CBSE Class 10 Mathematics Standard Question Paper 2020 Set 30-4 Solutions

Strictly Confidential - (For Internal and Restricted Use Only)
Secondary School Examination-2020
Marking Scheme - MATHEMATICS STANDARD
Subject Code: 041 Paper Code: 30/4/1, 30/4/2, 30/4/3
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1

30/4/1

QUESTION PAPER CODE 30/4/1
EXPECTED ANSWER/VALUE POINTS
SECTION A
Question numers 1 to 20 carry 1 mark each.
Question numbers 1 to 10 are multiple choice questions. Choose the correct option.
1. The number of zeroes for a polynomial p(x) where graph of y = p(x) is given in Figure-1, is
(A) 3 (B) 4 (C) 0 (D) 5
y
y = p(x)
x
0

Fig. 1
Sol. (A) 3 1

2. The first term of an A.P. is 5 and the last term is 45. If the sum of all the terms is 400, the
number of terms is
(A) 20 (B) 8 (C) 10 (D) 16
Sol. (D) 16 1
OR
th
The 9 term of the A.P. – 15, –11, –7, ..., 49 is
(A) 32 (B) 0 (C) 17 (D) 13
Sol. (C) 17 1
3. It is being given that the points A(l, 2), B(0, 0) and C(a, b) are collinear. Which of the following
relations between a and b is true?
(A) a = 2b (B) 2a = b (C) a + b = 0 (D) a – b = 0
Sol. (B) 2a = b 1
4. In Figure-2, TP and TQ are tangents drawn to the circle with centre at O. If ∠POQ = 115° then
∠PTQ is
P T

115°
O
Q

Fig. 2
(A) 115° (B) 57.5° (C) 55° (D) 65°
Sol. (D) 65º 1

(3)

30/4/1

OR
From an external point Q, the length of the tangent to a circle is 5 cm and the distance of Q
from the centre is 8 cm. The radius of the circle is

(A) 39 cm (B) 3 cm (C) 39 cm (D) 7 cm

Sol. (C) 39 cm 1

5. The value of θ for which cos (10° + θ) = sin 30°, is
(A) 50° (B) 40° (C) 80° (D) 20°
Sol. (A) 50° 1

6. A bag contains 3 red, 5 black and 7 white balls. A ball is drawn from the bag at random. The
probability that the drawn is not black, is

1 9 5 2
(A) (B) (C) (D)
3 15 10 3
Sol. (D) 2/3 1

7. The pair of linear equations y = 0 and y = –6 has
(A) a unique solution (B) no solution
(C) infinetly many solutions (D) only solution (0, 0)
Sol. (B) No solution 1

8. The mean and median of a distribution are 14 and 15 respectively. The value of mode is
(A) 16 (B) 17 (C) 18 (D) 13
Sol. (B) 17 1

9. The quadratic equation x2 – 4x + k = 0 has distinct real roots if
(A) k = 4 (B) k > 4 (C) k =16 (D) k < 4
Sol. (D) K < 4 1

⎛a ⎞
10. Point P ⎜ , 4 ⎟ is the mid-point of the line segment joining the points A(– 5, 2) and B(4, 6). The
⎝8 ⎠
value of ‘a’ is
(A) –4 (B) 4 (C) –8 (D) –2
Sol. (A) –4 1

(4)

30/4/1

Fill in the blanks in question numbers 11 to 15.

⎛2+ 5 ⎞
11. ⎜ ⎟ is _______ number..
⎝ 3 ⎠
Sol. irrational 1

12. Let ΔABC ~ ΔDEF and their areas be respectively 81 cm2 and 144 cm2. If EF = 24 cm, then
length of side BC is _______ cm.
Sol. 18 1

13. The distance between the points (a, b) and (– a, – b) is _______.

Sol. 2 a 2 + b2 1

14. If tan A = 1, then 2 sin A cos A = _______.
Sol. 1 1

15. A spherical metal ball of radius 8 cm is melted to make 8 smaller identical balls. The radius of
each new ball is _______ cm.
Sol. 4 1

Answer the following question numbers 16 to 20.
16. Given that HCF (135, 225) = 45, find the LCM (135, 225).

135 × 225 1
Sol. LCM =
45 2
1
= 675
2

17. In Figure-3, a tightly stretched rope of length 20 m is tied from the top of a vertical pole to the
ground. Find the height of the pole if the angle made by the rope with the ground is 30°.
B

20 m

30°
C
A
Fig. 3
AB 1
Sol. sin 30° =
20 2
1
AB = 10 m
2

(5)

30/4/1

18. Two dice are thrown simultaneously. What is the probability that the sum of the two numbers
appearing on the top is 13?
Sol. P(E) = 0 1

229
19. After how many decimal places will the decimal representation of the rational number
2 × 57
2

terminate?
Sol. After 7 decimal places 1

20. In Figure-4, AB and CD are common tangents to circle which touch each other at D. If AB =
8 cm, then find the length of CD.

B
C
A
D

Fig. 4

1
Sol. AC = CD = BC
2

1
CD = 4 cm
2

SECTION B
Question numbers 21 to 26 carry 2 marks each.
21. Solve for x:
6x2 + 11x + 3 = 0
Sol. 6x2 + 11x + 3 = 0
6x2 + 9x + 2x + 3 = 0 1

1
(2x + 3) (3x + 1) = 0
2

1
x = –3/2, x = –1/3
2

(6)

30/4/1

22. The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first
triangle is 9 cm long, find the length of the corresponding side of the second triangle.
Sol. Let the side of other triangle be x cm

1
Q Ratio of perimeters of two similar triangles is equal to ratio of their corresponding sides
2

9 30
∴ = 1
x 20

1
x = 6 cm
2

OR
In Figure-5, ΔPQR is right-angled at P. M is a point on QR such that PM is perpendicular to
QR. Show that PQ2 = QM × QR.

