CBSE Class 10 Mathematics Standard Question Paper 2020 Set 30-2 Solutions

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

QUESTION PAPER CODE 30/2/1
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Question numbers 1 to 10 are multiple choice questions of 1 mark each.
You have to select the correct choice :
Q.No. Marks

1. The sum of exponents of prime factors in the prime-factorisation of 196 is
(a) 3 (b) 4 (c) 5 (d) 2
Ans: (b) 4 1
2. Euclid’s division Lemma states that for two positive integers a and b,
there exists unique integer q and r satisfying a = bq + r, and
(a) 0 < r < b (b) 0 < r ≤ b (c) 0 ≤ r < b (d) 0 ≤ r ≤ b
Ans: (c) 0 ≤ r < b 1
3. The zeroes of the polynomial x2 – 3x – m (m + 3) are
(a) m, m + 3 (b) – m, m + 3 (c) m, – (m + 3) (d) – m, – (m + 3)
Ans: (b) – m, m + 3 1
4. The value of k for which the system of linear equations x + 2y = 3,
5x + ky + 7 = 0 is inconsistent is
14 2
(a) − (b) (c) 5 (d) 10
3 5
Ans: (d) 10 1
5. The roots of the quadratic equation x2 – 0.04 = 0 are
(a) ± 0.2 (b) ± 0.02 (c) 0.4 (d) 2
Ans: (a) ± 0.2 1
1 1 − p 1 − 2p
6. The common difference of the A.P. , , , … is
p p p
1 1
(a) 1 (b) (c) –1 (d) −
p p
Ans: (c) –1 1
7. The nth term of the A.P. a, 3a, 5a, …… is
(a) na (b) (2n – 1)a (c) (2n + 1) a (d) 2na
Ans: (b) (2n – 1)a 1
8. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is
(a) (2, 0) (b) (0, 2) (c) (3, 0) (d) (2, 2)
Ans: (a) (2, 0) 1
9. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis
are
(a) (3, 5) (b) (3, –5) (c) (–3, –5) (d) (–3, 5)
Ans: (c) (–3, –5) 1

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10. If the point P (6, 2) divides the line segment joining A(6, 5) and B(4, y) in
the ratio 3 : 1, then the value of y is
(a) 4 (b) 3 (c) 2 (d) 1
Ans: 1 mark be awarded to everyone 1
In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark.
ar(ΔAMN)
11. In fig. 1, MN || BC and AM : MB = 1 : 2, then = _________.
ar( ΔABC)
A

M N

B C
Fig. 1

1
Ans: 1
9
12. In given Fig. 2, the length PB = _________ cm.

A 5 cm
O
3 cm
P

B
Fig. 2

Ans: 4 1
13. In ΔABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm, then ∠B = _________.
Ans: 90° 1
OR
Two triangles are similar if their corresponding sides are _________.
Ans: proportional 1
14. The value of (tan 1º tan 2º …… tan 89º) is equal to _________.
Ans: 1 1
15. In Fig. 3, the angles of depressions from the observing positions O1 and O2
respectively of the object A are ______________ , _______________.
O2 O1

60°

45°
A
B C
Fig. 3

1 1
Ans: 30°, 45° +
2 2

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Q. Nos. 16 to 20 are short answer type questions of 1 mark each.
16. If sin A + sin2 A = 1, then find the value of the expression (cos2 A + cos4 A).

Ans: sin A = 1 − sin A ⎫
2
1/2
⎬
sin A = cos 2 A ⎭
cos2A + cos4 A = sin A + sin2 A = 1 1/2
17. In Fig. 4 is a sector of circle of radius 10.5 cm. Find the perimeter of
⎛ 22 ⎞
the sector. ⎜ Take π = ⎟
⎝ 7 ⎠

A B

60°

O
Fig. 4

πrθ
Ans: Perimeter = 2r +
180°
22 60°
= 2 × 10.5 + × 10.5 × 1/2
7 180°
= 21 + 11 = 32 cm 1/2
18. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3,
then find the probability of x2 < 4.
3
Ans: Number of Favourable outcomes = 3 i.e., {–1, 0, 1} ∴ P(x2 < 4) = 1/2+1/2
7
OR
What is the probability that a randomly taken leap year has 52 Sundays ?
5
Ans: P(52 sundays) = 1
7
19. Find the class-marks of the classes 10-25 and 35-55.
10 + 25 35 + 55
Ans: Class Marks = 17.5; = 45 1/2+1/2
2 2
20. A die is thrown once. What is the probability of getting a prime number.
Ans: Number of prime numbers = 3 i.e. ; {2, 3, 5} 1/2
3 1
P(Prime Number) = or 1/2
6 2

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SECTION – B
Q. Nos. 21 to 26 carry 2 marks each
21. A teacher asked 10 of his students to write a polynomial in one variable
on a paper and then to handover the paper. The following were the
answers given by the students:
2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3 + 3x + 7, 7x +
3
7 , 5x – 7x + 2,
5 1 1
2x2 + 3 – , 5x – , ax3 + bx2 + cx + d, x + .
x 2 x
Answer the following questions :
(i) How many of the above ten, are not polynomials ?
(ii) How many of the above ten, are quadratic polynomials ?
Ans: (i) 3 1
(ii) 1 1
A
22. In Fig. 5, ABC and DBC are two triangles on
the same base BC. If AD intersects BC
at O, show that C
B
ar ( ΔABC ) AO
=
ar ( ΔDBC ) DO D
Fig. 5
Ans: Draw AX ⊥ BC, DY ⊥ BC 1/2
A ΔAOX ~ ΔDOY

Y AX AO
C = …(i) 1/2
B XO DY DO
1
D ar ( ΔABC ) 2 ×BC×AX
=
ar ( ΔDBC ) 1 ×BC×DY 1/2
2
AX AO
= (From (1)) 1/2
DY DO
OR
In Fig. 6, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2.

