cbse math x board paper set 3 (delhi) (sec a) math x board paper – set 3 (delhi) (sec a) section a...

CBSE Math X Board Paper – SET 3 (Delhi) (Sec A)

Section A

1. If the sum of first p terms of an Α.P., is ap2 + bp, find its common difference.

Solution:

Let A be the first term and D be the common difference of the A.P.

It is given that the sum of p terms of an A.P. is 2ap bp .

2

1

2

2

(1) (1)

(2) (2) 4 2

S a b a b

S a b a b

Now, S1 is equal to the first term of the A.P.

A = a + b … (1)

A + A + D = 4a + 2b … (2)

On multiplying (1) with 2 and subtracting it from (2), we obtain

2A + D – 2A = 4a + 2b – 2(a + b)

D = 4a + 2b – 2a – 2b = 2a

Thus, the common difference of the A.P. is 2a.

2. In fig. 1, S and T are points on the sides PQ and PR, respectively of PQR, such that PT = 2 cm, TR =

4 cm and ST is parallel to QR. Find the ratio of the areas of PST and PQR.

Solution:

In PST and PQR:

SPT = QPR (Common)

PST = PQR (Corresponding angles)

PST PQR (By AA similarity criterion)

It is known that the ratio of areas of similar triangles is equal to the ratio of the squares of their

corresponding sides. 22 2 2

2 2 2

Area of PST (PT) (PT) (2 cm) 2 1

Area of PQR (PR) (PT TR) (2 cm 4 cm) 6 9

Thus, the ratio of the areas of PST and PQR is 1: 9.

3. In fig. 2, AHK is similar to ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC.

CR

PF P

UBLI

C SC

HOO

L


Solution:

It is given that AHK is similar to ABC.

It is known that the corresponding sides of similar triangles are in proportion.

AH HK AK

AB BC AC

HK AK

BC AC

7 cm 10 cm

3.5 cm AC

7AC 35 cm

AC 5 cm

Thus, the length of AC is 5 cm.

4. If α, β are the zeroes of a polynomial, such that α + β = 6 and αβ = 4, then write the polynomial.

Solution:

It is given that + β = 6 and β = 4

It is known that a polynomial is of the form k [x2 – (sum of zeroes) x + (product of zeroes)], where k is a

real number.

Thus, the required polynomial is of the form k (x2 – 6x + 4).

For k = 1, the polynomial is x2 – 6x + 4.

Thus, the required polynomial is x2 – 6x + 4.

5. Has the rational number2 7 2

441

2 5 7a terminating or a non-terminating decimal representation?

Solution:

A rational number in simplest form has a terminating decimal expansion if its denominator is of the form

2n5

m, where n and m are non-negative integers.

The given rational number is7 2

441

2 5 7 and is not in simplest form.

Now,2 2 2

7 2 7 2 7

441 7 3 3

2 5 7 2 5 7 2 5

It can be seen that its denominator is of the form 2n5

m, where n and m are non-negative integers.

Thus, the given rational number has a terminating decimal expansion.

6. If cosec = 2x and cot θ2

x, find the value of 2

2

12 x

x

CRPF

PUB

LIC

SCHO

OL


Solution:

2 22

2

2 2

2 2

cosec 2

cosec...(1)

2

2cot

1 cot...(2)

2

1 cosec cot2 2

4 4

2(cosec cot )

4

21 [ 1 cot cosec ]

4

1

2

x

x

x

x

xx

Thus, the value of 2

2

12 x

x is

1

2.

7. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of

getting a red face card.

Solution:

There are a total of 52 playing cards in the pack.

The number of red face cards in the pack is only 6.

Thus, the probability of getting a red face card =6 3

52 26

8. The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends

are 18 cm and 6 cm. Find the curved surface area of the frustum.

22Use π

7

Solution:

Slant height of the frustum of cone, l = 4 cm

Let r1 and r2 be the radii of the circular ends of the frustum of the cone and let h be its height.

The circumferences of its circular ends are given as 18 cm and 16 cm.

2 r1 = 18 cm and 2 r2 = 6 cm

Curved surface area of the frustum of cone 1 2π ( )r r l

1 2

2

1(2π 2π )

2

1(18 cm 6 cm) 4 cm

2

48 cm

r r l

CRPF

PUB

LIC

SCHO

OL


9. If A(1, 2), B(4, 3) and C(6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of

the fourth vertex D.

Solution:

Let the coordinates of the fourth vertex D be (x, y).

Since ABCD is a parallelogram, diagonals AC and BD bisect each other.