R

M

P Q

Fig. 5
Sol. ΔPQM ~ ΔRQP [By AA similarity] 1

PQ QM
∴ =
RQ PQ

⇒ PQ2 = QM × QR 1

23. Evaluate:

2 2
⎛ sin 47° ⎞ ⎛ cos 30° ⎞
⎟ +⎜ ⎟ − (sin 60° )
2
⎜
⎝ cos43° ⎠ ⎝ cot 30° ⎠

2 2 2
⎡ cos (90° – 47°) ⎤ ⎛ 3/2 ⎞ ⎛ 3 ⎞
Sol. ⎢⎣ cos 43° ⎥⎦ + ⎜ 3 ⎟ − ⎜⎝ 2 ⎟⎠ 1
⎝ ⎠

1 3 1
= 1+ − = 1
4 4 2

(7)

30/4/1

24. Find the mode of the following distribution:

Classes: 10 – 20 20 – 40 40 – 60 60 – 80 80 – 100
Frequency: 10 8 12 16 4

1
Sol. Modal class = 60 – 80
2

⎛ f1 − f 0 ⎞ ⎛ 16 − 12 ⎞
Mode = l + ⎜ ⎟ × h = 60 + ⎜ ⎟ × 20 1
⎝ 2f1 − f 0 − f 2 ⎠ ⎝ 32 − 12 − 4 ⎠

1
= 65
2

OR
From the following distribution, find the median:

Classes: 500 – 600 600 – 700 700 – 800 800 –900 900 – 1000
Frequency: 36 32 32 20 30

1
Sol. Median class: 700 – 800
2

⎛N ⎞
⎜ − cf ⎟
⎝2 ⎠×h
Median = l +
f

75 − 68
= 700 + × 100 1
32

1
= 721.88
2

25. In Figure-6, a tent is in the shape of a cylinder surmounted by a conical top. The cylindrical part
is 2.1 m high and conical part has slant height 2.8 m. Both the parts have same radius 2 m. Find
22
the area of the canvas used to make the tent. (Use π = )
7

2.8 m

2.1 m
2m

Fig. 6

(8)

30/4/1

Sol. Area of canvas = πr(2h + l)

22
= × 2 (2 × 2.1 + 2.8) 1
7
= 44 m2 1

26. Tree Plantation Drive
A group Housing Society has 600 members, who have their houses in the campus and decided
to hold a Tree Plantation Drive on the occasion of New Year. Each household was given he
choice of planting a sampling of its choice. The number of different types of sampings planted
were:
(i) Neem – 125
(ii) Peepal – 165
(iii) Creepers – 50
(iv) Fruit plants – 150
(v) Flowering plants – 110
On the opening ceremony, one of the plants is selected randomly for a prize. After reading the
above passage, answer the following questions.
What is the probability that the selected plant is
(i) A fruit plant or a flowering plant?
(ii) Either a Neem plant or a Peepal plant?
Sol. Total outcomes = 600

260 13
(i) P(Fruit plant or a flowering plant) = or 1
600 30

290 29
(ii) P(either neem plant or a peepal plant) = or 1
600 60

SECTION C
Question numbers 27 to 34 carry 3 marks each.

27. Prove that 5 is an irrational number..

Sol. Let 5 be a rational number

a 1
5= b≠0 HCF (a, b) = 1
b 2

(9)

30/4/1

a2
⇒ 5= 2
, a 2 = 5b 2
b
5 divides a 1

Put a = 5c (for some integer c)
⇒ 25c2 = 5b2 ⇒ b2 = 5c2

1
then we get, 5 divides b
2

Contradiction arises as HCF (a, b) = 1
∴ Our assumption is wrong

∴ 5 is irrational number 1

28. The sum of the First 30 terms of an A.P. is 1920. If the fourth term is 18, find its 11th term.

30
Sol. [2a + 29d] = 1920
2
⇒ 2a + 29d = 128 ...(i) 1

1
Also, a4 = 18 ⇒ a + 3d = 18 ...(ii)
2

From equation (i) & (ii)

a=6 d=4 1

1
∴ a11 = a + 10d = 46
2

29. Find the co-ordinates of the points of trisection of the line segment joining the points (3, – 1)
and (6,8).
Sol. Case I: If C and D trisect AB
C D
A (3, –1) B (6, 8) 1
then C divides AB in the ratio 1 : 2
2

1× 6 + 2 × 3 1
Co-ordinates of C: x = =4
3 2

1 × 8 + 2 (–1) 1
and y = =2
3 2
∴ Co-ordinates of C(4, 2)

(10)

30/4/1

1
Case II: Coordinates of D if D divides AB in the ratio 2 : 1
2

2 × 6 + 1× 3 1
Co-ordinates of D: x′ = =5
3 2

2 × 8 + 1 × (–1) 1
y′ = =5
3 2

Coordinates of D = (5, 5)
OR
Find the area of a quadrilateral ABCD having vertices at A(l, 2), B(l, 0), C(4, 0) and D(4, 4).

1
C (4, 0) ar (ΔABC) = [1(0 – 0) + 1(0 – 2) + 4(2 – 0)]
(4, 4) 2
D
1
= 3 sq. units 1
B (1, 0) 2

A (1, 2) 1
ar (ΔACD) = [1(0 – 4) + 4(4 – 2) + 4(2 – 0)]
2
= 6 sq. units 1

1
∴ Area of quadrialteral = 3 + 6 = 9 sq. units
2

30. In Figure-7, XY and MN are two parallel tangents to a circle with centre O and another tangent
AB with point of contact C intersecting XY at A and MN at B. Prove that ∠AOB = 90°.

X P A Y

O
C

M Q B N

Fig. 7
Sol. X P A Y Join OC (In given figure)
ΔAPO ≅ ΔACO [By RHS congruence] 1
O C
1
∴ ∠OAP = ∠OAC = x (let)
2
M Q B N
Similarily, ΔOQB ≅ ΔOCB

(11)

30/4/1

∴ ∠OBC = ∠OBQ = y (let)
Q XY || MN
∴ ∠PAB + ∠ABQ = 180° 1

⇒ x + y = 90°
∴ ∠AOB = 180° – (x + y) = 90°

1
∴ ∠AOB = 90° 2

31. Solve the pair of equations:

2 3 5 4
+ = 11, − = –7
x y x y
Hence, find the value of 5x – 3y.

2 3
Sol. + = 11 ...(i)
x y

5 4
− =−7 ...(ii)
x y
On solving equation (i) & (ii)
x=1 1+1
⎫
& y = 1/3 ⎪
⎬
⎪
∴ 5x – 3y = 4 ⎭ 1

OR
Taxi charges in a city consist of fixed charges and the remainings charges depend upon the
distance travelled. For a journey of 10 km, the charge paid is ` 75 and for a journey of 15 km,
the charge paid is ` 110. Find the fixed charge and charges per km. Hence, find the charge of
covering a distance of 35 km.