C
D

B
A
Fig. 6

Ans: In rt ΔABD AB2 = BD2 + AD2 … (i) 1/2
In rt ΔADC CD2 = AC2 – AD2 … (ii) 1/2
Adding (i) & (ii)
AB2 + CD2 = BD2 + AC2 1

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cot 2 α
23. Prove that 1 + = cos ec α
1 + cos ec α

Ans: L.H.S = 1 + cos ec α − 1
2
1/2
1 + cos ec α

( cos ec α − 1) ( cos ec α + 1)
= 1+ 1
cos ec α + 1
= cosec α = R.H.S 1/2
OR
Show that tan θ + tan θ = sec θ – sec θ
4 2 4 2

Ans: L.H.S = tan4θ + tan2θ
= tan2θ (tan2θ + 1) 1/2

= (sec2θ – 1) (sec2θ) = sec4θ – sec2θ = R.H.S 1+1/2

24. The volume of a right circular cylinder with its height equal to the radius
1 ⎛ 22 ⎞
is 25 cm3. Find the height of the cylinder. ⎜ Use π = ⎟
7 ⎝ 7 ⎠
Ans: Let height and radius of cylinder = x cm 1/2
176 3
V= cm
7
22 2 176
×x ×x= 1/2
7 7
x =8 ⇒x=2
3
1/2
∴ height of cylinder = 2 cm 1/2
25. A child has a die whose six faces show the letters as shown below :
A B C D E A
The die is thrown once. What is the probability of getting (i) A, (ii) D ?
2 1 1
Ans: (i) P(A) = or (ii) P(D) = 1+1
6 3 6
26. Compute the mode for the following frequency distribution :
Size of items
(in cm) 0 – 4 4 – 8 8 – 12 12 – 16 16 – 20 20 – 24 24 – 28
Frequency 5 7 9 17 12 10 6
Ans: l = 12 f0 = 9 f1 = 17 f2 = 12 h = 4 1/2

17 − 9 1
Mode = 12 + × 4 = 14.46 cm (Approx) 1+
34 − 9 − 12 2

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SECTION – C

Question numbers 27 to 34 carry 3 marks each.

⎛y ⎞
27. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and ⎜ − 2 ⎟
⎝x ⎠
Ans: 2x + y = 23 , 4x – y = 19
Solving, we get x = 7, y = 9 1+1
y −5
5y – 2x = 31, −2= 1/2+1/2
x 7
OR
1 1 11
Solve for x : x + 4 − x + 7 = 30 , x # – 4, 7

1 1 11 −11 11
Ans: − = ⇒ (x + 4)(x − 7) = 30 1
x + 4 x − 7 30
⇒ x2 – 3x + 2 = 0 1
⇒ (x – 2) (x – 1) = 0 1/2
⇒ x = 2, 1 1/2
The Following solution should also be accepted
1 1 11 x + 7 − x − 4 11
− = ⇒ = 1
x + 4 x + 7 30 (x + 4)(x − 7) 30

1
⇒ 11 x2 + 121x + 218 = 0 1
2
Here, D = 5049

−121 ± 5049
x= 1/2
22
28. Show that the sum of all terms of an A.P. whose first term is a, the
(a + c)(b + c − 2a)
second term is b and the last term is c is equal to
2(b − a)
Ans: Here d = b – a 1/2
th
Let c be the n term
∴ c = a + (n – 1) (b – a) 1/2
c + b − 2a
⇒ n= 1
b−a
c + b − 2a
⇒ Sn = (a + c) 1
2(b − a)

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OR
Solve the equation : 1 + 4 + 7 + 10 + … + x = 287.
Ans: Let sum of n terms = 287
n
[ 2 ×1 + (n − 1)3] = 287 1/2
2
3n2 – n – 574 = 0 1/2
(3n + 41) (n – 14) = 0 1/2
⎛ −41 ⎞
n = 14 ⎜ Reject n = ⎟ 1/2
⎝ 3 ⎠
x = a14 = 1 + 13 × 3 = 40 1
29. In a flight of 600 km, an aircraft was slowed down due to bad weather.
The average speed of the trip was reduced by 200 km/hr and the time
of flight increased by 30 minutes. Find the duration of flight.
Ans: Let actual speed = x km/hr
A.T.Q
600 600 1
− = 1
x − 200 x 2

x2 – 200x – 240000 = 0
(x – 600) (x + 400) = 0 1
x = 600 (x = – 400 Rejected) 1/2
600
Duration of flight = = 1 hr 1/2
600
30. If the mid-point of the line segment joining the points A(3, 4) and
B(k, 6) is P (x, y) and x + y – 10 = 0, find the value of k.
P
Ans: A (3, 4) B
(x, y) (K, 6)

3+ k
x= y=5 1/2+1/2
2
3+ k
x + y – 10 = 0 ⇒ + 5 – 10 = 0 1
2
⇒ k=7 1
OR
Find the area of triangle ABC with A (1, – 4) and the mid-points A(1, -4)
of sides through A being (2, –1) and (0, –1).
(2, -1) (0, -1)
Ans: B(3, 2), C (–1, 2) 1/2+1/2
B C
1
Area = 1(2 − 2) + 3(2 + 4) − 1(− 4 − 2) = 12 sq.units 1+1
2

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31. In Fig. 7, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked
along them, then find the lengths of sides of each triangle.
A D
2x–1 3x 6x
18

B 2x+2 C E 3x+9 F
Fig. 7
Ans: As ΔABC ~ ΔDEF
2x − 1 3x
= 1
18 6x
x=5 1
AB = 9 cm DE = 18 cm
BC = 12 cm EF = 24 cm 1/2+1/2
CA = 15 cm FD = 30 cm
32. If a circle touches the side BC of a triangle ABC at P and extended sides AB
and AC at Q and R, respectively, prove that
1
AQ = (BC + CA + AB)
2
Ans: Correct Fig 1/2

1
Q
AQ = (2AQ) 1/2
2
B
1
P A
= (AQ + AQ)
2

C 1
= (AQ + AR)
R 2
1
= (AB + BQ + AC + CR) 1
2
1
= (AB + BC + CA) 1
2
Q [BQ = BP, CR = CP]

33. If sin θ + cos θ = 2 , prove that tan θ + cot θ = 2.
Ans: sin θ + cos θ = 2 1
tan θ + 1 = 2 sec θ
Sq. both sides
tan2 θ + 1 + 2 tan θ = 2sec2 θ
tan2 θ + 1 + 2 tan θ = 2(1 + tan2 θ) 1
tan2 θ + 1 + 2 tan θ = 2 + 2tan2 θ
2 tan θ = tan2 θ +1 1
2 = tan θ + cot θ

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34. The area of a circular play ground is 22176 cm2. Find the cost of fencing
this ground at the rate of `50 per metre.
Ans: Let the radius of playground be r cm
πr2 = 22176 cm2 1
r = 84 cm
22
Circumference = 2πr = 2 × × 84 = 528 cm 1
7
50
Cost of fencing = × 528 = `264 1
100
SECTION – D
Question numbers 35 to 40 carry 4 marks each.