Mid point of BD = Mid-point of AC

4 3 1 6 2 6, ,

2 2 2 2

4 3 7 8, ,

2 2 2 2

x y

x y

4 + x = 7 and 3 + y = 8

x = 7 – 4 = 3 and y = 8 – 3 = 5

Thus, the coordinates of the fourth vertex, i.e., of vertex D are (3, 5).

10. If P(2, p) is the mid-point of the line segment joining the points A(6, – 5) and B(– 2, 11), find the

value of p.

Solution:

It is given that P (2, p) is the mid-point of the line segment joining the points A (6, –5) and B (–2, 11).

6 ( 2) 5 11(2, ) ,

2 2

4 6(2, ) ,

2 2

p

p

(2, p) = (2, 3)

p = 3

Thus, the required value of p is 3.

CRPF

PUB

LIC

SCHO

OL

CBSE Math X Board Paper – SET 3 (Delhi) (Sec B)

Section B

11. If 5 and 5 are two zeroes of the polynomial x3 + 3x

2 – 5x – 15, find its third zero.

Solution:

It is given that 5 and 5 are the two zeroes of p(x) = x3 + 3x

2 – 5x – 15.

Therefore, ( 5)x and ( 5)x are the two factors of p(x). Thus, 2( 5)( 5) 5x x x is also a factor

of p(x).

Now, on dividing p(x) by (x2 – 5), we obtain

2 3 2

3

2

2

3

5 3 5 15

5

3 15

3 15

0

x

x x x x

x x

x

x

Therefore, the quotient so obtained is (x + 3).

(x + 3) is the third factor of p(x).

On substituting x + 3 = 0, we obtain,

x = –3

Thus, –3 is the third zero of p(x) = x3 + 3x

2 – 5x – 15.

12. If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.

Solution:

It is given that all the sides of the parallelogram touch the circle.

Let the parallelogram be ABCD and let the circle touch the sides AB, BC, CD and AD at P, Q, R and S

respectively.

Now, ABCD is a parallelogram.

AB = CD … (1)

BC = AD … (2)

It is known that lengths of tangents drawn from an external point to the circle are equal.

DR = DS … (3)

CR = CQ … (4)

BP = BQ … (5)

AP = AS … (6) CRPF

PUB

LIC

SCHO

OL


On adding equations (3), (4), (5) and (6), we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

From equations (1) and (2), we obtain

2AB = 2BC

AB = BC

AB = BC = CD = DA

Therefore, ABCD is a parallelogram all of whose sides are equal.

Thus, ABCD is a rhombus.

13. Without using trigonometric tables, find the value of the following expression: 2 2sec(90 ) cosec tan(90 ) cot cos 25 cos 65

3tan 27 tan 63

Solution: 2 2

2 2

2 2 2 2

sec(90 ) cosec tan(90 ) cot cos 25 cos 65

3 tan 27 tan 63

cosec cosec cot cot cos (90 65) cos 65

3 tan(90 63) tan 63

[ sec(90 ) cosec , tan(90 ) cot ]

(cos ec cot ) (sin 65 cos 65 )

3cot 6

2 2 2 2

3 tan 63

1 1

13 tan 63

tan 63

[ cosec cot 1, sin cos 1]

2

3

Thus, the value of the given expression is2

3.

OR

Find the value of cosec 30 , geometrically.

Solution:

Consider an equilateral triangle ABC.

Each angle in an equilateral triangle is 60°. CRPF

PUB

LIC

SCHO

OL


Therefore, A = B = C = 60°

Let the perpendicular AD be drawn from A to side BC.

Now, in ABD and ACD

ADB = ADC (Each 90°)

AB = AC (Given)

AD = AD (Common)

ABD ACD (By RHS congruency criterion)

BD = DC (CPCT)

BAD = CAD = 30º

Let the length of AB be 2a.

1BD BC

2a

Now, sin 30°BD 1

AB 2 2

a

a

cosec 30°1

2sin 30

Thus, the value of cosec 30° is 2.

14. Find the value of k for which the following pair of linear equation have infinitely many solutions:

2x + 3y = 7; (k – 1) x + (k + 2)y = 3k

Solution:

The given pair of linear equations is:

2x + 3y = 7 or 2x + 3y – 7 = 0 … (1)

(k – 1)x + (k + 2)y = 3k ... (2)

Equation (2) can be written as:

(k – 1)x + (k + 2)y – 3k = 0 … (3)

It is known that the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has infinitely many

solutions if 1 1 1

2 2 2

a b c

a b c.

