Class X

Mathematics

CBSE Question Paper, 2009

Maximum Marks: 80 Time: 3 Hrs

1. All questions are compulsory.

2. The question paper consist of 30questions divided into four sections A, B,C and D. Section A comprises of 10 questions of one mark each, section

B comprises of 5 questions of two marks each ,section C comprises of 10 questions of three marks each and section D comprises of 5 questions of six marks each.

3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.

4. In question on construction, the drawing should be neat and exactly as per the given measurements.

5. Use of calculators is not permitted. You may ask for mathematical tables,

if required.

6. There is no overall choice. However, internal choice has been provided in

one question of 02 marks each, three questions of 03 marks each and two questions of 06 marks each. You have to attempt only one of the alternatives in all such questions.

Section A

Q.1 The decimal expansion of the rational number 4 3

43

2 .5, will terminate

after how many places of decimals?

Q.2 For what value of k, (-4) is a zero of the polynomial x² - x – (2k+ 2)?

Q.3 For what value of p, are 2p – 1, 7 and 3p three consecutive terms of

an A.P.?

Q.4 In Fig. 1, CP and CQ are tangents to a circle with centre O. ARB is

another tangent touching the circle at R. If CP = 11 cm, and BC = 7

cm, then find the length of BR.

Fig1

Q.5 In Fig. 2, ∠M = ∠N = 46°k. Express x in terms of a, b and c where a,

b and c are length of LM, MN and NK respectively.

Fig 2

Q.6 If sin θ =1

3, then find the value of (2 cot² θ + 2).

Q.7 Find the value of a, so that the point (3, a) lies on the line represented

by 2x – 3y = 5.

Q.8 A cylinder and a cone are of same base radius and of same height.

Find the ratio of the volume of cylinder to that of the cone.

Q.9 Find the distance between the points 8 2,2 and ,2

5 5

−

.

A

B

R CO

P

Q

A

B

R CO

P

Q

Q.10 Write the median class of the following distribution:

Classes Frequency

0 – 10 4

10 – 20 4

20 – 30 8

30 – 40 10

40 – 50 12

50 – 60 8

60 – 70 4

Section B

Q.11 If the polynomial 6x4 + 8x³ + 17x² + 21x + 7 is divided by another

polynomial 3x² + 4x + 1, the remainder comes out to be (ax + b), find

a and b.

Q.12 Find the value(s) of k for which the pair of linear equations

kx + 3y = k – 2 and 12x + ky = k has no solution.

Q.13 If Sn, the sum of first n terms of an A.P. is given by Sn = 3n²-4n, then

find its nth term.

Q.14 Two tangents PA and PB are drawn to a circle with centre O from an

external point P. Prove that ∠APB = 2∠OAB.

Fig 3

OR

Prove that the parallelogram circumscribing a circle is a circle.

A

B

OP

A

B

OP

Q.15 Simplify: 3 3sin cos

sin cossin cos

θ + θ+ θ θ

θ + θ

Section C

Q.16 Prove that 5 is an irrational number.

Q.17 Solve the following pair of equations:

5 1

2x 1 y 2

+ =− −

6 3

1x 1 y 2

− =− −

Q.18 The sum of 4th and 8th terms of an A.P. is 24 and sum of 6th and 10th

terms is 44. Find A.P.

Q.19 Construct a ∆ABC in which BC = 6.5 cm, AB = 4.5 cm and ∠ABC =

60°. Construct a triangle similar to this triangle whose sides are 3

4 of

the corresponding sides of the triangle ABC.

Q.20 In fig. 4, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ ABC ~ ∆

ADE and hence find the lengths of AE and DE.

Fig 4

Or

A

D

CB

E

12 cm

3 cm

2 cm

A

D

CB

E

12 cm

3 cm

2 cm

In Fig. 5, DEFG is a square and ∠BAC = 90°. Show that DE²=BD x EC.

Fig 5

Q.21 Find the value of sin 30° geometrically.

