −

= −

=

[ ]

n(T' H' I') n(H (T I)')

n(H) (n(T I) (H))

n(H) (n(T H) n(I H))

n(H) n(T H) n(I H) n(T H I)

25 {11 9 3}

25 {17}

8

∩ ∩ = ∩ ∪

= − ∪ ∩

= − ∩ ∪ ∩

= − ∩ + ∩ − ∩ ∩

= − + −

= −

=

n(T' H' I) n(T' H I') n(T H' I')

12 10 8 30

∴ ∩ ∩ + ∩ ∩ + ∩ ∩

= + + =

Or

[ ]n(T) n(H) n(I) 2{n(T H) n(H I) n(I I)} 3n (T H I)⇒ + + − ∩ + ∩ + ∩ + ∩ ∩

26 26 25 2{11 9 8} 3(3)

77 56 9

30

⇒ + + − + + +

= − +

=

25. If sec α

(θ α) sec(θ α) 2secθ,prove that cosθ 2cos2

+ + − = = ±

Sol.

1 1 2

cos(θ α) cos(θ α) cosθ+ =

+ −

1 1 2

cosθ cosα sinθ sinα cosθ cosα sinθ sinα cosθ⇒ + =

− +




23232323

2 2 2 2

2cosθ cosα 2

cos θcos α sin θsin α cosθ⇒ =

−

2 2 2 2 2

2 cosα 2

cos θ cos α sin θ sin α cos θ⇒ =

−

⇒ 22 2 2 2

cos α 1

cos θcos θ cos α (1 cos θ)(1 cos α)=

− − −

⇒ 22 2 2 2 2 2

cos α 1

cos θcos θ cos α 1 cos α cos θ cos θcos α=

− − − +

⇒ 2 2 2

cosα 1

cos α cos θ 1 cos θ=

−

⇒ 2 2 2cosα cos θ cos α cos θ 1= + −

⇒ 2 2 2cosα cos θ cos α 1 cos θ 0− + − =

⇒ 2 2cos θ (cosα 1) (1 cos α) 0− + − =

⇒ 2 2cos θ(cosα 1) 1 cos α− =− +

22

2

2

2 2

2 2

cos α 1cos θ

cosα 1

(cos α 1)(cos α 1)cos θ

(cosα 1)

cos θ cosα 1

αcos θ 2cos 1 1

2

cos θ cos α

2 2

α cosθcos

2 2

−⇒ =

−

+ −⇒ =

−

⇒ = +

⇒ = − +

⇒ =

⇒ = ±

or

If the arcs of the same length in two circles subtend angles 65 o and 110 o at the

centre, find the ratio of their radii.




24242424

Sol: Let 1 2r andr be the radii of the given circles and let this of same length (l)

subtends angles of 065 and 0110 at their centers

0 0

0 65π 13π65

180 36

= =

0 0

0 110π 11π110

180 18

= =

1 2

13π 11πr S; r S

36 18= =

cs

θr

=

∵

1 2

13π 11πr r

36 18⇒ =

1

2

r 22

r 13⇒ =

26. 7 5 3 2n n n n 37

n7 5 3 2 210

+ + + − is positive integer for all n N∈ .

Sol:

7 5 3 2n n n n 37n

7 5 3 2 210+ + + − is an integer

1 1 1 1 37P(1)

7 5 3 2 210= + + + −

30 42 70 105 37

210

+ + + −=

1= which is an integer.

P(1)∴ is true, Let P (k) be true. Then

7 5 3 2k k k k 37k

7 5 3 2 210+ + + − is an integer

Let 7 5 3 2k k k k 37

k7 5 3 2 210

+ + + − λ;λ Z= ∈ .

We small now show that P(k 1)+ is true for which we have show that




25252525

7 5 3 2(k 1) (k 1) (k 1) (k 1) 37(k 1)

7 5 3 2 210

+ + + + ++ + + − is an integer.

( )7 6 5 4 3 21k 7k 21k 35k 35k 21k 7k 1

7⇒ + + + + + + +

( ) ( )5 4 3 2 3 21 1k 5k 10k 10k 5k 1 k 3k 3k 1

5 3+ + + − + + + + + + ( )

( )2 37k 371k 1 2k

2 210

++ + + −

7 5 3 26 5 4 3 2k k k k 37k

k 3k 6k 7k 7k 4k 17 5 3 2 210

= + + + − + + + + + + +

6 5 4 3 2int eger k 3k 6k 7k 7k 4k 1= + + + + + + + from ….(i)

Which is an integer.

P(m) is true ⇒ (m +1) is true

Hence by the principle of mathematical induction P(n) is true for

All n N∈ . i.e. 7 5 3n n n 62

n7 5 3 165

+ + + is an integer.

27. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this

be done when the committee consists of

(i) Exactly 3 girls ?

(ii) At least 3 girls

(iii) At most 3 girls?

Sol.

(i)

A committee consisting of 3 girls and 4 boys can be formed in 4 9

3 4C C× ways

= 4 9

1 4

4! 9 8 7 6C C

1 1.2.3.4

× × ×× = ×

= 504 ways.

(ii)

A committee having at least 3 girls will consists of (a) 3 girls 4 boys (b) 4 girls 3 boys.




26262626

This can be done in 4 9 4 9

3 4 4 3C C C C ways× + ×

= 4 9 8 7 6 9 8 70

1 ways1 1 2 3 4 1 2 3

× × × × ×× + ×

× × × × ×

= 504 + 84 ways

= 588 ways.