Thus, from equations (1) and (3), we obtain

2 3 7

1 2 3

2 7

1 3

k k k

k k

6k = 7(k – 1)

6k = 7k – 7

7 = 7k – 6k

k = 7

Thus, the value of k is 7.

CRPF

PUB

LIC

SCHO

OL


15. In an A.P., first term is 2, the last term is 29 and sum of the terms is 155. Find the common difference

of the A.P.

Solution:

Let a and d be the first term and the common difference of the A.P. respectively.

First term, a = 2

Last term, l = 29

It is known that the sum of n terms of an A.P. is given by ( )2

n

nS a l .

155 (2 29)2

31155

2

155 210

31

n

n

n

It is known that the nth

term of an A.P. is given by an = a + (n – 1) d.

29 = 2 + (10 – 1) × d

9d = 29 – 2 = 27

273

9d

Thus, the common difference of the A.P. is 3.

CRPF

PUB

LIC

SCHO

OL

CBSE Math X Board Paper – SET 3 (Delhi) (Sec C)

Section C

16. Prove that 2 3 1is an irrational number.

Solution:

We first prove that 3 is irrational.

Let 3 be a rational number.

Therefore, we can find two integers a, b (b ≠ 0) such that 3a

b.

Let us assume that a and b are co-prime.

2 2

3

3

a b

a b

Therefore, a2 is divisible by 3. Hence, a is divisible by 3.

Let a = 3k, where k is an integer. 2 2

2 2

(3 ) 3

3

k b

b k

This means that b2 is divisible by 3. Hence, b is divisible by 3.

This implies that a and b have 3 as a common factor.

This is in contradiction to the fact that a and b are co-prime.

Hence, 3 cannot be expressed as a

bor it can be said that 3 is irrational.

We now prove that 2 3 1 is irrational.

For this, we assume that 2 3 1 is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

2 3 1

2 3 1

13 1

2

a

b

a

b

a

b

Since a and b are integers, 1

12

a

bwill also be rational. Therefore, 3 is rational.

This contradicts the fact that 3 is irrational. Hence, our assumption that 2 3 1 is rational is false.

Therefore, 2 3 1 is irrational.

Hence proved

17. In figure 3, ABC is a right triangle, right angled at C and D is the mid-point of BC. Prove that AB2 =

4AD2 – 3AC

2.

CRPF

PUB

LIC

SCHO

OL


Solution:

In ABC, D is the mid-point of BC.

BD = CD = 1

2BC

On using Pythagoras theorem in ACD, we obtain

AD2 = AC

2 + CD

2

AD2 = AC

2 +

21

BC2

AD2 = AC

2 +

1

4BC

2

4AD2 = 4AC

2 + BC

2

BC2 = 4AD

2 – 4AC

2 ... (1)

On using Pythagoras theorem in ABC, we obtain

AB2 = AC

2 + BC

2

AB2 = AC

2 + (4AD

2 – 4AC

2) [Using equation (1)]

AB2 = 4AD

2 – 3AC

2

Thus, the given result is proved.

18. Prove the following

tan cot1 tan cot

1 cot 1 tan

A AA A

A A

Solution:

L.H.S. =tan cot

1 cot 1 tan

A A

A A

2

3

23 3 2 2

2

1

tan 1tan cot1 1 tan tan

1tan

tan 1

tan 1 tan (1 tan )

tan

tan 1

tan 1 tan (tan 1)

tan 1

tan (tan 1)

(tan 1)(tan tan 1)[ ( )( )]

tan (tan 1)

tan tan 1

ta

A A AA A

A

A

A A A

A

A

A A A

A

A A

A A Aa b a b a ab b

A A

A A

n

1tan 1

tan

tan 1 cot

A

AA

A A

CRPF

PUB

LIC

SCHO

OL


2

3

23 3 2 2

2

1

tan 1tan cot1 1 tan tan

1tan

tan 1

tan 1 tan (1 tan )

tan

tan 1

tan 1 tan (tan 1)

tan 1

tan (tan 1)

(tan 1)(tan tan 1)[ ( )( ]

tan (tan 1)

tan tan 1

tan

A A AA A

A

A

A A A

A

A

A A A

A

A A

A A Aa b a b a ab b

A A

A A

1tan 1

tan

tan 1 cot

1 tan cot R.H.S.