Or

Without using trigonometrical tables, evaluate:

cos58 sin22 cos38 cosec52

sin32 cos68 tan18 .tan35 tan60 tan72 tan55

° ° ° °+ −

° ° ° ° ° ° °

Q.22 Find the point on y – axis which is equidistant from the points (5, -2)

and (-3, 2).

Or

The line segment joining the points A (2, 1) and B(5, -8) is trisected at

the points P and Q such that P is nearer to A. If P also lies on the line

given by 2x – y + k = 0, find the value of k.

Q.23 If P (x, y) is any point on the line joining the points A(a, 0) and B(0, b)

then show that x y

1a b

+ = .

A

C

F

GB

D

E

A

C

F

GB

D

E

Q.24 In Fig. 6, PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Find the area of shaded region (take π = 3.14).

Fig 6

Q.25 The king, queen and jack of clubs are removed from a deck of 52

playing cards and the remaining cards are shuffled. A card is drawn

from the remaining cards. Find the probability of getting a card of:

(i) heart (ii) queen (iii) clubs.

Section D

Q.26 The sum of the squares of two consecutive odd numbers is 394. Find

the numbers.

Or

Places A and B are 100 km apart on a highway. One car starts from A

and B at the same time. If the cars travel in the same direction at

different speeds, they meet in 5 hours. If they travel towards other,

they meet in 1 hour. What are the speeds of the two cars?

Q.27 Prove that, if a line is drawn parallel to one side of a triangle to

intersect the other two sides in distinct points, the other two sides are

divided in the same ratio.

P

Q

R

O

P

Q

R

O

Using the above result, do the following:

In Fig. 7, DE||BC and BD = CE. Prove that ∆ ABC is an isosceles

triangle

Fig 7

Q.28 A straight highway leads to the foot of a tower. A man standing at the

top of the tower observes a car at an angle of depression of 30°, which

is approaching the foot of the tower with a uniform speed. Six second

later the angle of depression of the car is found to be 60°. Find the

time taken by the car to reach the foot of the tower from this point.

Q.29 From a solid cylinder whose height is 8 cm and radius 6 cm, a conical

cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find

the volume of the remaining solid correct to two places of decimals.

Also find the total surface area of the remaining solids.(take π=3.1416)

Or

In Fig 8, ABC is a right triangle right angled at A. Find the area of

shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the

incircle of ∆ ABC. (take π = 3.14)

A

E

C

D

B

A

E

C

D

B

Fig 8

Q.30 The following table gives the daily income of 50 workers of a factor:

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

Number of workers 12 14 8 6 10

Find the Mean, Mode and Median of the above data.

A B

C

O

Set 1

CBSE Past Year 2009

Solutions

X Math

Ans1. .. .

×= = =

4 3 4 4 4

43 43 5 2150215

2 5 2 5 10

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

(1 mark)

Ans 2.

( )

( )

is a zero of polynomialx x ( k )

( ) ( k )

k

k k

− − − +

⇒ − − − − + =

⇒ + − − =

⇒ = ⇒ =

2

2

4 2 2

4 4 2 2 0

16 4 2 2 0

2 18 9

(1 mark)

Ans3.

p , , p are three consecutive terms of an AP

( p ) p

p p

p p

−

⇒ − − = −

⇒ − = −

⇒ = ⇒ =

2 1 7 3

7 2 1 3 7

8 2 3 7

5 15 3

(1 mark)

Ans 4. We know that lengths of tangents from an external point to a circle are equal Hence, CP = CQ and BQ = BR

CQ = 11 cm, BC = 7 cm BQ = CQ – BC = 4 cm

⇒ BR = 4 cm.

