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

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Secondary School Examination-2020
Marking Scheme - MATHEMATICS STANDARD
Subject Code: 041 Paper Code: 30 (B)
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QUESTION PAPER CODE 30/B
EXPECTED ANSWER/VALUE POINTS
SECTION A
Question numbers 1 to 10 are multiple choice questions. Choose the correct option.
117
1. The decimal representation of 3 4 2 will
25 3
(A) terminate after 3 decimal places (B) terminate after 2 decimal places
(C) terminate after 4 decimal places (D) not terminate
Sol. (C) Terminate after 4 decimal places. 1

2. For what value(s) of 'a' will the equations 2x + 3y = 13 and 3x + ay = 18 have no solution?

(A) 2 (B) 4.5 (C) ≤ 10 (D) –4

Sol. (B) 4.5 1

3. If in ΔABC, ∠A = 90°, then the value of cosec (B + C) is

2 1 3
(A) (B) (C) (D) 1
3 2 2
Sol. (D) 1 1

4. Given in the table below are the marks obtained by 50 students in a class test:

Marks 1 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of Students: 4 7 19 12 8

From this data, the lower limit of median class is
(A) 10 (B) 20 (C) 25 (D) 30
Sol. (B) 20 1

5. The LCM of the smallest two-digit number and the largest multiple of 6 which is less than 50
is
(A) 2 (B) 48 (C) 120 (D) 240
Sol. (D) 240 1

6. The value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is

1
(A) 0 (B) –1 (C) 1 (D)
2
Sol. (C) 1 1

7. The distance between the points (3, 4) and (–2, –1) is

(A) 5 2 (B) 30 (C) 6 3 (D) 10

Sol. (A) 5 2 1

k
8. If P (6, ) is the mid-point of the line segment joining the points A(8, 5) and B (4, 3), then the
2
value of k is
(A) –8 (B) 16 (C) –6 (D) 8
Sol. (D) 8 1

3 5 sin θ − 2 cos θ
9. If sin θ = , then the value of the expression is
5 tan θ

15 28 23 92
(A) (B) (C) (D)
28 15 15 15

28
Sol. (B) 1
15
10. The point which divides the line segment joining the points A(2, 3) and B (–3, 4) in the ratio
3 : 4 internally lies in which quadrant?
(A) I (B) II (C) III (D) IV
Sol. (B) II 1

In question numbers 11 to 15, in each question blanks are to be filled correctly:

1 22
11. The diameter of a sphere of volume 1437 Cu. cm is _______ cm. (Use π = )
3 7
Sol. 14 1

12. If 5, b, c, 14 are the consecutive terms of an A.P., then b + c = _______.
Sol. 19 1
OR

1 1 − p 1 – 2p
The next term of A.P. , , , ... is _______
p p p

1 − 3p ⎛ 1 ⎞
or − 3
p ⎜⎝ p ⎟⎠
1

13. If – 2 is a root of the quadratic equation 3x2 – 5x + k = 0, then the value of k is _______.
Sol. k = –22 1

14. A number is selected at random from the numbers 1 to 20. The probability that the selected
number is a multiple of 3 is _______.

6 3
Sol. or 1
20 10

15. If two triangles ABC and DEF are similar and ∠A = 67°, ∠E = 63°, then the measure of ∠C
is _______.
Sol. 50° 1

Answer the question numbers from 16 to 20:
16. Write one irrational number between 0.15 and 0.21.
Sol. 0.15010010001 ... (Any relevant answer) 1

OR
Find the HCF of 12, 18 and 30.

1 1
HCF(12, 18) = 6 HCF(6, 30) = 6 +
2 2

17. If a – b, k, a + b and x are four consecutive terms of an A.P., then find the ratio between k and
x in terms of a and b.

1
Sol. Getting K = a, x = a + 2b
2

1
K : x = a : a + 2b
2

18. Two concentric circles are of radii 5 cm and 3 cm. Find the length of that chord of the larger
circle which touches the smaller circle.

1
Sol. BC = 4 cm
2
O
1
A C B AB = 8 cm
2

19. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. If AB =
3 CD, find the ratio of the areas of triangles AOB and COD.