1
Let fixed charge be ` x and charges per km be ` y
2
x + 10y = 75 ...(i) ⎫⎪
⎬
x + 15y = 110 ...(ii) ⎪ 1
⎭
Solve equation (i) & (ii)

x =5 ⎤ 1 1
& y = 7 ⎥⎦
+
2 2

1
∴ Total charge for 35 km = x + 35y = ` 250
2

(12)

30/4/1

32. Prove that:

sin θ − cos θ + 1 1
=
cos θ + sin θ − 1 sec θ − tan θ

sin θ − cos θ + 1
Sol. L.H.S =
cos θ + sin θ − 1
Dividing Nr and Dr by cos θ

tan θ − 1 + sec θ
= 1
1 + tan θ − sec θ

tan θ + sec θ − 1
= 1
(sec θ − tan 2 θ) + tan θ − sec θ
2

tan θ + sec θ − 1
=
(sec θ − tan θ) (sec θ + tan θ − 1)

1
= = R.H.S 1
sec θ − tan θ

33. In Figure-8, find the area of the shaded region where a circular arc of radius 7 cm has been
22
drawn with vertex O of an equilateral traiangle OAB of side 14 cm as centre. (Use π = and
7
3 = 1.73)

O
m
7c

A 14 cm B

Fig. 8

π r 2θ 3 2
Sol. Area of shaded ragion = + a 1
360° 4

π × 7 2 × 300° 3
= + × 142 1
360° 4
= 213.1 cm2 1

(13)

30/4/1

34. Construct a triangle with sides 5 cm, 6 cm and 7 cm. Now construct another triangle whose sides
2
are times the corresponding sides of the first triangle.
3
Sol. Correct construction of given triangle 1

Correct constriction of similar triangle with scale 2/3. 2

OR
Draw a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle
of 60°.
Sol. Correct construction of circle with radius 3 cm. 1

Correct constrcution of two tangents. 2

SECTION D
Question numbers 35 to 40 carry 4 marks each.
35. In a flight of 600 km, the speed of the aircraft was slowed down due to bad weather. The
average speed of the trip was decreased by 200 km/hr and thus the time of flight increased by
30 minutes. Find the average speed of the aircraft originally.
Sol. Let average speed of aircraft be x km/h

600 600 1
− = 2
x − 200 x 2
x2 – 200x – 240000 = 0 1

(x – 600) (x + 400) = 0
x = 600 km/h 1

∴ Original speed = 600 km/h
OR
` 9,000 were divided equally among a certain number of persons. Had there been 20 more
persons, each would have got ` 160 less. Find the original number of persons.
Let original number of persons be x

9000 9000
− = 160 2
x x + 20
x2 + 20x – 1125 = 0 1

(x + 45) (x – 25) = 0
x = 25
∴ Number of persons = 25 1

(14)

30/4/1

36. Draw a 'more than' cumulative frequency curve for the following distribution. Also, find the
median from the graph.

Weight (in kg): 40 – 44 44 – 48 48 – 52 52 – 56 56 – 60 60 – 64 64 – 68
Number of 7 12 33 47 20 11 5
Students:

Sol. Points to be plotted for more than ogive are
(40, 135), (44, 128), (48, 116), (52, 83), (56, 36), (60, 16), (64, 5) 2

1
For drawing correct ogive 1
2

1
For correct median = 53.3 (Approx.)
2

37. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct
points, then prove that the other two sides are divided in the same ratio.

1
Sol. For correct given, To prove, Construction and figure 4× =2
2

For correct proof 2

OR
In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the
squares of the other two sides.

1
Sol. For correct given, To prove, construction & figure 4× =2
2

For correct proof 2

38. 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. After covering a distance of 50 m, the angle of depression of the car becomes 60°. Find
the height of the tower. (Use 3 = 1-73).
Sol. D Let height of tower be h m and BC = x m Correct figure 1
30°
60°
h
h tan 60° =
x
30° 60°
A 50 m B x C
⇒h= 3x ...(i) 1

h
tan 30° =
x + 50

x + 50 = 3h ...(ii) 1

(15)

30/4/1

From equation (i) & (ii)

x = 25 m, h = 25 3 m

= 43.25 m 1

39. A bucket open at the top has top and bottom radii of circular ends as 40 cm and 20 cm
respectively. Find the volume of the bucket if its depth is 21 cm. Also find the area of the tin
22
sheet required for making the bucket. (Use π = )
7

πh 2 2
Sol. Volume = [R + r + Rr]
3

22 21 2
= × [40 + 202 + 40 × 20] 1
7 3

1
= 61600 cm3
2

l= h 2 + (R − r ) 2 = 29 cm 1

Area of tin = πl(R + r) + πr2
= π[29 × 60 + 400] 1

1
= 6725.7 cm2
2

40. Obtain other zeroes of the polynomial
f(x) = 2x4 + 3x3 – 5x2 – 9x – 3

if two of its zeroes are 3 and – 3
Sol. f(x) = 2x4 + 3x3 – 5x2 – 9x – 3

Q 3 and − 3 and zeroes of f(x)

1
∴ (x − 3) and (x + 3) are factors of f(x)
2

1
∴ x2 – 3 is a factor of f(x)
2

2x 4 + 3x 3 − 5x 2 − 9 x − 3
q(x) = 2
x2 − 3
= 2x2 + 3x + 1

(16)

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For zeroes q(x) = 0
∴ 2x2 + 3x + 1 = 0

1
(x + 1) (2x + 1) = 0
2

x = –1, –1/2

1
∴ Remaining zeroes are –1 & –1/2
2

OR
Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are recipro-
cals of the zeroes of the polynomial 5x2 + 2x – 3.
Let zeroes of given quadratic polynomial be α and β

−2 ⎤
α+β= ⎥
5
⎥
−3 ⎥ 1
αβ =
5 ⎥⎦

Now,

−2
1 1 β+α 5 2
+ = = = 1
α β αβ −3 3
5

1 −5
= 1
αβ 3
Required Polynomial is

2 5
x2 − x− 1
3 3
or
3x2 – 2x – 5

(17)

30/4/2

QUESTION PAPER CODE 30/4/2
EXPECTED ANSWER/VALUE POINTS
SECTION A
Question numers 1 to 20 carry 1 mark each.
Question numbers 1 to 10 are multiple choice questions. Choose the correct option.
1. It is being given that the points A(l, 2), B(0, 0) and C(a, b) are collinear. Which of the following
relations between a and b is true?
(A) a = 2b (B) 2a = b (C) a + b = 0 (D) a – b = 0
Sol. (B) 2a = b 1

2. In Figure-2, TP and TQ are tangents drawn to the circle with centre at O. If ∠POQ = 115° then
∠PTQ is
P T

115°
O
Q

Fig. 2
(A) 115° (B) 57.5° (C) 55° (D) 65°
Sol. (D) 65º 1

OR
From an external point Q, the length of the tangent to a circle is 5 cm and the distance of Q
from the centre is 8 cm. The radius of the circle is

(A) 39 cm (B) 3 cm (C) 39 cm (D) 7 cm

Sol. (C) 39 cm 1

3. The mean and median of a distribution are 14 and 15 respectively. The value of mode is
(A) 16 (B) 17 (C) 18 (D) 13
Sol. (B) 17 1

4. The equation x2 – 8x + k = 0 has real and distinct roots if
(A) k = 16 (B) k > 16 (C) k = 8 (D) k < 16
Sol. (D) k < 16 1

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30/4/2

5. The first term of an A.P. is 5 and the last term is 45. If the sum of all the terms is 400, the
number of terms is
(A) 20 (B) 8 (C) 10 (D) 16
Sol. (D) 16 1

