Class 10 Mathematics: CBSE Sample Question Paper- Term I (2021-22) - 3 | CBSE Sample Papers For Class 10 PDF Download

Class-X
Time: 90 Minutes
M.M: 40

General Instructions:

1. The question paper contains three parts A, B and C. 
2. Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted
3. Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted
4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions.
5. There is no negative marking.

Q.1: The smallest number which when divided by 28 and 42 leaves remainders 8 and 22 respectively is:
(a) 104
(b) 84
(c) 64
(d) 74

Correct Answer is Option (c)

∵ 28 – 8 = 20 and 42 – 22 = 20

Now, 28 = 2² × 7

and 42 = 2 × 3 × 7

∴ LCM (28, 42) = 2² × 3 × 7

= 84

Required number = LCM (28, 42) – 20

= 84 – 20

= 64.

Q.2: If tan θ – cot θ = 8, then the value of tan²θ + cot²θ is:
(a) 68
(b) 69
(c) 66
(d) 70

Correct Answer is Option (c)

Given,

tan θ – cot θ = 8

Taking square on both sides, we get

tan² θ + cot² θ – 2 tan ^q . cot θ = 64

Q.3: In the given figure,  and ∠AED = ∠ABC, then :

(a) ∠B = ∠C
(b) ∠A = ∠D
(c) ∠A = ∠E
(d) ∠A = ∠B

Correct Answer is Option (a)

Here,

So, DE || BC   [By the converse of Thales theorem]

∴ ∠AED = ∠ACB

(corresponding angles)

But ∠AED = ∠ABC (given)

∴ ∠ABC = ∠ACB.

Q.4: In a given DABC, DE || BC and AB/DB = 3/4. If AC = 4.9 cm, then AE is :

(a) 2.5 cm
(b) 2.8 cm
(c) 2.1 cm
(d) 2.6 cm

Correct Answer is Option (c)

Let AE = x cm,

then EC = (4.9 – x) cm

and DE || BC (given)

⇒  3(4.9 – x) = 4x

⇒  14.7 – 3x = 4x

⇒  7x = 14.7

⇒  x = 2.1

Q.5: If p(– 3) = 0, then one factor of the polynomial p(x) is:
(a) x – 3
(b) x + 3
(c) 3x
(d) x/3

Correct Answer is Option (b)

∵ p(– 3) = 0

∴p(x) = x + 3.

Q.6: The decimal expansion of the rational number  will terminate after:

(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places

Correct Answer is Option (c)

Q.7: If the short and long hands of a clock are 5 cm and 6 cm long respectively, then the distance travelled by long hand in 60 hours is:
(a) 32π cm
(b) 28π cm
(c) 60π cm
(d) 96π cm

Correct Answer is Option (c)

∵ Long hand makes 5 rounds in 60 hours and radius of the circle formed by long hand = 6 cm
∴ Distance travelled by long hand in 5 rounds
= 5 × circumference of the circle
= 5 × 2πr
= 5 × 2π × 6
= 60π cm.

Q.8: as a decimal fraction is:
(a) 5.025
(b) 4.485
(c) 4.575
(d) 4.285

Correct Answer is Option (a)

Q.9: There is a circular playground having radius 21cm to be constructed. Then the engineers decided to construct a minor sector for open gym which as shown in the below figure, which is OACBO. Then find the area of sector OACBO.

(a) 221 cm²
(b) 231 cm²
(c) 866 cm²
(d) 235 cm²

Correct Answer is Option ()

Here, r = 21 cm, θ = 60°

∴ Area of the sector OACBO

Q.10: The graphs of the equations 5x – 15y = 8 and 3x – 9y = 24/5 are two lines which are:
(a) coincident
(b) parallel
(c) perpendicular to each other
(d) intersecting exactly at one point

Correct Answer is Option (a)

Hence, both lines are coincident.

Q.11: From a pack of 52 cards, Jacks, queens, kings and aces of black colour are removed. From the remaining, a card is drawn at random, then the probability that the card drawn is a ten; is:
(a) 1/22
(b) 3/22
(c) 1/11
(d) 10/11

Correct Answer is Option (c)

No. of removed cards = 2 + 2 + 2 + 2
∴ Remaining cords = 52 – 8 = 44
i.e., n(S) = 44
and no. of tens cards = 4
i.e., favourable outcomes

n(E) = 4

Q.12: If x = (2² × 3² × 7³) and y = (2³ × 3³ × 7²), then HCF (x, y) =
(a) 1760
(b) 1752
(c) 1754
(d) 1764

Correct Answer is Option (d)

HCF (x, y) = Product of common terms

with lowest power

= 2² × 3² × 7²

= 1764

Q.13: If in an equilateral triangle of side 12 cm, then the length of the altitude is:
(a) 4√3 cm
(b) 6√3 cm
(c) 5√3 cm
(d) 7√3 cm

Correct Answer is Option (b)

Here, ABC is an equilateral triangle of side 12 cm.