35. Prove that 5 is an irrational number..
Ans: Let 5 be a rational number..
p
5 = , p & q are coprimes & q ≠ 0 1
q
5q2 = p2 ⇒ 5 divides p2 ⇒ 5 divides p also Let p = 5a, for some integer a 1
5q2 = 25a 2 ⇒ q2 = 5a2 ⇒ 5 divides q2 ⇒ 5 divides q also
∴ 5 is a common factor of p, q, which is not possible as 1
p, q are coprimes.
Hence assumption is wrong 5 is irrational no. 1
36. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of
larger diameter is used for four hours and the pipe of smaller diameter for
9 hours, only half of the pool can be filled. How long would it take for each
pipe to fill the pool separately ?
Ans: Let time taken by pipe of larger diameter to fill the tank be x hr
Let time taken by pipe of smaller diameter to fill the tank be y hr
A.T.Q
1 1 1 4 9 1
+ = , + = 1+1
x y 12 x y 2
Solving we get x = 20 hr y = 30 hr 1+1
37. Draw a circle of radius 2 cm with centre O and take a point P outside the
circle such that OP = 6.5 cm. From P, draw two tangents to the circle.
Ans: Correct construction of circle of radius 2 cm 1
Correct construction of tangents. 3
OR
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then construct another
3
triangle whose sides are times the corresponding sides of the first triangle.
4
Ans: Correct construction of given triangle 1
Construction of Similar triangle 3

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38. From a point on the ground, the angles of elevation of the bottom and
the top of a tower fixed at the top of a 20 m high building are 45° and
60° respectively. Find the height of the tower.
Ans: Let height of tower = h m
BC A
In rt. ΔBCD tan 45° = corr fig. 1
CD
hm
20
1 =
CD ⎫⎪
B
CD = 20 m ⎬ 1
⎪
AC ⎭
In rt. ΔACD tan 60° = 20 m
CD
20 + h 60°
3 = 20 C
45° 1
D

h = 20 ( 3 − 1) m 1
39. Find the area of the shaded region in Fig. 8, if PQ = 24 cm, PR = 7 cm
and O is the centre of the circle.
Q

O

R P
Fig. 8

25 1
Ans: ∠P = 90° RQ = (24) + 7 = 25 cm, r =
2 2
cm 1
2 2
Area of shaded portion = Area of semi circle – ar (Δ PQR) ⎫
⎪
1 22 ⎛ 25 ⎞
2
⎬ 2
= × × ⎜ ⎟ − 84 ⎪
2 7 ⎝ 2 ⎠ ⎭
= 161.54 cm2 1/2
OR
Find the curved surface area of the frustum of a cone, the diameters of
whose circular ends are 20 m and 6 m and its height is 24 m.
Ans: R = 10 m r = 3 m h = 24 m 1/2+1/2
l= (24) 2 + (10 − 3) 2 = 25 m 1
CSA = π(10 + 3)25 = 325 π m2 1+1
40. The mean of the following frequency distribution is 18. The frequency
f in the class interval 19 – 21 is missing. Determine f.
Class interval 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25
Frequency 3 6 9 13 f 5 4

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Ans: C.I f x xf
11-13 3 12 36
13-15 6 14 84
15-17 9 16 144
17-19 13 18 234
19-21 f 20 20f
21-23 5 22 110
23-25 4 24 96
40+f 704 + 20f 2

∑ xf 704 + 20f
Mean = ⇒ 18 = ⇒ f=8 2
∑f 40 + f
OR
The following table gives production yield per hectare of wheat of 100
farms of a village :
Production yield 40-45 45-50 50-55 55-60 60-65 65-70
No. of farms 4 6 16 20 30 24
Change the distribution to a ‘more than’ type distribution and draw its ogive.
Ans:
Production yield Number of farms
More than or equal to 40 100
More than or equal to 45 96
More than or equal to 50 90
More than or equal to 55 74
More than or equal to 60 54
More than or equal to 65 24 2

Plotting of points (40, 100) (45, 96) (50, 90) (55, 74) (60, 54) (65, 24)
join to get ogive. 2

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QUESTION PAPER CODE 30/2/2
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Question numbers 1 to 10 are multiple choice questions of 1 mark each.
You have to select the correct choice :
Q.No. Marks
1. The value of k for which the system of linear equations x + 2y = 3,
5x + ky + 7 = 0 is inconsistent is
14 2
(a) − (b) (c) 5 (d) 10
3 5
Ans: (d) 10 1
2. The zeroes of the polynomial x2 – 3x – m (m + 3) are
(a) m, m + 3 (b) – m, m + 3 (c) m, – (m + 3) (d) – m, – (m + 3)
Ans: (b) – m, m + 3 1
3. Euclid’s division Lemma states that for two positive integers a and b,
there exists unique integer q and r satisfying a = bq + r, and
(a) 0 < r < b (b) 0 < r ≤ b (c) 0 ≤ r < b (d) 0 ≤ r ≤ b
Ans: (c) 0 ≤ r < b 1
4. The sum of exponents of prime factors in the prime-factorisation of 196 is
(a) 3 (b) 4 (c) 5 (d) 2
Ans: (b) 4 1
5. If the point P (6, 2) divides the line segment joining A(6, 5) and B(4, y) in
the ratio 3 : 1, then the value of y is
(a) 4 (b) 3 (c) 2 (d) 1
Ans: 1 mark be awarded to everyone 1
6. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis
are
(a) (3, 5) (b) (3, –5) (c) (–3, –5) (d) (–3, 5)
Ans: (c) (–3, –5) 1
7. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is
(a) (2, 0) (b) (0, 2) (c) (3, 0) (d) (2, 2)
Ans: (a) (2, 0) 1
8. The nth term of the A.P. a, 3a, 5a, …… is
(a) na (b) (2n – 1)a (c) (2n + 1) a (d) 2na
Ans: (b) (2n – 1)a 1
1 1 − p 1 − 2p
9. The common difference of the A.P. , , , … is
p p p
1 1
(a) 1 (b) (c) –1 (d) −
p p
Ans: (c) –1 1