OR
The 9th term of the A.P. – 15, –11, –7, ..., 49 is
(A) 32 (B) 0 (C) 17 (D) 13
Sol. (C) 17 1

6. The number of zeroes for a polynomial p(x) where graph of y = p(x) is tgiven in Figure-1, is
(A) 3 (B) 4 (C) 0 (D) 5
y
y = p(x)

x
0

Fig. 1
Sol. (A) 3 1

7. A bag contains 3 red, 5 black and 7 white balls. A ball is drawn from the bag at random. The
probability that the drawn is not black, is

1 9 5 2
(A) (B) (C) (D)
3 15 10 3
Sol. (D) 2/3 1

8. The value of θ for which cos (10° + θ) = sin 30°, is
(A) 50° (B) 40° (C) 80° (D) 20°
Sol. (A) 50° 1

⎛a ⎞
9. Point P ⎜ , 4 ⎟ is the mid-point of the line segment joining the points A(– 5, 2) and B(4, 6). The
⎝8 ⎠
value of ‘a’ is
(A) –4 (B) 4 (C) –8 (D) –2
Sol. (A) –4 1

(19)

30/4/2

10. The pair of equations, x = 0 and x = –4 has
(A) a unique solution (B) no solution
(C) infinitely many solutions (D) only solution (0, 0)
Sol. (B) No solution 1

Fill in the blanks in question numbers 11 to 15.
11. The distance between the points (a, b) and (– a, – b) is _______.

Sol. 2 a 2 + b2 1

12. If tan A = 1, then 2 sin A cos A = _______.
Sol. 1 1

⎛2+ 5 ⎞
13. ⎜ ⎟ is _______ number..
⎝ 3 ⎠
Sol. irrational 1

14. A spherical metal ball of radius 8 cm is melted to make 8 smaller identical balls. The radius of
each new ball is _______ cm.
Sol. 4 1

15. Let ΔABC ~ ΔDEF and their areas be respectively 81 cm2 and 144 cm2. If EF = 24 cm, then
length of side BC is _______ cm.
Sol. 18 1

Answer the following question numbers 16 to 20.
229
16. After how many decimal places will the decimal representation of the rational number
2 × 57
2

terminate?
Sol. After 7 decimal places 1

17. Given that HCF (120, 160) = 40, find LCM (120, 160).
120 × 160 1
Sol. LCM =
40 2
1
= 480
2

(20)

30/4/2

18. In Figure-4, AB and CD are common tangents to circle which touch each other at D. If AB =
8 cm, then find the length of CD.

B
C
A
D

Fig. 4

1
Sol. AC = CD = BC
2

1
CD = 4 cm
2

19. Two dice are thrown simultaneously. What is the probability that the sum of the two numbers
appearing on the top is 13?
Sol. P(E) = 0 1
20. In Figure-3, a tightly stretched rope of length 20 m is tied from the top of a vertical pole to the
ground. Find the height of the pole if the angle made by the rope with the ground is 30°.
B

20 m

30°
C
A
Fig. 3
AB 1
Sol. sin 30° =
20 2
1
AB = 10 m
2

SECTION B
Question numbers 21 to 26 carry 2 marks each.
21. Tree Plantation Drive
A group Housing Society has 600 members, who have their houses in the campus and decided
to hold a Tree Plantation Drive on the occasion of New Year. Each household was given he
choice of planting a sampling of its choice. The number of different types of sampings planted
were:

(21)

30/4/2

(i) Neem – 125
(ii) Peepal – 165
(iii) Creepers – 50
(iv) Fruit plants – 150
(v) Flowering plants – 110
On the opening ceremony, one of the plants is selected randomly for a prize. After reading the
above passage, answer the following questions.
What is the probability that the selected plant is
(i) A fruit plant or a flowering plant?
(ii) Either a Neem plant or a Peepal plant?
Sol. Total outcomes = 600

260 13
(i) P(Fruit plant or a flowering plant) = or 1
600 30

290 29
(ii) P(either neem plant or a peepal plant) = or 1
600 60
22. Find the mode of the following distribution:

Classes: 10 – 20 20 – 40 40 – 60 60 – 80 80 – 100
Frequency: 10 8 12 16 4

1
Sol. Model class = 60 – 80
2

⎛ f1 − f 0 ⎞ ⎛ 16 − 12 ⎞
Mode = l + ⎜ ⎟ × h = 60 + ⎜ ⎟ × 20 1
⎝ 2f1 − f 0 − f 2 ⎠ ⎝ 32 − 12 − 4 ⎠

1
= 65
2

OR
From the following distribution, find the median:

Classes: 500 – 600 600 – 700 700 – 800 800 –900 900 – 1000
Frequency: 36 32 32 20 30

1
Median class: 700 – 800
2

(22)

30/4/2

⎛N ⎞
⎜ − cf ⎟
Median = l + ⎝ 2 ⎠×h
f

75 − 68
= 700 + × 100 1
32

1
= 721.88
2

23. In Figure-6, a tent is in the shape of a cylinder surmounted by a conical top. The cylindrical part
is 2.1 m high and conical part has slant height 2.8 m. Both the parts have same radius 2 m. Find
22
the area of the canvas used to make the tent. (Use π = )
7

2.8 m

2.1 m
2m

Fig. 6
Sol. Area of canvas = πr(2h + l)

22
= × 2 (2 × 2.1 + 2.8) 1
7
= 44 m2 1

24. Solve for x:
8x2 – 2x – 3 = 0
Sol. 8x2 – 6x + 4x – 3 = 0 1

1
(4x – 3) (2x + 1) = 0
2

3 1 1
x= , x= −
4 2 2

(23)

30/4/2

25. The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first
triangle is 9 cm long, find the length of the corresponding side of the second triangle.
Sol. Let the side of other triangle be x cm

1
Q Ratio of perimeters of two similar triangles is equal to ratio of their corresponding sides
2

9 30
∴ = 1
x 20

1
x = 6 cm
2

OR
In Figure-5, ΔPQR is right-angled at P. M is a point on QR such that PM is perpendicular to
QR. Show that PQ2 = QM × QR.

R

M

P Q

Fig. 5
ΔPQM ~ ΔRQP [By AA similarity] 1

PQ QM
∴ =
RQ PQ

⇒ PQ2 = QM × QR 1

26. Evaluate:

cos 72° sin11°
+ – tan 15° tan 75°
sin 18° cos 79°

cos(90° − 18°) sin(90° − 79°) 1
Sol. + − tan(90° − 75°) . tan 75° 1
sin 18° cos 79° 2

sin 18° cos 79°
= + − cot 75° tan 75°
sin18° cos 79°

1
=1+1–1=1
2

(24)

30/4/2

SECTION C
Question numbers 27 to 34 carry 3 marks each.
27. In Figure-7, two tangents PA and PB are drawn to a circle with centre O from an external point
P. Prove that ∠APB = 2 ∠OAP.
P . P r o v e t h a t

A

O P

B

Fig. 7
Sol. ∠AOB = 180° – ∠APB 1

In ΔAOB, ∠AOB + ∠OAB + ∠OBA = 180° 1

⇒ 180° – ∠APB + ∠OAB + ∠OBA = 180°
⇒ ∠APB = 2∠AOB 1

28. Solve the pair of equations:

2 3 5 4
+ = 11, − = –7
x y x y
Hence, find the value of 5x – 3y.