Let the length of the altitude be h cm and BD is 6 cm.

In right angled DADB,
(AD)² = (AB)² – (BD)²
(By using Pythagoras theorem)
= (12)² – (6)²
= 144 – 36
= 108

Q.14: If 2x = sec A and 2/x = tan A, then the value of x² - 1/x² is:
(a) 1/2
(b) 1/4
(c) 1/3
(d) 1

Correct Answer is Option (b)

Q.15: A jar contains 45 marbles. Some of these are white and others are yellow. If a marble is drawn at random from the jar, the probability that it is white is 3/5,  then the number of yellow marbles in the jar is:
(a) 18
(b) 27
(c) 25
(d) 20

Correct Answer is Option (a)

Total outcomes,

n(S) = 45

Let the number of yellow marbles be x,

then the no. of white marbles = 45 – x.

Given, P(getting a white marble) = 3/5

Q.16: DABC ~ DDEF such that ar(DABC) = 81 cm² and ar(DDEF) = 121 cm², then the ratio of their corresponding sides is:
(a) 3 : √11
(b) 9 : 11
(c) 11 : 9
(d) √9 : √11

Correct Answer is Option (b)
Since DABC ~ DDEF, we have

Hence, the ratio of the corresponding sides is 9 : 11.

Q.17: The sum of the ages of a father and the son is 45 years. If father's age is four times that of his son, then son's age is:
(a) 7 years
(b) 8 years
(c) 9 years 
(d) none of these

Correct Answer is Option (c)

Let age of father and son be x and y respectively, then
x + y = 45 ...(i)

and x = 4y ...(ii)

Substituting x = 4y in eq. (i), we get

4y + y = 45

⇒ 5y = 45

⇒ y = 9

Q.18: If the distance between the points A(– 3, 2) and B(x, 10) is 10, then the possible values of x is:
(a) 3 and 9 
(b) - 3 and 9
(c) 3 and - 9
(d) - 3 and -9

Correct Answer is Option (c)

Taking positive sign, x = 6 – 3 = 3

Taking negative sign, x = – 6 – 3 = – 9.

Q.19: The probability of getting a bad orange in a lot of 1000 is 0.046. The number of bad oranges in the lot is:
(a) 46 
(b) 48
(c) 41
(d) 50

Correct Answer is Option (a)

Total oranges,

n(S) = 1000

Let the no. of bad oranges be x, then

n(E) = x

P(getting bad orange) = 0.046

⇒ x = 1000 × 0.046

⇒ x = 46.

Q.20: The value of tan² 30° – cot² 30° is:
(a) - 1
(b) - 2
(c) -8/3
(d) -1/3

Correct Answer is Option (c)

Q.21: If two positive integers p and q are written as p = a²b³ and q = a³b³, where a and b are prime numbers, then LCM (p, q) × HCF (p, q) is :
(a) a²b⁴ 
(b) a⁴b⁵
(c) a³b⁴
(d) a⁵b⁶

Correct Answer is Option (d)

In case of assertion:

p = a² b³

and q = a³ b³

Then LCM (p, q) = a³ b³

and HCF (p, q) = a² b³

∴LCM (p, q) × HCF (p, q)

= a³ b³ × a² b³

= a⁵ b⁶.

Q.22: If α and β are the zeroes of a polynomial x² – 5√3x + 6, then α + β:
(a) 4√3
(b) 5√3
(c) 2√3
(d) 3√3

Correct Answer is Option (b)

Let x² – 5√3x + 6= 0

Q.23: If cosec θ . cos θ = 1, then the value of q is:
(a) 30°
(b) 60°
(c) 90°
(d) 45°

Correct Answer is Option (d)

cosec θ . cos θ = 1

⇒ cot θ = 1

= cot 45°

⇒ θ = 45°

Q.24: 19 cards numbered 1, 2, 3, ....., 19 are put in a box and mixed thoroughly. A man draws a card from the box, then the probability (getting the no. on the card is odd) is:
(a) 9/19
(b) 8/19
(c) 7/19
(d) 10/19

Correct Answer is Option (d)

Total outcomes n(S) = 19

odd numbered cards = 1, 3, 5, 7, 9, 11, 13,

15, 17, 19

i.e., favourable outcomes

n(E) = 10

∴ P(getting a card is odd)

Q.25: One zero of the polynomial 2x³ - x² - 13x - 6 is:
(a) -8
(b) 3
(c) -3
(d) 1

Correct Answer is Option (b)

Let p(x) = 2x³ – x² – 13x – 6, then 3 is a

zero of p(x).