A

AA

A A

A A

tan cot1 tan cot

1 cot 1 tan

A AA A

A A

Hence proved

OR

Prove the following:

1(cosec sin ) (sec cos )

tan cotA A A A

A A

Solution:

To prove: (cosec A – sin A) (sec A – cos A)1

tan cotA A

L.H.S = (cosec A – sin A) (sec A – cos A)

2 2

2 22 2

1 1 1 1sin cos cosec , sec

sin cos sin cos

1 sin 1 cos

sin cos

cos sin[sin cos 1]

sin cos

sin cos ...(1)

A A A AA A A A

A A

A A

A AA A

A A

A A

R.H.S = 1

tan cotA A

2 2

2 2

1

sin cos

cos sin

1

sin cos

cos sin

1[sin cos 1]

1

cos sin

cos sin

1

sin cos ...(2)

A A

A A

A A

A A

A A

A A

A A

A A

Thus, from (1) and (2), (cosec A – sin A) (sec A – cos A)1

tan cotA A

Hence proved

CRPF

PUB

LIC

SCHO

OL


19. In an A.P., the sum of first ten terms is –150 and the sum of its next ten terms is –550. Find the A.P.

Solution:

Let a and d be the first term and the common difference of the Α.P.

It is given that the sum of the first 10 terms of the Α.P. is –150.

10

10{2 (10 1) } 150 [ {2 ( 1) }]

2 2n

nS a d S a n d

5(2a + 9d) = – 150

2a + 9d = – 30 ... (i)

It is also given that the sum of the next 10 terms of the Α.P. is –550.

Thus, the sum of the first 20 terms of the Α.P. is (–150) + (–550) = –700

20{2 (20 1) } 700

2a d

10 (2a + 19d) = –700

2a + 19d = –70 ... (ii)

Subtracting equation (i) from equation (ii):

10d = –70 – (–30)

10d = –40

d = –4

Substituting d = –4 in equation (i):

2a + 9 × (–4) = –30

2a = –30 + 36 = 6

a = 3

Now, the Α.P. is given by a, a + d, a + 2d, ...

a + d = 3 + (–4) = –1

a + 2d = 3 + 2(–4) = 3 – 8 = –5

a + 3d = 3 + 3(–4) = 3 – 12 = –9

Thus, the required Α.P. is 3, –1, –5, –9....

20. The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each of

the numerator and denominator is decreased by 1, the fraction becomes 1

2. Find the fraction.

Solution:

Let the numerator and denominator of the fraction be x and y respectively.

Thus, the fraction is x

y.

It is given that the sum of numerator and denominator of the fraction is 3 less than twice the denominator.

x + y = 2y – 3

x + y – 2y = –3

x – y = –3 … (1)

It is also given that if each of the numerator and denominator is decreased by 1, then the fraction

becomes1

2. CRPF

PUB

LIC

SCHO

OL


1 1

1 2

2( 1) 1

2 2 1

2 1 2

2 1 ...(2)

x

y

x y

x y

x y

x y

Subtracting equation (1) from equation (2):

x = 1 – (– 3)

x = 4

Substituting the value of x in equation (1):

4 – y = – 3

y = 4 + 3 = 7

Thus, the required fraction is 4

7.

OR

Solve the following pair of equations:

4 63 8; 4 5y y

x x

Solution:

The given pair of equations is:

43 8 ...(1)

64 5 ...(2)

yx

yx

Let 1

zx

.

Then, equations (1) and (2) become:

4z + 3y = 8 … (3)

6z – 4y = –5 … (4)

Multiplying equation (3) with 4 and equation (4) with 3 and then adding the obtained equations:

4(4 3 ) 3(6 4 ) 4 8 3( 5)

16 12 18 12 32 15

34 17

1

2

1 1

2

2

z y z y

z y z y

z

z

x

x

Substituting x = 2 in equation (1): CRPF

PUB

LIC

SCHO

OL


43 8

2

3 8 2 6

62

3

y

y

y

Thus, the solution of the given pair of equations is x = 2 and y = 2.

21. Construct a triangle PQR in which QR = 6 cm, Q = 60 and R = 45 . Construct another triangle

similar to PQR such that its side are 5

6of the corresponding sides of PQR.

Solution:

The steps of constructions are as follows:

Step 1

Draw a PQR with side QR = 6 cm, Q = 60 and R = 45 .

Step 2

Draw a ray QX making an acute angle with QR on the opposite side of vertex P.

Step 3

Locate 6 points (as 6 is greater between 5 and 6), Q1, Q2, Q3, Q4, Q5, and Q6 on QX.

Step 4

Join Q6R. Draw a line through Q5 parallel to Q6R such that it intersects QR at R'.

Step 5

Through R', draw a line parallel to PR such that it intersects QP at P'.

Thus, P'QR' is the required triangle.