2 3
Sol. + = 11 ...(i)
x y

5 4
− =−7 ...(ii)
x y
On solving equation (i) & (ii)
x=1 ⎫
1+1
⎪
& y = 1/3 ⎬
⎪
∴ 5x – 3y = 4 ⎭ 1

OR
Taxi charges in a city consist of fixed charges and the remainings charges depend upon the
distance travelled. For a journey of 10 km, the charge paid is ` 75 and for a journey of 15 km,
the charge paid is ` 110. Find the fixed charge and charges per km. Hence, find the charge of
covering a distance of 35 km.

(25)

30/4/2

1
Let fixed charge be ` x and charges per km be ` y
2
x + 10y = 75 ...(i) ⎫
⎪
⎬
x + 15y = 110 ...(ii) ⎪ 1
⎭
Solve equation (i) & (ii)

x =5 ⎤ 1 1
& y = 7 ⎥⎦
+
2 2

1
∴ Total charge for 35 km = x + 35y = ` 250
2

29. Construct a triangle with side 5 cm, 6 cm and 7 cm. Now construct another triangle whose side
2
are times the corresponding sides of the first triangle.
3
Sol. Correct construction of given triangle 1

Correct constriction of similar triangle with scale 2/3. 2

OR
Draw a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle
of 60°.
Sol. Correct construction of circle with radius 3 cm. 1

Correct constrcution of two tangents. 2

30. Prove that 5 is an irrational number..

Sol. Let 5 be a rational number

a 1
5= b≠0 HCF (a, b) = 1
b 2

a2
⇒ 5= 2
, a 2 = 5b 2
b
5 divides a 1

Put a = 5c (for some integer c)
⇒ 25c2 = 5b2 ⇒ b2 = 5c2

1
then we get, 5 divides b
2

Contradiction arises as HCF (a, b) = 1

(26)

30/4/2

∴ Our assumption is wrong

∴ 5 is irrational number 1

31. If the sum of the first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum
of the first 11 terms.
Sol. Let a be first term and d be common difference
Sum of first 6 terms = 36 ⇒ 2a = 12 – 5d 1

1
Sum of first 16 terms = 256 ⇒ 2a = 32 – 15d
2

Getting a = 1, d = 2 1

1
Getting the sum of first 11 terms = 121
2

32. Find the co-ordinates of the points of trisection of the line segment joining the points (3, – 1)
and (6,8).
Sol. Case I: If C and D trisect AB

C D 1
A (3, –1) B (6, 8) then C divides AB in the ratio 1 : 2
2

Co-ordinates of C: x = 1 × 6 + 2 × 3 = 4
1
3 2

1 × 8 + 2 (–1) 1
and y = =2
3 2

∴ Co-ordinates of C(4, 2)

1
Case II: Coordinates of D if D divides AB in the ratio 2 : 1
2

2 × 6 + 1× 3 1
Co-ordinates of D: x′ = =5
3 2

2 × 8 + 1 × (–1) 1
y′ = =5
3 2

Coordinates of D = (5, 5)

(27)

30/4/2

OR
Find the area of a quadrilateral ABCD having vertices at A(l, 2), B(l, 0), C(4, 0) and D(4, 4).

1
ar (ΔABC) = [1(0 – 0) + 1(0 – 2) + 4(2 – 0)]
2

1
= 3 sq. units 1
2

1
ar (ΔACD) = [1(0 – 4) + 4(4 – 2) + 4(2 – 0)]
2
= 6 sq. units 1

1
∴ Area of quadrialteral = 3 + 6 = 9 sq. units
2

33. Prove that:

cos A − sin A + 1
= cosec A + cot A
cos A + sin A – 1
Sol. Dividing Nr & Dr by sin A in LHS

cot A − 1 + cosec A
= 1
cot A + 1 − cosec A

cot A + cosec A − ( cosec2 A − cot 2 A)
= 1
cot A + 1 − coscec A
= cosec A + cot A 1

34. In Figure-8, find the area of the shaded region where a circular arc of radius 7 cm has been
22
drawn with vertex O of an equilateral traiangle OAB of side 14 cm as centre. (Use π = and
7
3 = 1.73)

O
m
7c

A 14 cm B

Fig. 8

(28)

30/4/2

πr 2θ 3 2
Sol. Area of shaded region = + a 1
360° 4

π × 7 2 × 300° 3
= + × 142 1
360° 4
= 213.1 cm2 1

SECTION D
Question numbers 35 to 40 carry 4 marks each.
35. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct
points, then prove that the other two sides are divided in the same ratio.
1
Sol. For correct given, To prove, Construction and figure 4× =2
2
For correct proof 2

OR
In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the
squares of the other two sides.

1
Sol. For correct given, To prove, construction & figure 4× =2
2
For correct proof 2

36. A bucket open at the top has top and bottom radii of ciecular ends as 40 cm and 20 cm
respectively. Find the volume of the bucket if its depth is 21 cm. Also find the area of the tin
22
sheet required for making the bucket. (Use π = )
7
πh 2 2
Sol. Volume = [R + r + Rr]
3
22 21 2
= × [40 + 202 + 40 × 20] 1
7 3
1
= 61600 cm3
2

l= h 2 + (R − r ) 2 = 29 cm 1

Area of tin = πl(R + r) + πr2
= π[29 × 60 + 400] 1

1
= 6725.7 cm2
2

(29)

30/4/2

37. Obtain other zeroes of the polynomial
f(x) = 2x4 + 3x3 – 5x2 – 9x – 3

if two of its zeroes are 3 and – 3
Sol. f(x) = 2x4 + 3x3 – 5x2 – 9x – 3

Q 3 and − 3 and zeroes of f(x)

1
∴ (x − 3) and (x + 3) are factors of f(x)
2

1
∴ x2 – 3 is a factor of f(x)
2

2x 4 + 3x 3 − 5x 2 − 9 x − 3
q(x) = 2
x2 − 3
= 2x2 + 3x + 1
For zeroes q(x) = 0
∴ 2x2 + 3x + 1 = 0

1
(x + 1) (2x + 1) = 0
2

x = –1, –1/2

1
∴ Remaining zeroes are –1 & –1/2
2

OR
Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are recipro-
cals of the zeroes of the polynomial 5x2 + 2x – 3.
Let zeroes of given quadratic polynomial be α and β