∴ p(x) = 0

i.e., p(3) = 2(3)³ – (3)² – 13(3) – 6

= 2 × 27 – 9 – 39 – 6

= 54 – 9 – 45 = 0

Q.26: Reema wanted to go her school after covid-19, before go to her school, She wanted to calculate the total distance between her house and school from the adjacent diagram.

(a) 61
(b) 61
(c) √62
(d) √62

Correct Answer is Option (c)

Here, x₁ = 0, y₁ = 5 and x₂ = 0, y₂ = - 6

So, distance between A(0, 5) and B(0, -6):

Q.27: The distance of the point (7, 5) from x-axis is:
(a) 7 
(b) - 7
(c) - 5 
(d) 5

Correct Answer is Option (d)

Since y-coordinate of a given point is the distance of point from x-axis, then the distance of the point (7, 5) from x-axis is 5.

Q.28: If the circumference of a circle is 14π, then the area of the circle is:
(a) 160 cm² 

(b) 154 cm²
(c) 170 cm² 
(d) 180 cm²

Correct Answer is Option (d)

Circumference = 14π
⇒ 2pr = 14π
⇒ r = 7 cm
Then area of the circle = πr²

Q.29: Mr. Shayam has recently shifted in his new house. When he entered in master bed room and observed its dimensions 8m × 6m × 4m. He decided to put one longest rod into it. The length of the longest rod will be.

(a) 4 m 
(b) 6 m
(c) 2 m 
(d) 12 m

Correct Answer is Option (c)

Length = 8 m, breadth = 6 m and height = 4 m

Since, the length of the longest rod is equal to HCF (8, 6 and 4), i.e.,

8 = 2³,

6 = 2 × 3

and 4 = 2²

Then, HCF (8, 6 and 4) = 2

Thus, the longest rod that can measure

the dimensions of the room exactly = 2m.

Q.30: The value of 10 sec² A – 10 tan² A is:
(a) 9
(b) 10
(c) 11
(d) 12

Correct Answer is Option (b)

10 sec² A – 10 tan² A = 10 (sec² A – tan² A)

= 10 × 1

[∵ sec² q – tan² q =1]

= 10.

Q.31: The value of the quadratic polynomial f(x) = 2x² - 3x - 2 at x = 1 is:
(a) - 2 
(b) - 1
(c) - 4 
(d) - 3

Correct Answer is Option (d)

Given,

f(x) = 2x² - 3x - 2

Then, f(1) = 2(1)² - 3(1) -– 2

= 2 - 3 - 2

= -3.

Q.32: In a lottery, there are 8 prizes and 16 blanks. The probability of getting a prize is:
(a) 1/2
(b) 2/3
(c) 1/3
(d) 5/3

Correct Answer is Option (c)

In case of assertion,

Total no. of lottery tickets

= 8 + 16 = 24
∴ P(getting a prize) = 8/24 = 1/3

Q.33: If one zero of the quadratic polynomial 5x² + p(x) + 8 = 0 is 1/2, then the value of p is:
(a) 33/2
(b) 35/2
(c) - 37/2
(d) 41/2

Correct Answer is Option (c)

5x² + p(x) + 8 = 0
∵ 1/2 is a zero, so it must satisfy the given equation,

Q.34: If the product of the zeros of the quadratic polynomial 6x² – 5x + k is 4, then the value of k is:
(a) 24 
(b) 20
(c) 22
(d) 26

Correct Answer is Option (a)

Let the zeroes of the given quadratic polynomial be α and β, so we have

Q.35: If cosec θ . cos θ = 1, then the value of q is:
(a) 30°
(b) 60°
(c) 90°
(d) 45°

Correct Answer is Option (d)

cosec θ . cos θ = 1

⇒ cot θ = 1

= cot 45°

⇒ θ = 45°

Q.36: If the circumference of a circle is 88 cm, then the area of a quadrant of a circle is:
(a) 77 cm² 
(b) 77/8 cm² 
(c) 154 cm²
(d) 308 cm²

Correct Answer is Option (c)

Circumference,

2πr = 88 cm

Q.37: If one zero of the quadratic polynomial x² – 3x – k is – 1, then the value of k is:
(a) 2 
(b) 3
(c) 4
(d) 6

Correct Answer is Option (c)

Let p(x) = x² – 3x – k

∵ -1 is a zero of p(x), then

p(–1) = 0

∴ (-1)² - 3(-1) - k = 0

⇒ 1 + 3 - k = 0

⇒ k = 4.

