solution cbse-board-class-10-previous-paper-2007

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CBSE CLASS 10th 2007: MATHEMATICS SOLVED PAPER

Time Allowed: 3 Hours Max Marks: 80

General Instruction:

(i) All the questions are compulsory

(ii) The question paper consists of 25 questions divided into three sections- A, B, and C.

Section A contains 7 questions of 2 marks each. Section B contains 12 questions of 3

marks each. Section C contains 6 questions of 5 marks each.

(iii) There is no overall choice. However, an internal choice has been provided in two

questions of two marks each, two questions of three marks each and two questions of

five mark each.

(iv) In questions on construction, the drawing should be neat and exactly as per the given

measurement.

(v) Use of the calculator is not permitted. However, you may ask for Mathematical tables.

Section A: Question No 1 to 7 carry 2 marks each.

Ques [1] : Find the GCD of the following polynomials

12x4 + 324x and 36x

3 + 90x

2 – 54x

Solution:

12x4 + 324x = 12x(x

3 + 27)

= 6 × 2x(x3 + (3)

3)

= 6 × 2x (x + 3)(x

2 + 9 –3x)

36x3 + 90x

2 – 54x = 18x(2x

2 + 5x – 3)

= 6 × 3x (2x2 + 6x – x – 3)

= 6 × 3x (2x (x+ 3) – 1(x + 3))

= 6 × 3x (x+ 3)(2x – 1)

Thus, the GCD of the given polynomials is 6x (x + 3).

Ques [2] : Solve for x and y:

2

2 and 4x y x y

a b a b

Solution:

We are given the equations as

2

2x y

a b …(1)

4x y

a b …(2)



From (1) + (2), we get

22 4

36

2

x y x y

a b a b

x

a

x a

Now, putting the value of x in the equation (1), we get

2 22

4 2

2

2

a y

a b

y

b

y

b

y b

Thus, the solution of the given equations are x = 2a and y = –2b.

OR

Solve for x and y:

31x + 29y = 33 and 29x + 31y = 27

Solution:


31x + 29y = 33 …(1)

29x + 31y = 27 …(2)


31x + 29y + 29x + 31y = 33 + 27

60x + 60y = 60

60(x + y) = 60

x + y = 1 …(3)

From (1) – (2), we get

31x + 29y – 29x – 31y = 33 – 27

2x – 2y = 6

2(x – y) = 6

x – y = 3 …(4)


2x = 4 x = 2

From (3) – (4), we get

2y = –2 y = –1

Thus, we get x = 2 and y = –1

Ques [3]: Find the sum of all the three digits whole numbers which are multiple of 7

Solution:



All the three-digits whole numbers are 100, 101, 102,…, 999.

Out of which, the multiples of 7 are 105, 112, 119, …, 994

Now,

994 = 105 + (n – 1)(7)

994 –105 = 7(n –1)

889 = 7(n –1)

889

1 1277

127 1 128

n

n

Thus, the required sum of the whole numbers which are multiples of 7 is

128

2

128105 994 64 1099 70336

2

n

nS a l

S

Ques [4]: In the figure (1), PQ || AB and PR || AC. Prove that QR || BC.

Solution:

Since PQ || AB in ∆OAB, we get

OP OQ

AP BQ …(1) [ By Basic Proportionally Theorem ]

Since PR || AC in ∆OAC, we get

OP OR

AP RC …(2) [ By Basic Proportionally Theorem ]

From the equations (1) and (2), we get

||OQ OR

QR BCBQ RC

[ By Converse of Basic Proportionally Theorem ]

OR

In the figure (2) incircle of ∆ABC touches its sides AB, BC and CA at D, E

and F respectively. If AB = AC, prove that BE = EC



Solution:

We know that the tangents drawn from an external point to the circle are equal in

length. Thus, we get

Now, we have

AB = AC x + y = x + z y = z BE = EC.