−2 ⎤
α+β= ⎥
5
⎥
−3 ⎥ 1
αβ =
5 ⎥⎦

Now,

−2
1 1 β+α 5 2
+ = = = 1
α β αβ −3 3
5
(30)

30/4/2

1 −5
= 1
αβ 3
Required Polynomial is

2 5
x2 − x− 1
3 3
or
3x2 – 2x – 5

38. Draw a ‘less than ogive for the following distribution. Hence, find median from the graph.

Marks Number of Students
0 – 10 2
10 – 20 8
20 – 30 12
30 – 40 10
40 – 50 16
50 – 60 8
60 – 70 3
70 – 80 1

Sol. Plotting the points (10, 2), (20, 10), (30, 22)
(40, 32), (50, 48), (60, 56), (70, 59), (80, 60) 2

1
Drawing the correct Ogive 1
2

1
Finding correct Median = 38
2

39. In a flight of 600 km, the speed of the aircraft was slowed down due to bad weather. The
average speed of the trip was decreased by 200 km/hr and thus the time of flight increased by
30 minutes. Find the average speed of the aircraft originally.
Sol. Let average speed of aircraft be x km/h

600 600 1
− = 2
x − 200 x 2
x2 – 200x – 240000 = 0 1

(x – 600) (x + 400) = 0
x = 600 km/h 1

∴ Original speed = 600 km/h

(31)

30/4/2

OR
` 9,000 were divided equally among a certain number of persons. Had there been 20 more
persons, each would have got ` 160 less. Find the original number of persons.
Let original number of persons be x

9000 9000
− = 160 2
x x + 20
x2 + 20x – 1125 = 0 1

(x + 45) (x – 25) = 0
x = 25
∴ Number of persons = 25 1

40. The angle of elevation of an airplane from point A on the ground is 60°. After a flight of 10
seconds, on the same height, the angle of elevation from point A becomes 30°. If the airplane
is flying at the speed of 720 km/hr, find the constant height at which the airplane is flying.
Sol. C E
Correct figure 1

h 1
60° h Distance travelled in 10 seconds = 2000 m
2
30°
A x B 2000 m D h
Getting x = (In ΔABC) 1
3

ED 1 h
In ΔEDA tan 30° = ⇒ = 1
AD 3 x + 2000

1
Getting correct value of h = 1000 3 m.
2

(32)

30/4/3

QUESTION PAPER CODE 30/4/3
EXPECTED ANSWER/VALUE POINTS
SECTION A
Question numers 1 to 20 carry 1 mark each.
Question numbers 1 to 10 are multiple choice questions. Choose the correct option.
1. The mean and median of a distribution are 14 and 15 respectively. The value of mode is
(A) 16 (B) 17 (C) 18 (D) 13
Sol. (B) 17 1

2. The quadratic equation x2 – 4x + k = 0 has distinct real roots if
(A) k = 4 (B) k > 4 (C) k =16 (D) k < 4
Sol. (D) K < 4 1

3. The first term of an A.P. is 5 and the last term is 45. If the sum of all the terms is 400, the
number of terms is
(A) 20 (B) 8 (C) 10 (D) 16
Sol. (D) 16 1
OR
th
The 9 term of the A.P. – 15, –11, –7, ..., 49 is
(A) 32 (B) 0 (C) 17 (D) 13
Sol. (C) 17 1

⎛a ⎞
4. Point P ⎜ , 4 ⎟ is the mid-point of the line segment joining the points A(– 5, 2) and B(4, 6). The
⎝8 ⎠
value of ‘a’ is
(A) –4 (B) 4 (C) –8 (D) –2
Sol. (A) –4 1

5. The number of zeroes for a polynomial p(x) whose graph is given in Figure-1, is
y
y = p(x)

x
0

Fig. 1
(A) 4 (B) 3 (C) 5 (D) 1
Sol. (B) 3 1

(33)

30/4/3

6. It is being given that the points A(l, 2), B(0, 0) and C(a, b) are collinear. Which of the following
relations between a and b is true?
(A) a = 2b (B) 2a = b (C) a + b = 0 (D) a – b = 0
Sol. (B) 2a = b 1

7. The value of θ for which sin (44° + θ) = cos 30°, is
(A) 46° (B) 60° (C) 16° (D) 90°
Sol. (C) 16° 1

8. The pair of linear equations y = 0 and y = –6 has
(A) a unique solution (B) no solution
(C) infinetly many solutions (D) only solution (0, 0)
Sol. (B) No solution 1

9. A bag contains 3 red, 5 black and 7 white balls. A ball is drawn from the bag at random. The
probability that the drawn is not black, is

1 9 5 2
(A) (B) (C) (D)
3 15 10 3
Sol. (D) 2/3 1

10. In Figure-2, TP and TQ are tangents drawn to the circle with centre at O. If ∠POQ = 115° then
∠PTQ is
P T

115°
O
Q

Fig. 2
(A) 115° (B) 57.5° (C) 55° (D) 65°
Sol. (D) 65º 1
OR
From an external point Q, the length of the tangent to a circle is 5 cm and the distance of Q
from the centre is 8 cm. The radius of the circle is

(A) 39 cm (B) 3 cm (C) 39 cm (D) 7 cm

Sol. (C) 39 cm 1

(34)

30/4/3

Fill in the blanks in question numbers 11 to 15.
11. The distance between the points (a, b) and (– a, – b) is _______.

Sol. 2 a 2 + b2 1

12. A spherical metal ball of radius 8 cm is melted to make 8 smaller identical balls. The radius of
each new ball is _______ cm.
Sol. 4 1

⎛2+ 5 ⎞
13. ⎜ ⎟ is _______ number..
⎝ 3 ⎠
Sol. irrational 1

14. Let ΔABC ~ ΔDEF and their areas be respectively 81 cm2 and 144 cm2. If EF = 24 cm, then
length of side BC is _______ cm.
Sol. 18 1

15. If tan A = 1, then 2 sin A cos A = _______.
Sol. 1 1

Answer the following question numbers 16 to 20.

229
16. After how many decimal places will the decimal representation of the rational number
2 × 57
2
terminate?
Sol. After 7 decimal place 1

17. In Figure-4, AB and CD are common tangents to circle which touch each other at D. If AB =
8 cm, then find the length of CD.

B
C
A
D

Fig. 4

1
Sol. AC = CD = BC
2

1
CD = 4 cm
2

(35)

30/4/3

18. Given that HCF (135, 225) = 45, find the LCM (135, 225).

135 × 225 1
Sol. LCM =
45 2
1
= 675
2

19. In Figure-3, a tightly stretched rope of length 20 m is tied from the top of a vertical pole to the
ground. Find the height of the pole if the angle made by the rope with the ground is 30°.
B

20 m

30°
C
A
Fig. 3
AB 1
Sol. sin 30° =
20 2
1
AB = 10 m
2

20. Two dice are thrown similtaneously. What is the probability that the product of the numbers
appearing on the top is 1?