(1 mark)

Ans 5. ∠M = ∠N = 46o and ∠ K =∠ K (common)

So ∆LMK ~ ∆PNK

(AA similarity)

LM MK

PN NK

LM PNHence,

MK NK

a x

b c c

acx

b c

=

=

⇒ =+

⇒ =+

(1 Mark) Ans6.

sin cosec

To find: cot

We know that cosec cot

cot cosec

cot (cosec )

cosec

θ = ⇒ θ =

θ +

θ − θ =

⇒ θ = θ −

θ + = θ − +

= θ

= × =

2

2 2

2 2

2 2

2

13

3

2 2

1

1

2 2 2 1 2

2

2 9 18

(1 Mark)

Ans7. ( ,a) lies on x y

a

a

a a

− =

⇒ × − × =

⇒ − =

⇒ = ⇒ =

3 2 3 5

2 3 3 5

6 3 5

13 1

3

(1 Mark)

Ans8. Let radius of cylinder and cone = r

Height = h

Volume of cylinder= r h2π

Volume of cone = h21

3π

Volume of cylinder r h:

Volume of coner h

π= =

π

2

2

3 11

3

(1 Mark)

Ans9. Given: Points

A , ,B ,

AB ( )

units

−

= + + −

= +

=

2

2

8 22 2

5 5

2 82 2

5 5

4 0

2

(1 Mark)

Ans10.

Classes Frequency cf

0-10 4 4

10-20 4 8

20-30 8 16

30-40 10 26

40-50 12 38

50-60 8 46

60-70 4 50

N=50 N/2 = 25 so the class with cumulative frequency just greater then 25 is the

class 30-40 Hence, median class is 30 – 40. (1 Mark)

Ans11.

x

x x x x x x

x x x

x x

x x

x

+

+ + + + + +

+ +

− − −

+ +

+ +

− − −

+

2

2 4 3 2

4 3 2

2

2

2 5

3 4 16 8 17 21 7

6 8 2

15 21 7

15 20 5

2

(1 Mark)

ax + b = x+ 2 Hence, a = 1 b = 2 (1 Mark)

Ans12. For the system of equations

k x + 3y = k-2 and 12x + ky = k to be inconsistent i.e no solution

i.e

a b c

a b c

k k

k k

k

k or

1 1 1

2 2 2

2

3 2

12

36

6 6

= ≠

−= ≠

=

= −

(1 Mark)

Hence, for k=6 and -6, the pair of linear equations will have no solution. (1 Mark)

Ans13. Sum of n terms of an AP is given by

th

n n

th

n[ a (n )d]

To find n term

S S

n n (n ) (n ) ( mark)

n n n n n

n

Hence n term of the AP is n ( mark)

−

= + −

=

= −

= − − − + −

= − − − + + −

= −

−

1

2 2

2 2

2 12

3 4 3 1 4 1 1

3 4 3 3 6 4 4

6 7

6 7 1

Ans14. Join OB

APB ( PAB PBA)

PA PB(tangents from an external point to a circle are equal)

PAB PBA

APB PAB ( mark)

OAP (radius is to tangent at the point of contact)

OAB BAP

APB ( OAB)

∠ = − ∠ + ∠

=

⇒ ∠ = ∠

∴ ∠ = − ∠

∠ = ⊥

⇒ ∠ + ∠ =

⇒ ∠ = − − ∠

= − +

�

�

�

�

� �

� �

180

180 2 1

90

90

180 2 90

180 180 2 OAB

APB OAB (mark)

∠

⇒ ∠ = ∠2 1

OR

Given: A circle with centre O and a Parallelogram ABCD circumscribing the

circle To show: ABCD is a rhombus.

Proof: We know that tangents drawn from external point to a circle are Equal.

Hence, AX = AT

BX = BY CY = CZ DZ = DT (1 mark)

ABCD is a � gram ⇒ AB=CD and BC=AD

AB = AX + BX = AT + BY = (AD – TD) +BY = (AD-DZ) + BY

= BC – DZ + BY

=BC – (CD – CZ) +BY =BC – (AB – CY) + BY =BC – AB + BC (1 mark)

⇒ 2 AB = 2 BC

⇒AB = BC

Hence, ABCD is a rhombus.

Ans15.