Ques [5]: If the mean of the following frequency distribution is 49, find the missing

frequency p:

Class Frequency

0-20 2

20-40 6

40-60 p

60-80 5

80-100 2

Solution:



Class Class-Mark

(x)

Frequency

(f)

x×f

0-20 10 2 20

20-40 30 6 180

40-60 50 p 50p

60-80 70 5 350

80-100 90 2 180

Total f = 15 + p xf =730 + 50p

Now, we know that the mean of the grouped frequency is

730 5049

15

49 15 730 50

735 49 730 50

5

xfM

f

p

p

p p

p p

p

Ques [6]: A wrist watch is available for Rs. 1000 cash or Rs. 500 as cash down payment

followed by three equal monthly instalments of Rs. 180. Calculate the rate of

interest charged under the installment plan.

Solution:

Here, we have

P = Rs. 1000,

A = Rs. (500 + 180 × 3) = Rs. (500 + 540) = Rs 1040

T = 3 Months = 3 1

12 4 years

SI = A – P = Rs. (1040 – 1000) = Rs. 40.

Thus, we need to find the rate of interest R = ?

We know that

100

1000 140

100 4

16%

PRTSI

R

R

Therefore, the rate of the interest is R = 16% per year.



Ques [7]: An unbiased die is tossed once. Find the probability of getting

(i) A multiple of 2 or 3

(ii) A prime number greater than 2.

Solution:

In an unbiased die, we have {1,2,3,4,5,6}

Thus, we have n(S) = 6

(i) A = multiple of 2 = {2,4,6} n(A) = 3

B = multiple of 3 = {3,6} n(B) = 2

A B = multiple of both 2 and 3 = {6} 1n A B

Thus, the required probability of getting a multiple of 2 or 3 is

3 2 1

6 6 6

4

6

2

3

P A B P A P B P A B

n A n B n A B

n S n S n S

(ii) A = A prime number greater than 2 = {3,5} n(A) = 2

Thus, the required probability of getting a prime number greater than 2 is

2

6

1

3

2

3

n AP A

n S

Section B: Question No 8 to 19 carry 3 marks each.

Ques [8] : Solve the following system of equations graphically

2x + y = 8 and x + 1 = 2y

Solution:


2x + y = 8 …(1)

x + 1 = 2y …(2)

From the equation (1), we get



y = 8 – 2x

On taking values of x, we get the corresponding values of y as given in the table.

From the equation (2), we get

y = 1

2

x

On taking values of x, we get the corresponding values of y as given in the table.

On plotting these points of the two tables, we get graph of the given equations as

below.

Now, we see that both the line intersect at the point (3, 2). Thus, the solution of the

given equation is x = 3 and y = 2

Ques [9] : Simplify the following rational expression in the lowest terms

2 3 2 2 2 2

3 3 2 2 4 4 4

2ax x a ax x a ax x

a x a x x a x

Solution:

2 3 2 2 2 2

3 3 2 2 4 4 4

2 22 4 4

2 22 2 2 2 2

2

2

ax x a ax x a ax x

a x a x x a x

a ax xx a x a x

a ax xa x a ax x x a x

x 1 2 3 4

y 6 4 2 0

x 1 2 3 4

y 1 1.5 2 2.5



2 2 2 2

22 2

2 2

2 2

1

1 1

1

a x a xa x

a x a x a x

a x a x a x

a x a xa x

Ques [10] : If the sum to first n terms of an AP is given by Sn = n(n + 1), find the 20th

term

of the AP.

Solution:

We know that if the sum of n terms of an AP is Sn, then the nth term of the AP is

Tn = Sn – S(n – 1)

Thus,

T20 = 20(20 + 1) – 19(19 + 1) = 20 × 21 – 19 × 20 = 20(21 – 19) = 20 × 2 = 40.

Ques [11] : In a cyclic quadrilateral ABCD, diagonal AC bisects angle C. Prove that the

tangent to circle at A is parallel to the diagonal BD.

Solution:

First, we draw the cyclic quadrilateral ABCD with

tangent AM at the point A.

In the given cyclic quadrilateral ABCD, the

diagonal AC bisects BCD.