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10. The roots of the quadratic equation x2 – 0.04 = 0 are
(a) ± 0.2 (b) ± 0.02 (c) 0.4 (d) 2
Ans: (a) ± 0.2 1
In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark.
11. In Fig. 1, the angles of depressions from the observing positions O1 and O2
respectively of the object A are ______________ , _______________.
O2 O1

60°

45°
A
B C
Fig. 1

1 1
Ans: 30°, 45° +
2 2
ar(ΔAMN)
12. In Fig. 2, MN || BC and AM : MB = 1 : 2, then = _________.
ar( ΔABC)
A

M N

B C
Fig. 2

1
Ans: 1
9
13. In given Fig. 3, the length PB = _________ cm.

A 5 cm
O
3 cm
P

B
Fig. 3

Ans: 4 1
14. In ΔABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm, then ∠B = _________.
Ans: 90° 1
OR
Two triangles are similar if their corresponding sides are _________.
Ans: proportional 1
15. The value of sin 23° cos 67° + cos 23° sin 67° is _________.
Ans: 1 1

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Q. Nos. 16 to 20 are short answer type questions of 1 mark each.
16. In Fig. 4 is a sector of circle of radius 10.5 cm. Find the perimeter of
⎛ 22 ⎞
the sector. ⎜ Take π = ⎟
⎝ 7 ⎠

A B

60°

O
Fig. 4

πrθ
Ans: Perimeter = 2r +
180°
22 60°
= 2 × 10.5 + × 10.5 × 1/2
7 180°
= 21 + 11 = 32 cm 1/2
17. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3,
then find the probability of x2 < 4.
3
Ans: Number of Favourable outcomes = 3 i.e., {–1, 0, 1} ∴ P(x2 < 4) = 1/2+1/2
7
OR
What is the probability that a randomly taken leap year has 52 Sundays ?
5
Ans: P(52 sundays) = 1
7
18. A die is thrown once. What is the probability of getting a prime number.
Ans: Number of prime numbers = 3 i.e. ; {2, 3, 5} 1/2
3 1
P(Prime Number) = or 1/2
6 2
19. If tan A = cot B, then find the value of (A + B).
Ans: tan A = tan (90° – B) 1/2
∴ A + B = 90° 1/2
20. Find the class marks of the classes 15 – 35 and 45 – 60.
15 + 35
Ans: = 25 1/2
2
45 + 60
= 52.5 1/2
2
SECTION – B
Q. Nos. 21 to 26 carry 2 marks each
21. A teacher asked 10 of his students to write a polynomial in one variable
on a paper and then to handover the paper. The following were the
answers given by the students:

.30/2/2. 15

2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3 + 3x + 7, 7x +
3
7 , 5x – 7x + 2,
5 1 1
2x2 + 3 – , 5x – , ax3 + bx2 + cx + d, x + .
x 2 x
Answer the following questions :
(i) How many of the above ten, are not polynomials ?
(ii) How many of the above ten, are quadratic polynomials ?
Ans: (i) 3 1
(ii) 1 1
22. Compute the mode for the following frequency distribution :
Size of items
(in cm) 0 – 4 4 – 8 8 – 12 12 – 16 16 – 20 20 – 24 24 – 28
Frequency 5 7 9 17 12 10 6
Ans: l = 12 f0 = 9 f1 = 17 f2 = 12 h = 4 1/2

17 − 9 1
Mode = 12 + × 4 = 14.46 cm (Approx) 1+
34 − 9 − 12 2
A
23. In Fig. 5, ABC and DBC are two triangles on
the same base BC. If AD intersects BC
at O, show that B
C

ar ( ΔABC ) AO
=
ar ( ΔDBC ) DO D
Fig. 5
Ans: Draw AX ⊥ BC, DY ⊥ BC 1/2
A
ΔAOX ~ ΔDOY
Y C AX AO
B XO = …(i) 1/2
DY DO

D
1
ar ( ΔABC ) 2 ×BC×AX
=
ar ( ΔDBC ) 1 ×BC×DY 1/2
2
AX AO
= (From (1)) 1/2
DY DO
OR
In Fig. 6, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2.

C
D

B
A
Fig. 6

.30/2/2. 16

Ans: In rt ΔABD AB2 = BD2 + AD2 … (i) 1/2
In rt ΔADC 2
CD = AC – AD 2 2
… (ii) 1/2
Adding (i) & (ii)
AB2 + CD2 = BD2 + AC2 1

cot 2 α
24. Prove that 1 + = cos ec α
1 + cos ec α

Ans: L.H.S = 1 + cos ec α − 1
2
1/2
1 + cos ec α

( cos ec α − 1) ( cos ec α + 1)
= 1+ 1
cos ec α + 1
= cosec α = R.H.S 1/2
OR
Show that tan θ + tan θ = sec θ – sec θ
4 2 4 2

Ans: L.H.S = tan4θ + tan2θ
= tan2θ (tan2θ + 1) 1/2

= (sec2θ – 1) (sec2θ) = sec4θ – sec2θ = R.H.S 1+1/2

25. A child has a die whose six faces show the letters as shown below :
A A B C C C
The die is thrown once. What is the probability of getting (i) A, (ii) D ?
2 1 3 1
Ans: (i) P(A) = or (ii) P(D) = or 1+1
6 3 6 2
26. A solid is in the shape of a cone mounted on a hemisphere of same
base radius. If the curved surface areas of the hemispherical part
and the conical part are equal, then find the ratio of the radius
and the height of the conical part.
Ans: CSA of conical part = CSA of hemispherical part
π r l = 2πr2 1/2

r 2 + h 2 = 2r 1/2
h2 = 3r2 1/2

r 1
= ⇒ ratio is 1: 3 1/2
h 3

.30/2/2. 17

SECTION – C

Question numbers 27 to 34 carry 3 marks each.
27. In Fig. 7, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked
along them, then find the lengths of sides of each triangle.
A D

2x–1 3x 6x
18

B 2x+2 C E 3x+9 F
Fig. 7
Ans: As ΔABC ~ ΔDEF
2x − 1 3x
= 1
18 6x
x=5 1
AB = 9 cm DE = 18 cm
BC = 12 cm EF = 24 cm 1/2+1/2
CA = 15 cm FD = 30 cm
28. If a circle touches the side BC of a triangle ABC at P and extended sides AB
and AC at Q and R, respectively, prove that
1
AQ = (BC + CA + AB)
2
Ans: Correct Fig 1/2