CRPF

PUB

LIC

SCHO

OL


22. Cards bearing numbers 1, 3, 5, ..., 35 are kept in a bag. A card is drawn at random from the bag. Find

the probability of getting a card bearing

(i) a prime number less than 15.

(ii) a number divisible by 3 and 5.

Solution: The cards are numbered as 1, 3, 5…35.

Total number of cards = 18

(i) The prime numbers less than 15 are 3, 5, 7, 11 and 13.

Required probability = 5

18

Thus, the probability of getting a card bearing a prime number less than 15 is5

18.

(ii) The numbers divisible by 3 and 5 are 15 and 30.

Required probability = 2 1

18 9

Thus, the probability of getting a card bearing a number divisible by 3 and 5 is1

9.

23. If the point P (m, 3) lies on the line segment joining the points 2

,65

A and B (2, 8), find the value

of m.

Solution:

It is given that point P (m, 3) lies on the line segment joining points 2

A , 65

and B (2, 8). Hence,

points A, P and B are collinear.

Area of APB = 0

It is known that the area of a triangle formed by points having vertices (x1, y1), (x2, y2) and (x3, y3) is given

by 1 2 3 2 3 1 3 1 2

1[ ( ) ( ) ( )]

2x y y x y y x y y .

1 2(3 8) (8 6) 2 (6 3) 0

2 5

2( 5) (2) 2 (3) 0

5

2 2 6 0

2 8

4

m

m

m

m

m

Thus, the value of m is – 4.

CRPF

PUB

LIC

SCHO

OL


24. Point P divides the line segment joining the points A (2, 1) and B (5, –8) such thatAP 1

AB 3. If P lies

on the line 2x – y + k = 0, find the value of k.

Solution:

It is given that point P divides the line segment joining A (2, 1) and B (5, –8) such thatAP 1

AB 3.

Hence, P divides AB in the ratio 1:2.

It is known that the coordinates of a point that divides the line segment joining points (x1, y1) and (x2, y2)

in the ratio m:n are given by 2 1 2 1,mx nx my ny

m n m n.

By section formula, the coordinates of point P are:

1 5 2 2 1,

1 2 1 2

5 4 8 2,

3 3

9 6,

3 3

(3, 2)

It is given that point P lies on the line 2x – y + k = 0.

2(3) – (–2) + k = 0

6 + 2 + k = 0

k = –8

Thus, the value of k is –8.

25. In figure 4, the boundary of shaded region consists of four semicircular arcs, two smallest being equal.

If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded

region. [Use =22

7]

Solution:

CRPF

PUB

LIC

SCHO

OL


Let the radii of the largest semicircle, the smallest semicircle and the circle with diameter BD be r1, r2 and

r3 respectively.

It is given that AE = 14 cm and DE = AB = 3.5 cm

1

14 cm 7 cm

2r

2

3.5 cm

2r

r3 = r1 – 2 r2 =3.5

7 cm 2 cm 7 cm 3.5 cm 3.5 cm2

The area of the shaded region is given by:

Area of semicircle with radius r1 + Area of semicircle with radius r3 – 2 × Area of semicircle with radius

r2

2 2 2

1 3 2

2 2 2

1 3 2

2

2 2

22 2

22

2

1 1 1( ) ( ) 2 ( )

2 2 2

1{( ) ( ) 2( ) }

2

1 22 3.5 cm(7 cm) (3.5 cm) 2

2 7 2

11 12.25 cm49 cm 12.25 cm

7 2

11 12.25 cm49 cm

7 2

11 12.2549 cm

7 2

11(49 6.

7

r r r

r r r

2

2

2

125) cm

1155.125 cm

7

86.625 cm

Thus, the area of the shaded region is 86.625 cm2.

OR

Find the area of shaded region in figure 5, if AC = 24 cm, BC = 10 cm and O is the centre of the circle.

[Use = 3.14]

CR

PF P

UBLI

C SC

HOO

L


Solution:

It can be seen that OACB is a semicircle.

ACB = 90

Therefore, ABC is a right-angled triangle.

On using Pythagoras theorem in ABC, we obtain

AB2 = AC

2 + BC

2

AB2 = (24 cm)

2 + (10 cm)

2

AB2 = 576 cm)

2 + 100 cm

2

AB2 = 676 cm

2

AB = 676 cm

AB = 26 cm

OA = 26

2cm = 13 cm

The area of the shaded region is given by: Area of semicircle – Area of ABC

2

2

2

2

2

2

1 1 Shaded area (OA) AC BC

2 2

1 22 1(13 cm) 24 cm 10 cm

2 7 2

3718120 cm

14

3718 1680 cm

14

2038 cm

14

145.57 cm

Thus, the area of the shaded region is 145.57 cm2.