1
Sol. Total outcomes = 36
2

Number of favourable outcomes = 1

1 1
Required probability =
36 2

SECTION B
Question numbers 21 to 26 carry 2 marks each.
21. Find the mode of the following distribution:

Classes: 10 – 20 20 – 40 40 – 60 60 – 80 80 – 100
Frequency: 10 8 12 16 4

1
Sol. Modal class = 60 – 80
2
⎛ f1 − f 0 ⎞ ⎛ 16 − 12 ⎞
Mode = l + ⎜ 2f − f − f ⎟ × h = 60 + ⎝⎜ 32 − 12 − 4 ⎠⎟ × 20 1
⎝ 1 0 2⎠
1
= 65
2

OR
(36)

30/4/3

From the following distribution, find the median:

Classes: 500 – 600 600 – 700 700 – 800 800 –900 900 – 1000
Frequency: 36 32 32 20 30

1
Median class: 700 – 800
2

⎛N ⎞
⎜ − cf ⎟
Median = l + ⎝ 2 ⎠×h
f
75 − 68
= 700 + × 100 1
32
1
= 721.88
2

22. In Figure-6, a tent is in the shape of a cylinder surmounted by a conical top. The cylindrical part
is 2.1 m high and conical part has slant height 2.8 m. Both the parts have same radius 2 m. Find
22
the area of the canvas used to make the tent. (Use π = )
7

2.8 m

2.1 m
2m

Fig. 6
Sol. Area of canvas = πr(2h + l)
22
= × 2 (2 × 2.1 + 2.8) 1
7
= 44 m2 1

23. Sovle for x:
14x2 + 17x – 6 = 0
Sol. 14x2 + 21x – 4x – 6 = 0 1

1
⇒ (2x + 3) (7x – 2) = 0
2

−3 2 1
⇒ x= ,x=
2 7 2

(37)

30/4/3

24. The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first
triangle is 9 cm long, find the length of the corresponding side of the second triangle.
Sol. Let the side of other triangle be x cm

1
Q Ratio of perimeters of two similar triangles is equal to ratio of their corresponding sides
2

9 30
∴ = 1
x 20

1
x = 6 cm
2

OR
In Figure-5, ΔPQR is right-angled at P. M is a point on QR such that PM is perpendicular
to QR. Show that PQ22 = QM × QR.
t o Q R . S h o w t h a t P Q

R

M

P Q

Fig. 5
ΔPQM ~ ΔRQP [By AA similarity] 1

PQ QM
∴ =
RQ PQ

⇒ PQ2 = QM × QR 1

25. Tree Plantation Drive
A group Housing Society has 600 members, who have their houses in the campus and decided
to hold a Tree Plantation Drive on the occasion of New Year. Each household was given he
choice of planting a sampling of its choice. The number of different types of sampings planted
were:
(i) Neem – 125
(ii) Peepal – 165
(iii) Creepers – 50
(iv) Fruit plants – 150
(v) Flowering plants – 110

(38)

30/4/3

On the opening ceremony, one of the plants is selected randomly for a prize. After reading the
above passage, answer the following questions.
What is the probability that the selected plant is
(i) A fruit plant or a flowering plant?
(ii) Either a Neem plant or a Peepal plant?
Sol. Total outcomes = 600

260 13
(i) P(Fruit plant or a flowering plant) = or 1
600 30

290 29
(ii) P(either neem plant or a peepal plant) = or 1
600 60
26. Evaluate:

2 sin 68° 2 cot 15°
− – 3 tan 40° tan 45° tan 50°
cos 22° tan 75°

2 sin(90° − 22°) 2 cot(90° − 75°) 1
Sol. − − 3 tan(90° − 50°) . tan 50° 1
cos 22° tan 75° 2

1
=–3
2

SECTION C
Question numbers 27 to 34 carry 3 marks each.
27. Solve the pair of equations:

2 3 5 4
+ = 11, − = –7
x y x y
Hence, find the value of 5x – 3y.

2 3
Sol. + = 11 ...(i)
x y

5 4
− =−7 ...(ii)
x y
On solving equation (i) & (ii)
x=1 ⎫ 1+1
⎪
& y = 1/3 ⎬
⎪
∴ 5x – 3y = 4 ⎭ 1

(39)

30/4/3

OR
Taxi charges in a city consist of fixed charges and the remainings charges depend upon the
distance travelled. For a journey of 10 km, the charge paid is ` 75 and for a journey of 15 km,
the charge paid is ` 110. Find the fixed charge and charges per km. Hence, find the charge of
covering a distance of 35 km.

1
Let fixed charge be ` x and charges per km be ` y
2
x + 10y = 75 ...(i)
x + 15y = 110 ...(ii) 1
Solve equation (i) & (ii)

x =5 ⎤ 1 1
& y = 7 ⎥⎦
+
2 2

1
∴ Total charge for 35 km = x + 35y = ` 250
2

28. In Figure-7, AB is the diameter of a circle with centre O and AC is its chord such that ∠BAC
= 30°. If the tangent drawn at C intersects extended AB at D, then show that BC = BD.

C

30°
A O B D

Fig. 7
Sol. OA = OC
1
⇒ ∠OCA = 30°
2
∠OCB = ∠ACB – ∠ACO
= 90° – 30° = 60° 1

∠BCD = 90° – ∠OCB

1
= 90° – 60° = 30° ...(i)
2

In ΔACD,
∠ACD + ∠CAD + ∠CDA = 180°
90° + 30° + 30° + ∠CDA = 180°

(40)

30/4/3

1
∠CDA = 30° ...(ii)
2

From (i) and (ii)
∠BCD = ∠CDA

1
⇒ BC = BD (In ΔCBD)
2

29. Prove that:

sin θ − cos θ + 1 1
=
cos θ + sin θ − 1 sec θ − tan θ

sin θ − cos θ + 1
Sol. L.H.S =
cos θ + sin θ − 1
Dividing Nr and Dr by cos θ

tan θ − 1 + sec θ
= 1
1 + tan θ − sec θ

tan θ + sec θ − 1
= 1
(sec θ − tan 2 θ) + tan θ − sec θ
2

tan θ + sec θ − 1
=
(sec θ − tan θ) (sec θ + tan θ − 1)

1
= = R.H.S 1
sec θ − tan θ

30. Construct a triangle with side 5 cm, 6 cm and 7 cm. Now construct another triangle whose side
2
are times the corresponding sides of the first triangle.
3
Sol. Correct construction of given triangle 1

Correct constriction of similar triangle with scale 2/3. 2

OR
Draw a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle
of 60°.
Sol. Correct construction of circle with radius 3 cm. 1

Correct constrcution of two tangents. 2

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31. Calculate the area of the shaded region common between two quadrants of circles of radius 7
cm each (as shown in Figure-8).