1
Q AQ = (2AQ) 1/2
B
2
1
P A = (AQ + AQ)
2
C 1
R = (AQ + AR)
2
1
= (AB + BQ + AC + CR) 1
2
1
= (AB + BC + CA) 1
2
Q [BQ = BP, CR = CP]
29. The area of a circular play ground is 22176 cm2. Find the cost of fencing
this ground at the rate of `50 per metre.
Ans: Let the radius of playground be r cm
πr2 = 22176 cm2 1
r = 84 cm
22
Circumference = 2πr = 2 × × 84 = 528 cm 1
7

.30/2/2. 18

50
Cost of fencing = × 528 = `264 1
100

⎛y ⎞
30. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and ⎜ − 2 ⎟
⎝x ⎠
Ans: 2x + y = 23 , 4x – y = 19
Solving, we get x = 7, y = 9 1+1
y −5
5y – 2x = 31, −2= 1/2+1/2
x 7
OR
1 1 11
Solve for x : x + 4 − x + 7 = 30 , x # – 4, 7

1 1 11 −11 11
Ans: x + 4 − x − 7 = 30 ⇒ (x + 4)(x − 7) = 30 1

⇒ x2 – 3x + 2 = 0 1
⇒ (x – 2) (x – 1) = 0 1/2
⇒ x = 2, 1 1/2
The Following solution should also be accepted
1 1 11 x + 7 − x − 4 11
− = ⇒ = 1
x + 4 x + 7 30 (x + 4)(x − 7) 30

1
⇒ 11 x2 + 121x + 218 = 0 1
2
Here, D = 5049

−121 ± 5049
x= 1/2
22
31. If the mid-point of the line segment joining the points A(3, 4) and
B(k, 6) is P (x, y) and x + y – 10 = 0, find the value of k.
P
Ans: A (3, 4) B
(x, y) (K, 6)

3+ k
x= y=5 1/2+1/2
2
3+ k
x + y – 10 = 0 ⇒ + 5 – 10 = 0 1
2
⇒ k=7 1
OR
Find the area of triangle ABC with A (1, –4) and the mid-points of
A(1, -4)
sides through A being (2, –1) and (0, –1).
Ans: B(3, 2), C(–1, 2) (2, -1) (0, -1) 1/2+1/2
1 B C
Area = 1(2 − 2) + 3(2 + 4) − 1(− 4 − 2) = 12 sq.units 1+1
2

.30/2/2. 19

32. If in an A.P., the sum of first m terms is n and the sum of its first n terms is
m, then prove that the sum of its first (m + n) terms is – (m + n).
Ans: Sm = n and Sn = m
2n 2m
2a + (m − 1)d = … (i) 2a + (n – 1)d = … (ii) 1
m n
m2 + n 2 + m n − n − m −2(n − m)
Solving (i) & (ii), a = & d= 1
mn mn

m + n ⎡ 2 × m 2 + n 2 + mn − n − m ⎤ ⎧ −2(n + m) ⎫
Sm +n = ⎢ ⎥ + (m + n − 1) ⎨ ⎬ 1/2
2 ⎣ mn ⎦ ⎩ mn ⎭
= (–1) (m + n) 1/2
OR
Find the sum of all 11 terms of an A.P. whose middle term is 30.
th
⎛ 11 + 1 ⎞
Ans: Middle term = ⎜ ⎟ term = a6 = 30 1
⎝ 2 ⎠
11
S11 = [ 2a + 10d ] 1/2
2
= 11(a + 5d) 1/2
= 11 a6 = 11 × 30 = 330 1
33. 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 10 km/h less than that of the fast train, find
the speed of each train.
Ans: Let the speeds of fast train & slow train be x km/hr
& (x – 10) km/hr respectively.
A.T.Q.
600 600
− =3 1
x − 10 x
x2 – 10x – 2000 = 0 1
(x – 50) (x + 40) = 0
x = 50 or – 40
Speed is always positive, So, x = 50 1/2
∴ Speed of fast train & slow train are 50 km/hr & 40 km/hr respectively. 1/2
1
34. If 1 + sin2θ = 3 sin θ cos θ, prove that tan θ = 1 or
2
1 + sin 2 θ 3sin θ ⋅ cos θ
Ans: = (Dividing both sides by cos 2θ) 1/2
2
cos θ cos 2 θ
sec2θ + tan2θ = 3 tan θ 1/2
(1 + tan θ) + tan θ = 3 tan θ
2 2
1/2
2 tan2θ – 3 tan θ + 1 = 0 1/2
(tan θ – 1) (2 tan θ – 1) = 0 1/2

.30/2/2. 20

1
tan θ = 1 or 1/2
2
SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. The mean of the following frequency distribution is 18. The frequency
f in the class interval 19 – 21 is missing. Determine f.
Class interval 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25
Frequency 3 6 9 13 f 5 4

Ans: C.I f x xf
11-13 3 12 36
13-15 6 14 84
15-17 9 16 144
17-19 13 18 234
19-21 f 20 20f
21-23 5 22 110
23-25 4 24 96
40+f 704 + 20f 2

∑ xf 704 + 20f
Mean = ⇒ 18 = ⇒ f=8 2
∑f 40 + f
OR
The following table gives production yield per hectare of wheat of 100
farms of a village :
Production yield 40-45 45-50 50-55 55-60 60-65 65-70
No. of farms 4 6 16 20 30 24
Change the distribution to a ‘more than’ type distribution and draw its ogive.
Ans:
Production yield Number of farms
More than or equal to 40 100
More than or equal to 45 96
More than or equal to 50 90
More than or equal to 55 74
More than or equal to 60 54
More than or equal to 65 24 2

Plotting of points (40, 100) (45, 96) (50, 90) (55, 74) (60, 54) (65, 24)
join to get ogive. 2