CRPF

PUB

LIC

SCHO

OL

CBSE Math X Board Paper – SET 3 (Delhi) (Sec D)

Section D

26. A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 33

10459 cm7

.

The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet

used in making the container at the rate of Rs. 1.40 per square centimetre. 22

Use π7

Solution:

The milk container can be diagrammatically represented as

Here, volume V = 3 33 73216

10459 cm cm7 7

The radii of the lower and upper circular ends are respectively given as:

r1 = 8 cm, r2 = 20 cm

Let the height of the container be h.

The volume (V) of a frustum of cone is given by, 2 2

1 2 1 2

1π ( )

3V h r r r r

3 2 2

2 3

73216 1 22 cm {(8 cm) (20 cm) (8 cm)(20 cm)}

7 3 7

73216 7 3(64 400 160) cm cm

7 22

624 3328 3 cm

3328 3 cm

624

16 cm

h

h

h

h

h

The metal sheet required for making the container is equal to the sum of the curved surface area of the

frustum of the cone and its base area.

Total surface area of the container is 2

1 2 1π ( ) πl r r r , where l is the slant height.

Now, 2 2

2 1

2 2

2 2

( )

(16 cm) (20 cm 8 cm)

256 cm 144 cm

400 cm

20 cm

l h r r

l

l

l

l

Thus, total surface area of the container 222 22

20 cm (8 cm 20 cm) (8 cm)7 7

2

2

22(20 cm 28 cm 64 cm)

7

22624 cm

7

CRPF

PUB

LIC

SCHO

OL


Cost of 1 cm2 of metal sheet = Rs 1.40

Cost of 222624 cm

7of metal sheet = Rs

21.40 624

7= Rs 2745.60

Thus, the cost of metal sheet required for making the container is Rs 2745.60.

Or

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of

the hemisphere. If the radius of base of the cone is 21 cm and its volume is 2

3 of the volume of the

hemisphere, calculate the height of the cone and the surface area of the toy. 22

Use π7

.

Solution:

Let the height of the conical part be h.

Radius of the cone = Radius of the hemisphere = r = 21 cm

The toy can be diagrammatically represented as

Volume of the cone = 21

π3

r h

Volume of the hemisphere = 32

π3

r

According to given information:

Volume of the cone 2

3 Volume of the hemisphere

2 3

3

2

1 2 2π π

3 3 3

2 2π

3 31π

3

4

3

421 cm 28 cm

3

r h r

r

h

r

h r

h

Thus, surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

CRPF

PUB

LIC

SCHO

OL


2

2 2 2

2 2

2 2

2

2

2

2

2

π 2π

π 2π

π ( 2 )

2221 cm( (28 cm) (21 cm) 2 21 cm)

7

66( 784 441 42) cm

66( 1225 42) cm

66(35 42) cm

66 77 cm

5082 cm

rl r

r h r r

r h r r

27. Prove that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other

two sides.

Using the above, prove the following:

Point D is the mid-point of the side BC of a right triangle ABC, right angled at C. Prove that 4AD2 = 4AC

2 +

BC2

Solution:

Let ABC be a triangle right-angled at B.

To prove: AC2 = AB

2 + BC

2

Construction: Draw BD AC

Proof:

In ABC and BDC:

ACB = DCB (Common)

ABC = BDC (Each 90°)

ABC BDC (By AA similarity criterion)

It is known that the corresponding sides of similar triangles are proportional.

AC BC

BC DC

BC2 = AC × DC … (1)

Now in ABC and ADB:

BAC = BAD (Common)

ABC = ADB (Each 90°) CRPF

PUB

LIC

SCHO

OL


ABC ADB (AA similarity criterion)

It is known that the corresponding sides of similar triangles are proportional.

AB AC

AD AB

AB2 = AC × AD … (2)

Adding equation (1) and (2):

BC2 + AB

2 = AC × DC + AC × AD

AB2 + BC

2 = AC × (DC + AD)

AB2 + BC

2 = AC × AC {AC = DC + AD}

AB2 + BC

2 = AC

2

Thus, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two

sides.

Now, using the theorem proved above, the relation to be proved is 4AD2 = 4AC

2 + BC

2

Using the theorem proved above in ACD:

AD2 = AC

2 + CD

2

AD2 = AC

2 +

2BC

2 [D is the mid-point of BC]

AD2 = AC

2 +

2BC

4

4AD2 = 4AC

2 + BC

2

Hence, the relation is proved.