D C

P
7 cm
Q

A 7 cm B

Fig. 8
Sol. Area of Shaded Region
= 2 (Area of one sector ABPD) – Area of square ABCD 1

⎛ 90° × π × 7 2 ⎞ 1
2
= ⎜ ⎟ − 7×7 1
⎝ 360° ⎠ 2

1
= 28 cm2
2

32. Prove that 5 is an irrational number..

Sol. Let 5 be a rational number

a 1
5= b≠0 HCF (a, b) = 1
b 2

a2
⇒ 5= , a 2 = 5b 2
b2
5 divides a 1

Put a = 5c (for some integer c)
⇒ 25c2 = 5b2 ⇒ b2 = 5c2

1
then we get, 5 divides b
2

Contradiction arises as HCF (a, b) = 1
∴ Our assumption is wrong

∴ 5 is irrational number 1

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33. If 6 times the 6th term of an A.P. is equal of 9 times the 9th term, show that its 15th term is zero.
Sol. Let a be the first term and d be the common difference

1
6(a + 5d) = 9(a + 8d) 1
2

a = –14d 1

1
a + 14 d = 0 ⇒ 15th term = 0
2

34. Find the co-ordinates of the points of trisection of the line segment joining the points (3, – 1)
and (6,8).
Sol. C D Case I: If C and D trisect AB
A (3, –1) B (6, 8)

1
then C divides AB in the ratio 1 : 2
2

1× 6 + 2 × 3 1
Co-ordinates of C: x = =4
3 2

1 × 8 + 2 (–1) 1
and y = =2
3 2

∴ Co-ordinates of C(4, 2)

1
Case II: Co-ordinates of D if D divides AB in the ratio 2 : 1
2

2 × 6 + 1× 3 1
Co-ordinates of D: x′ = =5
3 2

2 × 8 + 1 × (–1) 1
y′ = =5
3 2

Co-ordinates of D = (5, 5)
OR
Find the area of a quadrilateral ABCD having vertices at A(l, 2), B(l, 0), C(4, 0) and D(4, 4).

C (4, 0) 1
(4, 4) ar (ΔABC) = [1(0 – 0) + 1(0 – 2) + 4(2 – 0)]
D 2
1
B (1, 0) = 3 sq. units 1
2
A (1, 2)

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1
ar (ΔACD) = [1(0 – 4) + 4(4 – 2) + 4(2 – 0)]
2
= 6 sq. units 1

1
∴ Area of quadrialteral = 3 + 6 = 9 sq. units
2

SECTION D
Question numbers 35 to 40 carry 4 marks each.
35. From the top of a 7 m building, the angle of elevation of the top of a cable tower is 60° and the
angle of depression of its foot is 45°. Determine the height of the tower. (Use 3 = 1.73)

Sol. D For correct figure 1
h–7 7
60° tan 45° = ⇒x=7 1
45° x E x
h
h−7
7m 7m tan 60° = 1
x
45°
C 1
7( 3 + 1) = h 2
1
h = 7 × 2.73 = 19.11 m
2

36. Obtain other zeroes of the polynomial
f(x) = 2x4 + 3x3 – 5x2 – 9x – 3

if two of its zeroes are 3 and – 3
Sol. f(x) = 2x4 + 3x3 – 5x2 – 9x – 3
Q 3 and − 3 and zeroes of f(x)
1
∴ (x − 3) and (x + 3) are factors of f(x)
2

1
∴ x2 – 3 is a factor of f(x)
2

2x 4 + 3x 3 − 5x 2 − 9 x − 3
q(x) =
x2 − 3
= 2x2 + 3x + 1 2
For zeroes q(x) = 0
∴ 2x2 + 3x + 1 = 0

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1
(x + 1) (2x + 1) = 0
2
x = –1, –1/2
1
∴ Remaining zeroes are –1 & –1/2
2

OR
Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are recipro-
cals of the zeroes of the polynomial 5x2 + 2x – 3.
Let zeroes of given quadratic polynomial be α and β

−2 ⎤
α+β= ⎥
5
⎥
−3 ⎥ 1
αβ =
5 ⎥⎦

Now,

−2
1 1 β+α 5 2
+ = = = 1
α β αβ −3 3
5

1 −5
= 1
αβ 3
Required Polynomial is

2 5
x2 − x− 1
3 3
or
3x2 – 2x – 5

37. A bucket open at the top has top and bottom radii of circular ends as 40 cm and 20 cm
respectively. Find the volume of the bucket if its depth is 21 cm. Also find the area of the tin
22
sheet required for making the bucket. (Use π = )
7
πh 2 2
Sol. Volume = [R + r + Rr]
3
22 21 2
= × [40 + 202 + 40 × 20] 1
7 3
1
= 61600 cm3
2

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l= h 2 + (R − r ) 2 = 29 cm 1

Area of tin = πl(R + r) + πr2
= π[29 × 60 + 400] 1

1
= 6725.7 cm2
2

38. In a flight of 600 km, the speed of the aircraft was slowed down due to bad weather. The
average speed of the trip was decreased by 200 km/hr and thus the time of flight increased by
30 minutes. Find the average speed of the aircraft originally.
Sol. Let average speed of aircraft be x km/h
600 600 1
− = 2
x − 200 x 2
2
x – 200x – 240000 = 0 1
(x – 600) (x + 400) = 0
x = 600 km/h 1
∴ Original speed = 600 km/h
OR
` 9,000 were divided equally among a certain number of persons. Had there been 20 more
persons, each would have got ` 160 less. Find the original number of persons.
Sol. Let original number of persons be x
9000 9000
− = 160 2
x x + 20
x2 + 20x – 1125 = 0 1
(x + 45) (x – 25) = 0
x = 25
∴ Number of persons = 25 1

39. Change the following distribution into ‘less than’ type distribution and draw its ogive.
Hence find the median of the distribution.

Marks Number of Students
20 – 30 4
30 – 40 10
40 – 50 12
50 – 60 14
60 – 70 8
70 – 80 3
80 – 90 4
90 – 100 5

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Sol. Less than type distribution table is:

Marks fi cf
Less than 30 4 4
Less than 40 10 14
Less than 50 12 26
Less than 60 14 40 Correct Table 2
Less than 70 8 48
Less than 80 3 51
Less than 90 4 55
Less than 100 5 60

1
For Drawing the correct Ogive 1
2

1
Getting correct median = 52.86
2

40. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct
points, then prove that the other two sides are divided in the same ratio.

1
Sol. For correct given, To prove, Construction and figure 4× =2
2

For correct proof 2

OR
In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the
squares of the other two sides.

1
Sol. For correct given, To prove, construction & figure 4× =2
2

For correct proof 2

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