( )

sin cosConsider sin cos

sin cos

(sin cos )(sin cos sin cos )sin cos ( mark)

sin cos

Using a b (a b)(a b ab)

sin cos sin cos (using sin cos )

( mark)

θ + θ+ θ θ

θ + θ

θ + θ θ + θ − θ θ= + θ θ

θ + θ

+ = + + −

= − θ θ + θ θ θ + θ =

=

3 3

2 2

3 3 2 2

2 2

1

1 1

1 1

Ans16. Let 5 is a rational number

Than,a

b=5 a, b are integers

a and b are co-prime , b≠0 Squaring both sides, we get (1 mark)

a

b

a b

b divides a

b divides a

b divides a

=

=

⇒

⇒

⇒

2

2

2 2

2 2

2

5

5

⇒ a and b have a common factor other 1 (1 mark)

This leads in contradiction as a and b are co-prime Hence, our assumptions is wrong

Hence, 5 is irrational (1 mark)

Ans17. Let

a , bx y

a b

a b ( mark)

a b

a b

b ( mark)

b y y ( mark)y

a

a

a xx

x

Here, x ,y ( mark)

= =− −

⇒ + =

− =

+ =

− =

− + −

=

= = = ⇒ − = ⇒ =−

+ =

⇒ = − =

⇒ = ⇒ = ⇒ − =−

⇒ =

= =

1 1

1 2

5 2

16 3 1

2

30 6 12

30 15 5

21 7 1

1 1 1 12 3 5

3 2 3 2

15 2

3

1 55 2

3 3

1 1 11 3

3 1 3

4

4 5 1

Ans18. Let 1st term =a

Common difference=d

a a

a a

+ =

+ =

4 8

6 10

24

44

a+3d+a+7d=24 2a+10d=24_________(1) (1/2 mark)

a+5d+a+9d=44

2a+14d=44_________(2) (1 mark)

a d

a d

d

2 10 24

2 14 44

4 20

+ =

+ =

− − −

− = −

d=5 (1 mark) 2a+50=24 ⇒2a=-26

=a=-13

Hence, the AP is -13,-8, -3, 2, 7,… (½ mark)

Ans.19

(3 mark)

Ans20. In ABC and ADE∆ ∆

A A (Common)

ACB AED (given)

ABC ADE (by AA similarity)

AB BC AC( mark)

AD DE AE

AB

DE AE

AB BC AC

AB cm ( mark)

Hence, DE cm ( mark)

AE cm

∠ = ∠

∠ = ∠ =

∴ ∆ ∆

⇒ = =

⇒ = =

= +

= + =

⇒ =

=

=

�

∼

2 2 2

90

1

12 5

3

144 25 169

13 1

361

13

15

13

Or

To show:- DE BD EC2= ×

o

BDG BAC

B (Common)

BDG BAC

Similarly CEF CAB ( mark)

BDG FEC

DG BC BD( mark)

EC CF EF

DE BC BD

EC CF DE

DE BD EC ( mark)

∆ ∆

∠

∠ = ∠ =

∆ ∆

∴ ∆ ∆

= =

= =

= ×

∼

∼

∼

2

90

1

1

1

Ans21. Consider an equilateral triangle ABC with side a and AD as perpendicular bisector of ∠A

Hence, AD ⊥ BC

Now AD is also the angle bisector of ∠BAC.

Now in ∆ADB (1 mark)

∠BAD = 30o

o Perpendicular asin

Hypotenuse .a

Hence, sin

= = =

=�

130

2 2

130

2

(1 mark)

(1 mark)

OR

cos sin cos cosec

sin cos tan tan tan tan tan

cos cos( ) sin

sin sin( ) cos

cos cos( ) sin

tan tan( ) cot

tan tan( ) cot

+ −

= − =

= − =

= − =

= − =

= − =

� � � �

� � � � � � �

� � � �

� � � �

� � � �

� � � �

� � � �

58 22 38 52

32 68 18 35 60 72 55

58 90 32 32

22 90 68 68

38 90 52 52

18 90 72 72

35 90 35 55 ( marks)

sin sin sin cosec

sin sin tan tan tan cot cot

( marks). .