28. Three consecutive positive integers are such that the sum of the square of the first and the product of the

other two is 46, find the integers.

Solution:

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition:

x2

+ (x + 1) (x + 2) = 46

x2 + x

2 + 3x + 2 = 46

2x2

+ 3x – 44 = 0 CRPF

PUB

LIC

SCHO

OL


It is known that the roots of the quadratic equation ax2 + bx

2 + c = 0 are

2 4

2

b b ac

a

For equation 2x2 + 3x – 44 = 0:

23 (3) 4(2)( 44)

2(2)

3 9 352

4

3 361

4

3 19

4

3 19 3 19 or

4 4

224 or

4

x

Since x is a positive integer, x = 4

Thus, the required consecutive positive integers are 4, 5, and 6.

Or

The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number,

then find the two numbers.

Solution:

Let the smaller number be x.

Then, the larger number is 2x – 5.

According to the given condition:

(2x – 5)2 – x

2 = 88

4x2 + 25 – 20x – x

2 = 88

3x2 – 20x – 63 = 0

It is known that the roots of the quadratic equation ax2 + bx

2 + c = 0 are

2 4

2

b b ac

a .

For equation 3x2 – 20x – 63 = 0:

CRPF

PUB

LIC

SCHO

OL


2( 20) ( 20) 4(3)( 63)

2(3)

20 400 756

6

20 1156

6

20 34

6

20 34 20 34 or

6 6

149 or

6

x

x

x

x

x

x

When x = 9, the smaller number is 9, and the larger number is 2 (9) – 5 = 13

When 14

,6

x the smaller number is

14,

6 and the larger number is

14 292 5

6 3

Here, 29 14

3 6

The smaller number cannot be 14

6

Thus, the required numbers are 9 and 13.

29. From the top of a 7 m high building, the angle of elevation of the top of a tower is 60 and the angle of

depression of the foot of the tower is 30 . Find the height of the tower.

Solution:

The given situation can be represented as:

Here AB represents the building and CD represents the tower.

Now, AB = DE = 7 m

Also, BD = AE CRPF

PUB

LIC

SCHO

OL


In ABD:

ABtan 30

BD

7 m 1

BD 3

BD 7 3 m

AE 7 3 m

In ACE:

CEtan 60

AE

CE3

7 3 m

CE 3 7 3 m 21 m

CD = CE + AB = (21 + 7) m = 28 m

Thus, the height of the tower is CD = 28 m.

30. Find the mean, mode and median of the following frequency distribution:

Class: 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Frequency: 8 7 15 20 12 8 10

Solution:

The frequency distribution table for the given data can be drawn as:

Class Frequency (fi) Class marks (xi) fixi

0–10 8 5 40

10–20 7 15 105

20–30 15 25 375

30–40 20 35 700

40–50 12 45 540

50–60 8 55 440

60–70 10 65 650

80if 2850

2850Mean 35.625

80

i i

i

f x

f

The maximum class frequency is 20. Therefore, modal class is 30–40.

l = 30, h = 10, f1 = 20, f0 = 15, f2 = 12 CRPF

PUB

LIC

SCHO

OL


1 0

1 0 2

Mode2

20 1530 10

2 20 15 12

530 10

40 27

530 10

13

f fl h

f f f

= 30 + 3.85 (approx.)

= 33.85 (approx.)

Now, the cumulative frequency distribution table for the given data can be drawn as:

Class Frequency Cumulative

frequency

0–10 8 8

10–20 7 15

20–30 15 30

30–40 20 50

40–50 12 62

50–60 8 70

60–70 10 80

n = 80

8040

2 2

nlies in the class interval 30–40

Median class is 30–40.

l = 30, cf = 30, f = 20, h = 10

Median 2

ncf

l hf

40 30Median 30 10

20

1030 10

20

30 5

35

Thus, the mean, mode and median of the given frequency distribution are 35.625, 33.85 and 35 respectively.

CRPF

PUB

LIC

SCHO

OL

CBSE Math X Board Paper – SET 2 (Delhi)

Section A - Different Question(s)

6. If 2x = sec A and2

x= tan A, find the value of 2

2

12 x

x.