.30/2/2. 21

36. Find the area of the shaded region in Fig. 8, if PQ = 24 cm, PR = 7 cm
and O is the centre of the circle.
Q

O

R P
Fig. 8

25 1
Ans: ∠P = 90° RQ = (24) + 7 = 25 cm, r =
2 2
cm 1
2 2
Area of shaded portion = Area of semi circle – ar (Δ PQR) ⎫
⎪
1 22 ⎛ 25 ⎞
2
⎬ 2
= × × ⎜ ⎟ − 84 ⎪
2 7 ⎝ 2 ⎠ ⎭
= 161.54 cm2 1/2
OR
Find the curved surface area of the frustum of a cone, the diameters of
whose circular ends are 20 m and 6 m and its height is 24 m.
Ans: R = 10 m r = 3 m h = 24 m 1/2+1/2
l= (24) 2 + (10 − 3) 2 = 25 m 1
CSA = π(10 + 3)25 = 325 π m2 1+1

37. Prove that 5 is an irrational number..
Ans: Let 5 be a rational number..
p
5 = , p & q are coprimes & q ≠ 0 1
q
5q2 = p2 ⇒ 5 divides p2 ⇒ 5 divides p also Let p = 5a, for some integer a 1
5q2 = 25a 2 ⇒ q2 = 5a2 ⇒ 5 divides q2 ⇒ 5 divides q also
∴ 5 is a common factor of p, q, which is not possible as 1
p, q are coprimes.
Hence assumption is wrong 5 is irrational no. 1
38. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of
larger diameter is used for four hours and the pipe of smaller diameter for
9 hours, only half of the pool can be filled. How long would it take for each
pipe to fill the pool separately ?
Ans: Let time taken by pipe of larger diameter to fill the tank be x hr
Let time taken by pipe of smaller diameter to fill the tank be y hr
A.T.Q
1 1 1 4 9 1
+ = , + = 1+1
x y 12 x y 2
Solving we get x = 20 hr y = 30 hr 1+1

.30/2/2. 22

39. Draw two tangents to a circle of radius 4 cm, which are inclined to each
other at an angle of 60°.
Ans: Correct construction of circle of radius 4 cm 1
Correct construction of tangents 3
OR
Construct a triangle ABC with sides 3 cm, 4 cm and 5 cm. Now, construct
4
another triangle whose sides are times the corresponding sides of ΔABC.
5
Ans: Correct construction of triangle with sides 3 cm, 4 cm & 5 cm 1
Correct construction of similar triangle 3
40. The angle of elevation of the top of a building from the foot of a tower
is 30° and the angle of elevation of the top of a tower from the foot of
the building is 60°. If the tower is 50 m high, then find the height of
the building.
Ans: Correct figure 1
Let the height of building be h m
50
In rt. ΔBCD, tan 60° =
BC
50
⇒ BC = … (i) 1
3 D

h A
In rt. ΔABC, tan 30° =
BC
50 m
1 h hm
⇒ = (from (i)) 1
3 50/ 3 60° 30°
B C
50 2
∴ h = or 16 or 16.67 m 1
3 3

.30/2/2. 23

QUESTION PAPER CODE 30/2/3
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Question numbers 1 to 10 are multiple choice questions of 1 mark each.
You have to select the correct choice :
Q.No. Marks

1. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is
(a) (2, 0) (b) (0, 2) (c) (3, 0) (d) (2, 2)
Ans: (a) (2, 0) 1
2. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis
are
(a) (3, 5) (b) (3, –5) (c) (–3, –5) (d) (–3, 5)
Ans: (c) (–3, –5) 1
3. If the point P (6, 2) divides the line segment joining A(6, 5) and B(4, y) in
the ratio 3 : 1, then the value of y is
(a) 4 (b) 3 (c) 2 (d) 1
Ans: 1 mark be awarded to everyone 1
4. The sum of exponents of prime factors in the prime-factorisation of 196 is
(a) 3 (b) 4 (c) 5 (d) 2
Ans: (b) 4 1
5. Euclid’s division Lemma states that for two positive integers a and b,
there exists unique integer q and r satisfying a = bq + r, and
(a) 0 < r < b (b) 0 < r ≤ b (c) 0 ≤ r < b (d) 0 ≤ r ≤ b
Ans: (c) 0 ≤ r < b 1
6. The zeroes of the polynomial x2 – 3x – m (m + 3) are
(a) m, m + 3 (b) – m, m + 3 (c) m, – (m + 3) (d) – m, – (m + 3)
Ans: (b) – m, m + 3 1
7. The value of k for which the system of linear equations x + 2y = 3,
5x + ky + 7 = 0 is inconsistent is
14 2
(a) − (b) (c) 5 (d) 10
3 5
Ans: (d) 10 1
8. The roots of the quadratic equation x2 – 0.04 = 0 are
(a) ± 0.2 (b) ± 0.02 (c) 0.4 (d) 2
Ans: (a) ± 0.2 1
1 1 − p 1 − 2p
9. The common difference of the A.P. , , , … is
p p p
1 1
(a) 1 (b) (c) –1 (d) −
p p
Ans: (c) –1 1

.30/2/3. 24

10. The nth term of the A.P. a, 3a, 5a, …… is
(a) na (b) (2n – 1)a (c) (2n + 1) a (d) 2na
Ans: (b) (2n – 1)a 1
In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark.
11. In Fig. 1, the angles of depressions from the observing positions O1 and O2
respectively of the object A are ______________ , _______________.
O2 O1

60°

45°
A
B C
Fig. 1

1 1
Ans: 30°, 45° +
2 2

12. In ΔABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm, then ∠B = _________.
Ans: 90° 1
OR
Two triangles are similar if their corresponding sides are _________.
Ans: proportional 1
13. In given Fig. 2, the length PB = _________ cm.

A 5 cm
O
3 cm
P

B
Fig. 2

Ans: 4 1
ar(ΔAMN)
14. In Fig. 3, MN || BC and AM : MB = 1 : 2, then = _________.
ar( ΔABC)
A

M N

B C
Fig. 3

1
Ans: 1
9
15. The value of sin 32° cos 58° + cos 32° sin 58° is
Ans: 1 1

.30/2/3. 25

OR
tan 35° cot 78°
The value of tan 55° + tan 12° is –––––––––.

Ans: 2 1
Q. Nos. 16 to 20 are short answer type questions of 1 mark each.
16. A die is thrown once. What is the probability of getting a prime number.
Ans: Number of prime numbers = 3 i.e. ; {2, 3, 5} 1/2
3 1
P(Prime Number) = or 1/2
6 2
17. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3,
then find the probability of x2 < 4.
3
Ans: Number of Favourable outcomes = 3 i.e., {–1, 0, 1} ∴ P(x2 < 4) = 1/2+1/2
7
OR
What is the probability that a randomly taken leap year has 52 Sundays ?
5
Ans: P(52 sunday) = 1
7
18. If sin A + sin2 A = 1, then find the value of the expression (cos2 A + cos4 A).