= + −

= + − = −

� � � �

� � � � � � �

112

32 22 52 52

32 22 18 35 60 18 35

1 1 11 1 2 1

21 1 3 3

Ans22. Let the point on y-axis be (0, y) (1 mark)

( y) ( ) ( y) ( )

y y y y ( mark)

y

y ( mark)

− − + − = − + − −

+ + + = + − +

= −

= −

2 2 2 2

2 2

2 5 0 2 3 0

4 4 25 4 4 9 1

8 16

2 1

The point on y-axis which is equidistant from (5,-2) & (-3, 2) is (0.-2)

OR

P divides AB in the ratio 1:2

Thus, coordinates of P =

,

, marks

× + × × + ×

=

1 5 2 2 1 8 2 1

3 3

10 13 13 2

P lies on 2x-y+k=0

× − + =

− + = ⇒ = −

102 3 k 0

3

10 8 16 k 0 k 1 marks

3 3 2

Ans23.

Since P (x, y) is a point on line segment AB so P divides AB in certain ratio say k: 1

Applying section formula

a. k. kb .x ,y

k k

aie x ....(i),

k

kby .....(ii)

k

x aFrom(i) k marks

a k x

+ += =

+ +

=+

=+

= ⇒ = −

+

1 0 1 0

1 1

1

1

1 11 1

1 2

Using this in (ii) we get

ab

xy

a

x

ay abb

x x

ay bx ab

x x

ay bx ab

y x

b a

x ymarks

a b

−

=

⇒ = −

+⇒ =

⇒ + =

⇒ + =

⇒ + =

1

1

11 1

2

which is the required condition.

Ans24. PQ = 24 cm PR = 7 cm

Angle is a semicircle is 90�

RPQ

PR PQ QR

QR

QR

QR cm

Radius of the circle cm ( mark)

∴ ∠ =

∴ + =

+ =

= + =

=

∴ =

�

2 2 2

2 2 2

2

90

7 24

576 49 625

25

251

2

Area of shaded region =Area of semicircle – are of ∆PQR (1/2 mark0

( )

.

.( mark)

.

( . )

. cm ( mark)

= × π × × − × ×

× × = −

= −

= −

=

=2

1 25 25 17 24

2 2 2 2

1 3 14 25 25168

2 4

1 1962 5168 1

2 4

1490 825 168

2

1322 825

2

1161 4125

2

Ans25. Total = 52 cards After removing king, queen, jack of clubs, total cards = 49 cards

(i) P (Heart) = 13

49

(ii) P (Queen) =3

49

(iii) P (clubs) =10

49

(1 mark each)

Section – D

Ans26. Let X and X + 2 be two consecutive odd integers (1 mark) According to question,

X (X ) ( mark)

X X X

X X

X X ( mark)

X X X

(X )(X )

X , ( marks)

+ + =

+ + + =

⇒ + − =

⇒ + − =

⇒ + − − =

⇒ + − =

= −

2 2

2 2

2

2

2

2 394 1

4 4 394

2 4 390 0

2 195 0 1

15 13 195 0

15 13 0

15 13 2

So the consecutive integers are 13, 15 or -15, -13 (1 mark)

OR

Let speed of the two cars be x km/h and y km/h

dis tanceSpeed

time

dis tancetime

speed

=

⇒ =

(1 mark)

If cars travel towards each other than relative speed will be (x +y)

Km/h. If cars travel in same direction then relative speed will be (x-y) km/h (1 mark)

According to question,

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

x y ....(ii)x y

= ⇒ + =+

= ⇒ − =−

1001 100

1005 20

(2 mark)

Solving (i) and (ii) we get

x= 60 km/h y = 40 km/h (2 marks)

Ans27. Given: ∆PQR, in which XY||QR intersects other two sides PQ and PR at X and Y respectively

To prove:- Y

Q YR

ΡΧ Ρ=

Χ

Construction: QY and RX and then draw PRΧΜ ⊥ and YN ⊥ PQ (1/2 mark

each)

Proof : Area of ∆ PXY = 1 1

2 2base height YN× × = × ΡΧ ×

Area of ∆ PXY = 1

2PY × ΧΜ

Similarly Area of ∆ QXY = 1

2QX NY×

Area of ∆ RXY = 1

2YR × ΧΜ

Therefore

( )