Thus, BCA = DCA = θ …(1)

We know that the angles in the same segment of

the circle are equal.

Thus, BDA = BCA = θ …(2)

DBA = DCA = θ …(3)

Now, according to angle segment theorem, the

angle made by the tangent with the chord is equal

to the angle made in the opposite segment of the circle. Thus,

BAM = BDA = θ = DBA [From equation (3)]

Since BAM = DBA, we get the tangent to circle at A is parallel to the

diagonal BD. Proved.

OR

In the figure 3, O is any point in the interior of ∆ABC. OD, OE and OF are

drawn perpendiculars to the sides BC, CA and AB respectively. Prove that



AF2 + BD

2 + CE

2 = OA

2 + OB

2 + OC

2 – OD

2 – OE

2 – OF

2

Solution:

First, we draw OA, OB and OC.

In right triangle OAF, by Pythagorus Theorem, we have

AF2 = OA

2– OF

2 …(1)

In right triangle OBD, by Pythagorus Theorem, we have

BD2 = OB

2– OD

2 …(2)

In right triangle OCE, by Pythagorus Theorem, we have

CE2 = OC

2– OE

2 …(1)

On adding the equations (1), (2) and (3), we get

AF2 + BD

2 + CE

2 = OA

2 + OB

2 + OC

2 – OD

2 – OE

2 – OF

2

Ques [12] : Construct a ∆ABC in which base BC = 6 cm, B = 45 degree and C = 60

degree. Draw a circumcircle of ∆ABC.

Solution:



Ques [13] : The diameter of solid copper sphere is 18 cm. It is melted and drawn into a

wire of uniform cross section. If the length of the wire is 108 m, find its diameter.

Solution:

We know that if a solid is melted and recast into another solid then its volume is

constant. Thus, we have

Volume of Solid Copper Sphere = Volume of Wire

3

2

2

4 18108 100

3 2

49 9 9 108 100

3

r

r

2

2 4 9 9 9 3

3 108 100 10

30.3

10

r

r

Thus, the diameter of the wire is d = 2r = 0.6.

Ques [14] : The expenditure (in rupees) of a family for a month is as follows:

Items Rent Food Education Electricity &

Water

Others

Expenditure 800 3000 1200 400 1800

Solution:

Items Expenditure Central Angle in Pie

Chart

Rent 800 800360 40

7200

Food 3000 3000360 150

7200

Education 1200 1200360 60

7200

Elec & Water 400 400360 20

7200

Others 1800 1800360 90

7200

Total 7200



Ques [15] : From a pack of 52 cards, red face cards are removed. After that a card is

drawn at random from the pack. Find the probability that the card drawn is

(i) A queen

(ii) A red card

(iii) A spade card

Solution:

In a pack of 52 cards, there are 6 red face cards, which are removed. Thus, there are

46 cards are left in the pack of the cards.

Thus, we have n(S) = 46

(i) Now, in the pack of 46 cards, we have 6 queen cards. Thus, we have n(Q) = 6

Thus, the required probability that the card drawn is a queen is

6 3

46 23

n QP Q

n S

(ii) In the pack of 46 cards, we have 20 red cards. Thus, we have n(R) = 20

Thus, the required probability that the card drawn is red card is

20 10

46 23

n RP R

n S

(iii) In the pack of 46 cards, we have 13 spade cards. Thus, we have n(Sp) = 13

Thus, the required probability that the card drawn is spade card is

13

46

n SpP Sp

n S

Ques [16] : Prove that 2 2 2 4

1 1 11 1

tan cot sin sin

Solution:

2 2

2 2

2 2

1 11 1

tan cot

1 cot 1 tan

cos sec

LHS

ec



2 2

2 2

2 4

1

sin cos

1

sin 1 sin

1

sin sinRHS

OR

If A, B and C are the interior angles of a triangle ABC, show that

cos sin2 2

B C A

Solution:

In ∆ABC, we have

A + B + C = 1800

B + C = 180 – A

180

2 2

902 2

B C A

B C A

Taking cos of both angles, we get

cos cos 902 2

cos sin2 2

B C A

B C A

Ques [17]: The coordinates of the mid-points of the sides of a triangle are (4, 3), (6, 0) and

(7, –2). Find the coordinates of the centroid of the triangle.