Ans: sin A = 1 − sin A ⎫
2
1/2
⎬
sin A = cos 2 A ⎭
cos2A + cos4 A = sin A + sin2 A = 1 1/2
19. Find the area of the sector of a circle of radius 6 cm whose central angle is 30°.
(Take π = 3.14)
30°
Ans: Area = 3.14 × (6)2 × 1/2
360°
= 9.42 cm2 1/2
20. Find the class marks of the classes 20 – 50 and 35 – 60.
20 + 50
Ans: = 35 1/2
2
35 + 60
= 47.5 1/2
2
SECTION – B
Q. Nos. 21 to 26 carry 2 marks each
21. A teacher asked 10 of his students to write a polynomial in one variable
on a paper and then to handover the paper. The following were the
answers given by the students:
2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3 + 3x + 7, 7x +
3
7 , 5x – 7x + 2,
5 1 1
2x2 + 3 – , 5x – , ax3 + bx2 + cx + d, x + .
x 2 x

.30/2/3. 26

Answer the following questions :
(i) How many of the above ten, are not polynomials ?
(ii) How many of the above ten, are quadratic polynomials ?
Ans: (i) 3 1
(ii) 1 1
22. A child has a die whose six faces show the letters as shown below :
A B C D E A
The die is thrown once. What is the probability of getting (i) A, (ii) D ?
2 1 1
Ans: (i) P(A) = or (ii) P(D) = 1+1
6 3 6 A
23. In Fig. 4, ABC and DBC are two triangles on
the same base BC. If AD intersects BC C
at O, show that B

ar ( ΔABC ) AO
= D
ar ( ΔDBC ) DO Fig. 4

Ans: Draw AX ⊥ BC, DY ⊥ BC 1/2
A
ΔAOX ~ ΔDOY

Y AX AO
B XO
C = …(i) 1/2
DY DO
1
D ar ( ΔABC ) 2 ×BC×AX
=
ar ( ΔDBC ) 1 ×BC×DY 1/2
2
AX AO
= (From (1)) 1/2
DY DO
OR
In Fig. 5, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2.

C
D

B
A
Fig. 5

Ans: In rt ΔABD AB2 = BD2 + AD2 … (i) 1/2
In rt ΔADC CD2 = AC2 – AD2 … (ii) 1/2
Adding (i) & (ii)
AB2 + CD2 = BD2 + AC2 1

.30/2/3. 27

Prove that 1 + cot 2 α
24. = cos ec α
1 + cos ec α

Ans: L.H.S = 1 + cos ec α − 1
2
1/2
1 + cos ec α

( cos ec α − 1) ( cos ec α + 1)
= 1+ 1
cos ec α + 1
= cosec α = R.H.S 1/2
OR
Show that tan θ + tan θ = sec θ – sec θ
4 2 4 2

Ans: L.H.S = tan4θ + tan2θ
= tan2θ (tan2θ + 1) 1/2

= (sec2θ – 1) (sec2θ) = sec4θ – sec2θ = R.H.S 1+1/2

25. Find the mode of the following frequency distribution :
Class 15-20 20-25 25-30 30-35 35-40 40-45
Frequency 3 8 9 10 3 2
Ans: Modal class = 30-35, l = 30, f 0 = 9, f 1 = 10, f 2 = 3, h = 5 1/2
10 − 9 ⎞
Mode = 30 + ⎛⎜ ⎟×5 1
⎝ 2 × 10 − 9 − 3 ⎠
= 30.625 or 30.62 or 30.63 1/2
26. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a
right circular cone of same height and same base radius is removed. Find the
volume of the remaining solid.
1
Ans: Volume of remaining solid = π(6)2 × 14 – π(6)2 × 14 1
3
= 336 π cm3 or 1056 cm3 1
SECTION – C

Question numbers 27 to 34 carry 3 marks each.
27. If a circle touches the side BC of a triangle ABC at P and extended sides AB
and AC at Q and R, respectively, prove that
1
AQ = (BC + CA + AB)
2

.30/2/3. 28

Ans: Correct Fig 1/2

1
Q
AQ = (2AQ) 1/2
2
B
1
P A
= (AQ + AQ)
2

C 1
= (AQ + AR)
R 2
1
= (AB + BQ + AC + CR) 1
2
1
= (AB + BC + CA) 1
2
Q [BQ = BP, CR = CP]
28. The area of a circular play ground is 22176 cm2. Find the cost of fencing
this ground at the rate of `50 per metre.
Ans: Let the radius of playground be r cm
πr2 = 22176 cm2 1
r = 84 cm
22
Circumference = 2πr = 2 × × 84 = 528 cm 1
7
50
Cost of fencing = × 528 = `264 1
100
29. If the mid-point of the line segment joining the points A(3, 4) and
B(k, 6) is P (x, y) and x + y – 10 = 0, find the value of k.
P
Ans: A (3, 4) B
(x, y) (K, 6)

3+ k
x= y=5 1/2+1/2
2
3+ k
x + y – 10 = 0 ⇒ + 5 – 10 = 0 1
2
⇒ k=7 1
OR
Find the area of triangle ABC with A (1, –4) and the A(1, -4)
mid-points of sides through A being (2, –1) and (0, –1). (2, -1) (0, -1)
Ans: B(3, 2), C(–1, 2) 1/2+1/2
B C
1
Area = 1(2 − 2) + 3(2 + 4) − 1(− 4 − 2) = 12 sq.units 1+1
2

.30/2/3. 29

30. In Fig. 6, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked
along them, then find the lengths of sides of each triangle.
A D
2x–1 3x 6x
18

B 2x+2 C E 3x+9 F
Fig. 6
Ans: As ΔABC ~ ΔDEF
2x − 1 3x
= 1
18 6x
x=5 1
AB = 9 cm DE = 18 cm
BC = 12 cm EF = 24 cm 1/2+1/2
CA = 15 cm FD = 30 cm

⎛y ⎞
31. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and ⎜ − 2 ⎟
⎝x ⎠
Ans: 2x + y = 23 , 4x – y = 19
Solving, we get x = 7, y = 9 1+1
y −5
5y – 2x = 31, −2= 1/2+1/2
x 7
OR
1 1 11
Solve for x : x + 4 − x + 7 = 30 , x # – 4, 7