1
Sol. ΔAOB ~ ΔCOD (by AA similarity)
2

ar ΔAOB AB2 9 1
= = or 9 : 1
ar ΔOCD CD2 1 2

20. Find the value(s) of k for which the roots of the quadratic equation 9x2 + 3kx + 4 = 0 are real
and equal.
1
Sol. For roots to be real & equal, 9k2 – 144 = 0
2
1
which gives k = ±4
2

SECTION B
Question numbers 21 to 26 carry 2 marks each.
21. If the areas of two similar triangles are equal, then prove that the triangles are congruent.

ar ΔABC AB2 BC 2 AC2
Sol. Let ΔABC ~ ΔPQR ∴ = = = =1 1
ar ΔPQR PQ 2 QR 2 PR 2
1
∴ AB = PQ, BC = QR, AC = PR
2

1
∴ ΔABC ≅ ΔPQR, (by SSS congruency)
2
OR
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares
of its diagonals.
Let ABCD be given rhombus whose diagonals intersect at O
Clearly AO2 + OB2 = AB2 1
D C
AC 2 BD 2 1
O ⇒ + = AB2
4 4 2
A B ⇒ AC2 + BD2 = 4AB2
1
AC2 + BD2 = AB2 + BC2 + CD2 + DA2
2

22. Find the number of numbers lying between 146 and 300 which are divisible by both 3 and 5.
Sol. Required numbers are 150, 165, ... 285 1
1
150 + (n – 1)15 = 285
2

1
⇒ n = 10
2

23. Prove that the parallelogram circumscribing a circle is a rhombus.

Sol. Let ABCD is a parallelogram circumscribing a circle
Proof: AP = AS ⎫
D R C ⎪
PB = BQ ⎪⎪
S Q
⎬ 1
DR = DS ⎪
A P B ⎪
CR = CQ ⎪⎭
adding we get

1
AB + CD = AD + BC Note: (Accept, if written directly)
2

1
⇒ 2AB = 2AD ⇒ AB = AD
2

Hence ABCD is a rhombus

24. 100 jugs of equal volume full of water are emptied in a conical flask and the level of water in
the flask is 75 cm. If each jug contains 3850 cm3 of water, then find the diameter of the level
of water in the flask.
Sol. Let radius of conical flask be r

1 22 2
× × r × 75 = 3850 × 100 1
3 7

1
gives r = 70 cm
2

1
So, diameter = 140 cm
2

25. On the edge AB of a wall of a house, a projection AP, which is perpendicular to the wall, is
erected and its edge is of length 27 cm. A point searchlight is fixed on the corner point P of
projection which sends a ray of light PQ such that AQ = 9 3 cm. If ∠APQ = θ, then find the
value of (i) θ, and (ii) tan θ + sec θ.

9 3 1 1
Sol. (i) tan q = ⇒ q = 30° +
27 2 2
27
θ
9 3 1 2 1
(ii) tan θ + sec θ = tan 30° + sec 30° = +
3 3 2

3 1
= or 3
3 2

26. Two friends A and B take their breakfast occasionally in a restaurant which prepares a speci-
ality dish on Monday, Wednesday and Sunday. Each is equally likely to visit the restaurant on
any day on which the speciality dish is made. Find the probability that both will enjoy taking the
speciality dish on the (i) same day, and (ii) different days.

1
Sol. (i) Total outcomes = 9
2

Favourable outcomes are (Monday, Monday)
(Wednesday, Wednesday)

1
(Sunday, Sunday) i.e. 3
2

3 1 1
P(same day) = or
9 3 2

1 2 1
(ii) P(different days) = 1 − =
3 3 2

OR
Two coins are tossed together. Find the probability of getting (i) both heads, and (ii) exactly one
head.
Total outcomes = 4 1

1 1
(i) P(Both heads) = ,{i.e. (HH)}
4 2

1 1
(ii) P(exactly one head) = {i.e. (HT), (TH)}
2 2

SECTION C
Question numbers 27 to 34 carry 3 marks each.
27. A number consists of two digits whose sum is 8. If 36 is added to the number, the digits
interchange their places. Find the number.
Sol. Let digit at unit place be x & digit at tens place be y