Solution:

It is given that:

2x = sec A

x =sec

...(1)2

A

2tan

1 tan...(2)

2

Ax

A

x

2 2

2

2

2 2

2 2

2 2

1 sec tan2 2

2 2

sec tan2

4 4

2[sec tan ]

4

1(1) [ 1 tan sec ]

2

1

2

A Ax

x

A A

A A

A A

CRPF

PUB

LIC

SCHO

OL


Section B - Different Question(s)

11. If 3 and 3 are two zeroes of the polynomial x3 – 5x

2 – 3x + 15, find its third zero.

Solution:

It is given that 3 and 3 are two zeroes of the polynomial x3 – 5x

2 – 3x + 15.

Thus, 2( 3)( 3) 3x x x is a factor of the polynomial x3 – 5x

2 – 3x + 15.

To find the other zero of the given polynomial, we are required to divide

x3 – 5x

2 – 3x + 15 by x

2 – 3

This can be done as:

2 3 2

3

2

2

3 2 2

5

3 5 3 15

3

5 15

5 15

0

5 3 15 ( 3)( 5)

x

x x x x

x x

x

x

x x x x x

Therefore, x – 5 is a factor of the given polynomial.

Hence, the third zero of the given polynomial is 5.

CRPF

PUB

LIC

SCHO

OL


Section C - Different Question(s)

16. Prove that 4 5 2 is an irrational number.

Solution:

Assume that 4 5 2 is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that:

4 5 2

5 2 4

12 4

5

a

b

a

b

a

b

Since a and b are integers, 1

45

a

bwill also be rational and so will 2 .

This contradicts the fact that 2 is irrational.

Hence, our assumption that 4 5 2 is rational is false.

Thus, 4 5 2 is irrational.

21. Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and AC = 7 cm. Construct another triangle

similar to ABC such that its sides are3

5of the corresponding sides of ABC.

Solution:

The steps of constructions are as follows:

Step 1

Draw a line segment BC = 6 cm. Taking point B as centre and radius equal to 5 cm, draw an arc. Similarly,

taking point C as centre and radius equal to 7 cm, draw an arc. These arcs will intersect each other at point A.

Join AB, AC. ABC is thus formed.

CRPF

PUB

LIC

SCHO

OL


Step 2

Draw a ray BX making an acute angle with line BC on the opposite side of vertex A.

Step 3

Mark 5 points B1, B2, B3, B4, B5 (as 5 is greater between 3 and 5) along line BX such that

BB1 = B1B2 = B2B3 = B3B4 = B4B5

Step 4

Join CB5. Through point B3, draw a line parallel to CB5 to intersect BC at point C'.

Step 5

Through point C', draw a line parallel to line AC to intersect AB at A'.

Thus, A'BC' is the required triangle.

23. Prove that the points P(a, b + c), Q(b, c + a) and R(c, a+ b) are collinear.

Solution:

It is known that points (x1, y1), (x2, y2) and (x3, y3) are collinear if

x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0

Therefore, to prove that points P(a, b + c), Q(b, c + a) and R (c, a+ b) are collinear, we are required to

prove that:

[( ) ( )] [( ) ( )] [( ) ( )] 0a c a a b b a b b c c b c c a

Consider L.H.S.

[( ) ( )] [( ) ( )] [( ) ( )]

( ) ( ) ( )

( ) ( ) ( )

0

a c a a b b a b b c c b c c a

a c a a b b a b b c c b c c a

a c b b a c c b a

ca ab ab bc bc ca

= R.H.S.

Thus, points P(a, b + c), Q(b, c + a) and R(c, a+ b) are collinear.

CRPF

PUB

LIC

SCHO

OL


Section D - Different Question(s)

27. Prove that the lengths of tangents drawn from an external point to a circle are equal.

Using the above prove the following:

In fig. 6, PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at

M. Prove that PL + LM = PN + MN.

Solution:

Let PA and PB be two tangents from an external point P to the circle with centre O.

We have to prove that PA = PB.

Join OP, OA and OB.

In OAP and OBP:

OAP = OBP (Each right angle)

OP = OP (Common side)

OA = OB (Radii of a circle)

Therefore, by R.H.S. congruence rule, OAP OBP

It is known that the corresponding sides of congruent triangles are equal.

PA = PB

Thus, the tangents drawn from an external point to a circle are equal in length.

Using the above result:

PA = PB ... (1) (Tangents from point P)

LA = LM ... (2) (Tangents from point L)

NB = NM ... (3) (Tangents from point N)

PA = PB {from equation (1)}

PL + LA = PN + NB

PL + LM = PN + NM {using equations (2) and (3)}

Hence, proved. CRPF

PUB

LIC

SCHO

OL

cbse math x board paper set 3 (delhi) (sec a) math x board paper – set 3 (delhi) (sec a) section a...

Documents