( )

1

21

2

YNarea Y

area Q Y QQ Y

× ΡΧ ×∆ΡΧ ΡΧ

= =∆ Χ Χ

× Χ × Ν

(1)

( )

( )

1

21

2

PYar XY PY

ar RXY YRYR

× × ΧΜ∆Ρ

= =∆

× × ΧΜ

(2)

∆QXY and ∆RXY are on same base XY and between the same parallels XY

and QR. ∴ area (∆ QXY) = area (∆ RXY) (3)

Therefore from (1) (2) and (3) we have

Y

Q YR

ΡΧ Ρ=

Χ

Hence, proved (21

2 marks)

(ii) Part

Given: DE BC, BD CE=�

To prove: AB = AC Proof In ∆ABC since DE BC using BPT�

So

=

⇒ = =

⇒ =

+ = +

⇒ =

⇒ ∆

AD AE

DB EC

AD DB1

AE EC

AD AE

So AD DB AE EC

AB AC

ABC is isosceles

(2 marks)

Ans28.

(1 mark) Let DC be the tower and initial position of car be A and after 6 sec, it

Reaches point B

In DAC

DCtan

AC

ACDC ...(i)

In DBC

DCtan

BC

DC BC ....(ii)

∆

= =

⇒ =

∆

= =

=

�

�

130

3

3

60 3

3

(1 mark)

(1 mark)

From (i) and (ii)

AC

BC33

=

AB BCBC

AB BC

33

2

+=

⇒ =

(2 marks)

Car is traveling with uniform speed it takes 6 sec. to travel distance AB. So

time taken to travel distance BC will be 3 sec. (1 mark)

Ans29.

Radius of cylinder = 6 cm

Height of cylinder = 8 cm Radius of Cone = 6 cm

Height of Cone = 8 cm (1 mark) Volume of cylinder =

r h

.

cm ( marks)

π

= π

= π

2

2

3

6 8

1288 1

2

Volume of Cone =

.

cm ( marks)

π

= π

2

3

16 8

3

196 1

2

Volume of Remaining solid

Volume of cylinder Volume of cone

( mark)

. . cm ( mark)

= −

= π − π

= π

= × =3

288 96

192 1

192 3 1416 603 19 1

OR

ABC∆ is right angled at A

Given AB = 6 cm BC = 10 cm

So BC AB AC

AC cm ( mark)

Area of ABC cm ( mark)

= +

⇒ =

∆ = × × =

2 2 2

2

8 1

18 6 24 1

2

Join OB, OC and OA

( ) ( ) ( )

{ }

Area ABC ar OBC ar OAB

r

r

r cm ( marks)

∆ = ∆ + ∆

= × × + +

=

⇒ =

124 10 6 8

2

48 24

2 2

So area of circle =4π

Area of shaded portion= Ar( ABC) Ar(circle)

.

. cm

∆ −

= − ×

=2

24 4 3 14

11 44

(2 marks)

Ans30.

Class Midpoint Frequency fixi cf

100-120 110 12 1320 12

120-140 130 14 1820 26

140-160 150 8 1200 34

160-180 170 6 1020 40

180-200 190 10 1900 50

N fi 50= ∑ = 7260

Mean= fixi

.fi

7260 726145 2

50 5

∑= = =

∑ (2 marks)

N = 50 , N

252

= so median class is 120 – 140

Median= 2

Ncf

l hf

− + ×

l= lower limit of median class = 120

225

12

14

N

cf

f

=

=

=

Class size = 20

Median =

− + ×

= + =

25 12120 20

14

120 18.57 138.57

(2 marks)

Mode = 1 0

1 0 2

f fl h

2f f f

− + × − −

Again highest frequency (14) is of class 120 – 140

1 o 2f 14 , f 12 f 8

14 12So mode 120 20

28 20

2120 20

8

mode 125 (2marks)

= = =

− = + × −

= + ×

=

So mean = 145.2 Median = 138.57

And mode = 125