Q.38: If a chord of the circle of radius 8 cm subtends a right angle at the centre, then the area of sector OACBO is: (Use, p = 3.14)
(a) 50.24 cm² 
(b) 45.24 cm²
(c) 48.25 cm² 
(d) 50.28 cm²

Correct Answer is Option (a)

Here, r = 8 cm and θ = 90°

Then, area of sector OACBO

Q.39: If x + 5 is a factor of p(x) = 2x³ – 4x² + ax + 5, then the value of a is:
(a) - 4 
(b) - 3
(c) 4 
(d) 3

Correct Answer is Option (a)

x + 5 = 0

⇒ x = – 5

∴ 2(- 5)³ - 4(- 5)² + a(- 5) + 5 = 0

⇒ 2 x (-125) - 100 - 5a + 5 = 0

⇒ - 350 + 5 = 5a

⇒ 5a = - 345

⇒ a = - 69

Q.40: If tan θ = cot θ, then the value of 2 tan θ + cos² θ is:
(a) 1 
(b) 2
(c) 3 
(d) 5/2

Correct Answer is Option (d)

∵ tan θ = cot θ

Here tan 45° = cot 45°

= 1.

∴ θ = 45°

Now 2 tan θ + cos² θ = 2 tan 45° + (cos45°)²

Direction: Q. 41-Q.45 are based on case study-1

Case Study - 1

A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.

Q.41: What will be the value of z?
(a) 22
(b) 23
(c) 17
(d) 19

Correct Answer is Option (d)

From question (1),
Putting x = 33 is eq. (i), we get
33 + y = 83
⇒ y = 83 – 33 = 50.

Q.42: What will be the value of x?
(a) 15005
(b) 13915
(c) 56920
(d) 17429

Correct Answer is Option (c)

As we know,
∠A + ∠C = 180°
⇒ 2x – 1 + 2y + 15 = 180°
⇒ x + y = 83 ...(i)
∠B + ∠D = 180°
y + 5 + 4x – 7 = 180°
4x + y = 182 ...(ii)
Subtracting eq. (i) from (ii), we get
3x = 99
⇒ x = 33.

Q.43: The prime factorisation of 13915 is
(a) 5x 11³ x 13²
(b) 5x 11³ x 23²
(c) 5x 11² x 2³
(d) 5x 11² x 23²

Correct Answer is Option (a)

∠B + ∠C = (y + 5)° + (2y + 15)°

= (3y + 20)°

= (3 × 50° + 20°)

= 170°.

Q.44: What will be the value of y?
(a) 23
(b) 22
(c) 11
(d) 19

Correct Answer is Option (b)

∠A + ∠D = (2x – 1)° + (4x – 7)°

= (6x – 8)°

= (6 × 33 – 8)°

[From question (i), x = 33]

Q.45: According to Fundamental Theorem of Arithmetic 13915 is a
(a) Composite number
(b) Prime number
(c) Neither prime nor composite
(d) Even number

Correct Answer is Option (c)

∠D = (4x – 7)°

= (4 × 33 – 7)°

= 125°.

Direction: Q. 46-Q.50 are based on case study-2

Case Study - 2

Due to heavy storm, many electric wire got bend and many broke. Mr. Prabhakar observed the shape of a bent wire in front of his house. He then draws its shape in a graph as shown below:

Q.46: The zeroes of the polynomial are:
(a) - 1, 5
(b) - 1, 3
(c) 3, 5
(d) - 4, 2

Correct Answer is Option (b)

By using Heron's formula,

∴ Total area of small pizza

= 4 × Area of one slice

= 4 × 5√39

= 20√39 cm²

Q.47: The shape in which wire is bent, is:
(a) spiral
(b) ellipse
(c) linear
(d) parabola.

Correct Answer is Option (c)

Let ABC and PQR represent the slice of smaller and bigger pizza, then in DABC and DPQR,

Q.48: What is the value of polynomial, if x = 2 ?
(a) 3
(b) 0
(c) - 6
(d) - 3

Correct Answer is Option (b)

Now, total area of bigger pizza

= 3 × area of one slice

= 3 × 20√39 cm²

= 60√39 cm²

Q.49: How many zeroes are there for the polynomial, shape of the wire ?
(a) 2
(b) 3
(c) 1
(d) 0

Correct Answer is Option ()

When two triangles are similar the reduced ratio of any two corresponding sides is called the scale factor, so scale factor

Q.50: What will be the expression of the polynomial ?

(a) x² + 2x + 3
(b) x² – 2x + 3
(c) x² – 2x – 3
(d) x² + 2x – 3

Correct Answer is Option (a)