(iii)

A committee having atmost 3 girls will consists of

(i) No girls, 7 boys (ii) 1 girl, 6 boys (iii) 2 girls,5 boys (iv) 3 girls, 4 boys.

This can be done in 9 4 9 4 9 4 9

7 1 6 2 5 3 4C C C C C C C ways+ × + × + ×

= 9 4 9 4 9 4 9

2 1 3 2 4 1 4C C C C C C C ways+ × + × + ×

= 9 8 4 9 8 7 4 3 9 8 7 6 4 9 8 7 6

1 2 1 1 2 3 1 2 1 2 3 4 1 1 2 3 4

× × × × × × × × × ×+ × + × + ×

× × × × × × × × × ×

= 36 + 4 ×84 + 6 ×126 + 4 ×126 ways

= 36 + 336 + 126(6 + 4 ) ways

= 36 + 336 + 1260 ways

= 1632.

28. A man arranges to pay off a debt of Rs 3600 by 40 annual installment which form an

arithmetic series. When 30 of the installments are paid, he dies leaving one third of the debt

unpaid, find the value of the first installment.

Sol:

Let the first installment be=Rs x

Difference in each installment=Rs d

Total depth =3600

No of installment=40

[ ]n

nS 2a (n 1)d ,

2= + −




27272727

[ ]40

3600 2a (40 1)d ,2

= + −

[ ]3600 20 2a 39d ,= +

180 2a 39d.............[1]= +

now according to question

Left loan=1\3 rd

paid loan=2\3rd

30

2s 3600 2400

3= × =

[ ]n

nS 2a (n 1)d ,

2= + −

[ ]30

2400 2a (30 1)d ,2

= + −

[ ]2400 15 2a 29d ,= +

2a 29d 160.............[2]+ =

solving (1) and (2)

2a 39d 180

2a 29d 160

..............................

10d 20

.................................

d 2

+ =

+ =

− − −

=

=

Put d=2 in equation [1]

2a 39(2) 180

2a 78 180

2a 102

a 51

+ =

+ =

=

=

the first instalment is Rs 51∴




28282828

or

There are 25 trees at equal distance of 5 meters in a line with a well, the distance of the

well from the nearest tree being 10 meters. A gardener waters all the trees separately

starting from the well and he returns to the well after watering each tree to get water

for the next. Find the total distance the gardener will cover in order to water all the

trees.

Sol:

we have,

1

2

3

25

d distance of the well form nearest tree 2 10m

d distance of the well form nearest tree 2(10 5)m

d distance of the well form nearest tree 2(10 2 5)m

.

.

.

.

d dis tance of the well form nearest tree 2(10 24 5)m

= = ×

= = +

= = + ×

= = + ×

Therefore, total distance the garden will cover in order to water

All the trees=1 2 3 25

d d d ............... d+ + +

{ }(2 10) 2(10 5) 2(10 2 5) ............ 2(10 24 5) meter× + + + + × + + + ×

2[10 (10 5) (10 2 5) ............ (10 25 5)meter= + + + + × + + + ×

2[(10 10.......... 10(25times)] [5 5 2 5 3 ..... 25 5]meter= + + + + × + × + + ×

2[250 5(1 2 3 ....... 25)

252[250 5 (1 25)]

2

1252[250 26]

2

= + + + + +

= + × +

= + ×




29292929

2[250 125 13]

2[250 1625]

2 1875

3750meters

= + ×

= +

= ×

=

29. From the data given below state which group is more variable 1 2

G orG ?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70 – 80

Group 1

G 9 17 32 33 40 10 9

Group2

G 10 20 30 25 43 15 7

Sol: For Group 1

G

Class-interval Frequency

if

Mid-values

ix

ui=i

x 45

10

−

i if u 2

i if u

10–20

20–30

30–40

40–50

50–60

60–70

70–80

9

17

32

33

40

10

9

15

25

35

45

55

65

75

–3

–2

–1

0

1

2

3

–27

–34

–32

0

40

20

27

81

68

32

0

40

40

81

N=150 –6 342

Here N=150 i i

f u 6= −∑ 2

i if u 342=∑




30303030

( )

i i

1Mean X A h f u

N

1X 45 10 6

150

45 4

41

∴ = = +

⇒ = + −

= −

=

∑

( )

( )( )

[ ]

2

2 2

i i i i

2

1 1Var X h f u f u

N N

1 61100 342

150 150

342 36100

150 150 150

100 2.28 0.0016

227.84

= −

− = −

= − ×

= −

=

∑ ∑

For Group 1

G

Class-interval Frequency

if

Mid-values

ix

ui=i

x 45

10

−

i if u 2

i if u

10–20

20–30

30–40

40–50

50–60

60–70

70–80

10

20

30

25

43

15

7

15

25

35

45

55

65

75

–3

–2

–1

0

1

2

3

–30

–40

–30

0

43

30

21

90

80

30

0

43

60

63

N=150 –6 366




31313131

Here N=150, i i

f u 6= −∑ , 2

i if u 366=∑

( )

i i

1Mean X A h f u

N

1X 45 10 6

150

45 4

41

∴ = = +

⇒ = + −

= −

=

∑

( )

( )( )

[ ]

2

2 2

i i i i

2

1 1Var X h f u f u

N N

1 61100 366

150 150

366 36100

150 150 150

100 2.28 0.0016

100 2.4384

243.84

= −

− = −

= − ×

= −

= ×

=

∑ ∑

∴ G2 is more variable.




32323232




33333333

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