Solution:

Let the coordinate of the vertices of the ∆ABC are A(x1, y1), B(x2, y2) and

C(x3, y3), whose the mid-points of the sides are D(4, 3), E(6, 0) and F(7, –2).

Thus, we have

2 32 34 8

2

x xx x

…(1)

3 13 16 12

2

x xx x

…(2)

1 21 27 14

2

x xx x

…(3)


1 2 3

1 2 3

3 8 12 14

34

3

x x x

x x x



Now, we have

2 32 33 6

2

y yy y

…(4)

3 13 10 0

2

y yy y

…(5)

1 21 22 4

2

y yy y

…(6)


1 2 3

1 2 3

3 6 0 4

2

3

y y y

y y y

We know that the coordinate of the centroid of ∆ABC is 1 2 3 1 2 3,3 3

x x x y y yG

.

Thus, the coordinate of the centroid of the given ∆ABC is

34 234 23 3, ,

3 3 9 9G G

Ques [18]: If the distance of P(x, y) from two points with coordinates (5, 1) and (–1, 5) is

equal, prove that 3x = 2y.

Solution:

We are given that the distance of the point P(x, y) from two points A(5, 1) and

B(–1, 5) are equal. Thus, we have

2 2

2 2 2 2

2 2 2 2

5 1 1 5

5 1 5 1

5 1 5 1 5 1 5 1

2 4 6 2 6 4

3 2

AP BP

AP BP

x y x y

x x y y

x x x x y y y y

x y

x y

Ques [19]: A loan of amount of Rs. 24600 is to be paid in two equal semi-annual

installment. If the interest is charged at 10% per annum, compounded semi-

annual, find the installment.

Solution:

If P is the amount that is to be paid in n equal installment x at R% per annum

compounded annually, then we have



1 2 3

...

1 1 1 1100 100 100 100

n

x x x xP

R R R R

Here we have, P = Rs. 24600, R = 5% as the amount is paid in two equal

installment semi-annually. Thus, we get

1 2

246005 5

1 1100 100

13230

x x

x

Thus, the installment is Rs. 13230.

Section C: Question No 20 to 25 carry 5 marks each.

Ques [20] : Prove that the angle subtended by an arc at the center is double the angle

subtended by it at any point on the remaining part of the circle.

Using the above, prove the following:

In figure 4, O is the center of the circle. If BAO = 30 degree and

BCO = 40 degree, find the value of AOC.

Solution:

Let ACB be the arc. Draw the radii

OA, OB and OC. Draw AC and BC.

In ∆OAC, OA = OC,

OAC = OCA = θ (let)

Thus, by exterior angle of ∆OAC,

AOD = θ + θ = 2θ

In ∆OBC, OB = OC,

OBC = OCB = α (let)

Thus, by exterior angle of ∆OBC,

BOD = α + α = 2α



Now, the angle made by the arc ACB at the center O is

AOB = AOD + BOD = 2θ + 2α = 2(θ + α) = 2(ACB) Proved.

In the given figure, first we draw OB.

In ∆OAB, OA = OB, we haveOAB = OBA = 30

In ∆OBC, OB = OC, we haveOBC = OCB = 40

Thus, OBC = 30 + 40 = 70

From the above proof, we get AOC = 2 × 70 = 140.

Ques [21] : State and prove Pythagoras Theorem.

Use the above to prove to prove the following:

ABC is an isosceles right triangle, right angled at C. Prove that AB2 = 2AC

2.

Solution:

Pythagoras Theorem:

In a right angled triangle ABC right angled at C, AB2 = AC

2 + BC

2.

Proof:

Draw CD AB.

In ∆ACD and ∆ABC, we have

∠CAD = ∠BAC = 90 – θ

∠ACD = ∠ABC = θ

Thus, we get ∆ACD and ∆ABC are similar.