1 1 11 −11 11
Ans: x + 4 − x − 7 = 30 ⇒ (x + 4)(x − 7) = 30 1

⇒ x2 – 3x + 2 = 0 1
⇒ (x – 2) (x – 1) = 0 1/2
⇒ x = 2, 1 1/2
The Following solution should also be accepted
1 1 11 x + 7 − x − 4 11
− = ⇒ = 1
x + 4 x + 7 30 (x + 4)(x − 7) 30

1
⇒ 11 x2 + 121x + 218 = 0 1
2
Here, D = 5049

−121 ± 5049
x= 1/2
22

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1 1 3
32. Which term of the A.P. 20,19 ,18 , 17 … is the first negative term.
4 2 4
1 3
Ans: a = 20 & d = 19 – 20 = − 1/2
4 4
an < 0 1/2

⎛ 3⎞
20+(n – 1) ⎜ − ⎟ < 0 1
⎝ 4⎠
2
n > 27 1/2
3
∴ 28 term of the given A. P. is first negative term
th
1/2
OR
Find the middle term of the A.P. 7, 13, 19, …., 247.
Ans: a = 7 & d = 13 – 7 = 6 1/2
247 = 7 + (n – 1)6 1
n = 41 1/2
th
⎛ 41 + 1 ⎞ st
Middle term = ⎜ ⎟ = 21 term. 1/2
⎝ 2 ⎠
a21 = 7 + 20 × 6 = 127 1/2

33. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h.
How much area will it irrigate in 30 minutes, if 8 cm standing water is
required ?
Ans: Volume of water in canal in 1 hr = 10000 × 6 × 1.5 = 90000 m3 1
1
Volume of water in canal in 30 mins = × 90000 = 45000 m3 1/2
2
45000
Area = 1
8 / 100
= 562500 m2 1/2
34. Show that :

cos 2 (45° + θ) + cos 2 (45° − θ)
=1
tan (60° + θ) tan (30° − θ)

cos 2 (45° + θ) + sin 2 (90° − 45° + θ)
Ans: L.H.S = 1
tan (60° + θ) ⋅ cot (90° − 30° + θ)

cos 2 (45° + θ) + sin 2 (45° + θ)
= 1
tan (60° + θ) ⋅ cot (60° + θ)
1
= = 1 = R.H.S 1
1

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SECTION – D
Question numbers 35 to 40 carry 4 marks each.
35. The mean of the following frequency distribution is 18. The frequency
f in the class interval 19 – 21 is missing. Determine f.
Class interval 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25
Frequency 3 6 9 13 f 5 4

Ans: C.I f x xf
11-13 3 12 36
13-15 6 14 84
15-17 9 16 144
17-19 13 18 234
19-21 f 20 20f
21-23 5 22 110
23-25 4 24 96
40 + f 704 + 20f 2

∑ xf 704 + 20f
Mean = ⇒ 18 = ⇒ f=8 2
∑f 40 + f
OR
The following table gives production yield per hectare of wheat of 100
farms of a village :
Production yield 40-45 45-50 50-55 55-60 60-65 65-70
No. of farms 4 6 16 20 30 24
Change the distribution to a ‘more than’ type distribution and draw its ogive.
Ans:
Production yield Number of farms
More than or equal to 40 100
More than or equal to 45 96
More than or equal to 50 90
More than or equal to 55 74
More than or equal to 60 54
More than or equal to 65 24 2

Plotting of points (40, 100) (45, 96) (50, 90) (55, 74) (60, 54) (65, 24)
join to get ogive. 2
36. From a point on the ground, the angles of elevation of the bottom and
the top of a tower fixed at the top of a 20 m high building are 45° and
60° respectively. Find the height of the tower.
Ans: Let height of tower = h m

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A
BC
In rt. ΔBCD tan 45° = corr fig. 1
CD hm
20
1 =
CD ⎫⎬ B 1
CD = 20 m ⎭
AC 20 m
In rt. ΔACD tan 60° =
CD
60°
20 + h 45°
3 = 20 C D 1

h = 20 ( 3 − 1) m 1
37. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of
larger diameter is used for four hours and the pipe of smaller diameter for
9 hours, only half of the pool can be filled. How long would it take for each
pipe to fill the pool separately ?
Ans: Let time taken by pipe of larger diameter to fill the tank be x hr
Let time taken by pipe of smaller diameter to fill the tank be y hr
A.T.Q
1 1 1 4 9 1
+ = , + = 1+1
x y 12 x y 2
Solving we get x = 20 hr y = 30 hr 1+1

38. Prove that 5 is an irrational number..
Ans: Let 5 be a rational number..
p
5 = , p & q are coprimes & q ≠ 0 1
q
5q2 = p2 ⇒ 5 divides p2 ⇒ 5 divides p also Let p = 5a, for some
integer a 1
5q = 25a ⇒ q = 5a ⇒ 5 divides q ⇒ 5 divides q also
2 2 2 2 2

∴ 5 is a common factor of p, q, which is not possible as 1
p, q are coprimes.
Hence assumption is wrong 5 is irrational no. 1
39. Draw a circle of radius 3.5 cm. From a point P, 6 cm from its centre,
draw two tangents to the circle.
Ans: Correct construction of circle of radius 3.5 cm 1
Correct construction of tangents. 3
OR
Construct a ΔABC with AB = 6 cm, BC = 5 cm and ∠B = 60°.
2
Now construct another triangle whose sides are times the
3
corresponding sides of ΔABC.

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Ans: Correct construction of given triangle 1
Construction of Similar triangle 3
40. A solid is in the shape of a hemisphere surmounted by a cone. If the radius of
hemisphere and base radius of cone is 7 cm and height of cone is 3.5 cm, find
the volume of the solid.

⎛ 22 ⎞
⎜ Take π = ⎟
⎝ 7 ⎠

1 22 2 22
Ans: Volume of solid = × × (7) 2 × 3.5 + × × (7)3 2
3 7 3 7
22 ⎡ 3.5 2 ⎤
= × (7) 2 × ⎢ + ×7
⎣ 3 3 ⎥⎦ 3.5cm 1
7
7 cm
1
= 898 or 898.33 cm3 1
3

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