1
Acc. to Ques. x + y = 8 ...(i)
2

10y + x + 36 = 10x + y
gives x – y = 14 ...(ii) 1

Solving we get x = 6, y = 2 1

1
∴ number is 26
2

OR
Solve for x and y:

x y x y
+ = 6, + = 6
3 4 6 2
Given equation can be written as
4x + 3y = 72, x + 3y = 36 1+1

1 1
Solving we get x = 12, y = 8 +
2 2

28. The first and the last terms of an A.P. are 16 and 136 respectively. If the common difference
of the A.P. is 5, then find the number of terms in the A.P. Also find their sum.
Sol. d = 5, a = 16, last term = an = 136
16 + (n – 1)5 = 136 1

1
gives n = 25
2

25
S25 = (16 + 136) 1
2

1
= 1900
2

29. Assuming that 3 is an irrational number, prove that 5 3 – 7 is an irrational number..

Sol. Let 5 3 − 7 be rational

Let 5 3 − 7 = x 1

x+7
gives 3= 1
5
Clearly RHS is rational but LHS is irrational (given)

so our supposition is wrong, hence 5 3 − 7 is irrational 1

OR
If the HCF of 65 and 117 is written as 65m – 117, then find the value of m.
1
HCF (65, 117) = 13 1
2

1
65 m – 117 = 13
2
gives m = 2 1

30. If points A(0, 3), B(– 2, a) and C(– 1, 4) are the vertices of a right triangle right-angled at A,
then (i) find the value of 'a', (ii) find the length of the longest side, and (iii) find the area of
ΔABC.
Sol. (i) AB2 + AC2 = BC2 ⇒ (a – 3)2 + 4 + 2 = 1 + (a – 4)2 1

1
⇒ a=1
2

1
(ii) Length of longest side BC = 10 2

1 1
(iii) Area = AB × CA = 2 8=2 1
2 2

31. Find all the zeroes of the polynomial
2x4 – 5x3 - 11x2 + 20x + 12
if it is given that two of its zeroes are 2 and – 2.
Sol. As 2 & –2 are zeroes ⇒ x2 – 4 is a factor 1

( 2
x 2 − 4 2x 2 − 5x 3 − 11x 2 + 20x + 12 2x − 5x − 3
2x 4 − 8x 2
− +
− 5x − 3x 2 + 20x + 12
3

− 5x 3 + 20x
+ –
1
− 3x 2 + 12
2
− 3x + 12
+ –
0

Thus other two zeroes are given by 2x2 – 5x – 3 = 0
1
(2x + 1) (x – 3) = 0
2

−1
⇒ x= , 3
2
−1 1
All zeroes are 2, –2, , 3
2 2

32. Prove that:
(sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ

1
Sol. LHS = sin2 θ + cosec2 θ + 2 + cos2 θ + sec2 θ + 2 1
2

= sin2 θ + cos2 θ + 1 + cot2 θ + 1 + tan2 θ + 4 1

1
= cot2 θ + tan2 θ + 7 = RHS
2

OR
Find the value of

− cot(90° − θ) tan θ + sec θ cosec (90° – θ) + sin 2 35° + sin 2 55°
tan 10° tan 20° tan 30° tan 70° tan 80°

− tan θ ⋅ tan θ + sec θ ⋅ sec θ + sin 2 35° + cos 2 35° Correct numerator
1
1 Correct denominator
tan 10° tan 20° cot 20° cot 10° 1
3

1+1 1 1
= =2 3 +
1 2 2
3

33. The area of an equilateral triangle is 49 3 cm2. Taking each angular point as centre, circles
are drawn with radius equal to half the length of the side of the triangle. Find the area of that
part of the triangle which is not included in the circles.