We get

AC CD AD

AB BC AC

From first and last, we get AC2 = AB × AD …(1)

In ∆CBD and ∆ABC, we have

∠BCD = ∠BAC = 90 – θ

∠CBD = ∠ABC = θ

Thus, we get ∆CBD and ∆ABC are similar.

We get

BC CD BD

AB AC BC

From first and last, we get BC2 = AB × BD …(2)




AC2 + BC

2 = AB × AD + AB × BD = AB × (AD + BD) = AB × AB = AB

2

Thus, we get

In a right angled triangle ABC right angled at C, AB2 = AC

2 + BC

2.

Now, since ABC is an isosceles triangle right angled at C, we have

AC = BC.

Thus, by Pythagoras theorem, we get

AB2 = AC

2 + BC

2 = AC

2 + AC

2 = 2AC

2

Ques [22] : The side of the square exceeds the side of the another square by 4 cm and the

sum of areas of two squares is 400 sq cm. Find the dimension of the square.

Solution:

Let the two squares are ABCD and PQRS, whose sides are x cm and (x + 4) cm

respectively. Thus, we are given

x2 + (x + 4)

2 = 400

x2 + x

2 + 8x + 16 = 400

2x2 + 8x + 16 = 400

x2 + 4x – 192 = 0

x2 + 12x – 16x – 192 = 0

x(x + 12) – 16(x + 12) = 0

(x + 12)(x – 16) = 0

x = –12 and 16

The length of the side of the square cannot be negative. Thus, x = 16 cm

Therefore, the dimensions of the squares are 16 cm and 20 cm.

OR

A fast train takes 3 hours less than a slow train for a journey of 600 km. If the

speed of the slow train is 10km/h less than that of the fast train, find the

speeds of the two trains.

Solution:

Let the speed of the fast train is x km/h and the speed of the slow train is (x – 10)

km/h. Thus, we are given

600 600

310x x

x2 – 10x – 2000 = 0

x2 – 50x + 40x – 2000 = 0

x(x – 50) + 40(x – 50) = 0

(x + 40)(x – 50) = 0

x = –40 or 50.

Since x is the speed of the train it cannot be negative, x = 50



Thus, the speed of the fast train is 50 km/h and the speed of the slow train is 40

km/h.

Ques [23] : A hollow copper sphere of external and internal diameter 8 cm and 4 cm

respectively is melted into a solid cone of base diameter 8 cm. Find the height

of the cone.

Solution:

When a solid is melted to form another solid, then their volume remains constant.

Thus,

Vcone = Vsphere

2 3 31 8 4 8 4

3 2 3 2 2

116 4 56

3

4 56 342

16

h

h

h

Thus, the height of the cone is 42 cm.

OR

If the radii of the circular end of a bucket 45 cm high are 28 cm and 7 cm,

find the capacity and surface area of the bucket.

Solution:

The bucket is a form of frustum of the cone.

We are given R = 28 cm, r = 7 cm and h = 45 cm.

Thus, the volume of the bucket is

2 2

2 2

3

1

3

1 2245 28 28 7 7

3 7

1 2245 784 196 49

3 7

1 2245 1029

3 7

48510cm

V h R Rr r

V

V

V

V

The surface area of the bucket is

2 2A R r R r h



2 2

2

28 7 28 7 45

2235 49.65

7

5461.5 cm

A

A

A

Ques [24] : An observer in a lighthouse observes two ships on the same side of the

lighthouse and in the same straight line with the base of the lighthouse. The

angles of depression of the ships approaching it are 30 and 60 degrees. If the

height of the lighthouse is 150 m, find the distance between the ships.

Solution:

Let AB be the lighthouse and C and D be

the two ships in the same side of the

lighthouse.

Thus, in ∆ABC,

150tan 60

1503

15050 3

3

x

x

x

Now, in ∆ABD,

150tan 30

1 150

3 50 3

50 3 150 3

100 3

x d

d

d

d

Thus, the distance between the two ships is 100 3 m.