22
(Use 3 =1.73, π = 7 )

3 2 1
Sol. a = 49 3 ⇒ a = 14 cm +1
4 2

60° 22
Area of triangle not included in circle = 49 3 − 3 × ×7×7 1
360° 7

= 49 × 1.73 – 77

1
= 7.77 cm2
2

34. The number of patients attending a hospital in a month is given in the table below. Find the
mean number of patients attending the hospital in a day.

Number of patients Number of days
attending hospital
0 – 10 2
10 – 20 6
20 – 30 9
30 – 40 7
40 – 50 4
50 – 60 2

Sol. xi 5 15 A = 25 35 45 55 Total

fi 2 6 9 7 4 2 30
ui –2 –1 0 1 2 3 Correct Table 2

f iu i –4 –6 0 7 8 6 11

Σf i u i 11 1
x=A+ × h = 25 + × 10
Σf i 30 2

86 1
= or 28.67
3 2

SECTION D
Question numbers 35 to 40 carry 4 marks each.
35. Prove that in a right triangle, the square of the hypotenuse is equal to sum of the squares of
the remaining two sides.

1 1
Sol. For correct Given, to prove +
2 2

For correct proof 3

36. Write the steps of construction of a ΔABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°.
3
Then write the steps of construction of another triangle whose sides are times the corre-
4
sponding sides of ΔABC.

Sol. Writing steps for constructing ΔABC 2

Writing steps for constructing triangle similar to ΔABC 2

OR
Write the steps of construction of two tangents to a circle of radius 3 cm which are inclined to
each other at an angle of 60°.
Writing steps for contructing circle & constructing an angle of 120° at centre 2

Writing steps for constructing tangents at both radii 2

37. A bucket, open at the top, is in the form of a frustum of a cone with a capacity of 12308.8 cm3.
The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height
of the bucket. (Use π = 3.14)

1
Sol. × 3.14 × h × (400 + 144 + 240) = 12308.8 2
3

12308.8 × 3
⇒ h= 1
3.14 × 784
⇒ h = 15 cm 1
OR
A conical vessel whose internal radius is 5 cm and height 24 cm is full of water. This water is
emptied in a cylindrical vessel of internal radius 10 cm. Find the height to which water level
rises in the cylindrical vessel.

1 1 1
π × 52 × 24 = π × 102 × h 1 +1
3 2 2

⇒ h = 2 cm 1

38. A statue 2 m tall, stands on the top of a pedestal. From a point on the ground, the angle of
elevation of the top of the statue is 60° and from the same point the angle of elevation of the
top of the pedestal is 45°. Find the height of the pedestal. (Use 3 = 1.73)

h
Sol. 2 = tan 45° ⇒ x = h 1
x

h h+2
60° ∴ = tan 60° 1
45° h
x
⇒h+2= 3h

2
h= 1
3 −1

2 ( 3 − 1)
i.e. h = = 2.73 m 1
2

39. The sum of the areas of two squares is 640 m2. If the difference of their perimeters be 64 m,
find the sides of the two squares.
Sol. Let sides of larger & smaller squares be x & y respectively
x2 + y2 = 640 ...(i) 1

4x – 4y = 64 ⇒ x = 16 + y ...(ii) 1

By (i) and (ii) y2 + 16y – 192 = 0 1

1
Solving we get y = –24 or y = 8
2

1
Thus, side of smaller square = 8 cm & that of larger square = 24 cm
2

OR
Solve for x:

1 1 11
− = , x ≠ –4, 7
x+4 x–7 30

−11 11
2 = 2
x − 3x − 28 30
⇒ x2 – 3x + 2 = 0 1

Solving we get x = 2, x = 1 1

40. If the median of the following distribution is 28.5, then find the values of x and y:

Class Interval Frequency
0 – 10 5
10 – 20 x
20 – 30 20
30 – 40 15
40 – 50 y
50 – 60 5
Total 60

Sol. C.I. f cf
0-10 5 5
10-20 x 5+x
20-30 20 25 + x Correct Table 1

30-40 15 40 + x
40-50 y 40 + x + y
50-60 5 45 + x + y
Total 60
x + y = 15 1

30 − 5 − x
28.5 = 20 + × 10 1
20
1 1
give x = 8, y = 7 +
2 2