# Class 10 Mathematics: CBSE Sample Question Paper- Term I (2021-22) - 3 Notes | Study CBSE Sample Papers For Class 10 - Class 10

## Class 10: Class 10 Mathematics: CBSE Sample Question Paper- Term I (2021-22) - 3 Notes | Study CBSE Sample Papers For Class 10 - Class 10

The document Class 10 Mathematics: CBSE Sample Question Paper- Term I (2021-22) - 3 Notes | Study CBSE Sample Papers For Class 10 - Class 10 is a part of the Class 10 Course CBSE Sample Papers For Class 10.
All you need of Class 10 at this link: Class 10

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.
Section - A

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

∵ 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

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

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

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

∵ 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

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

∵ 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

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 cm
2

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

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

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

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

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

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

(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

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

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

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

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

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

Total oranges,

n(S) = 1000

Let the no. of bad oranges be x, then

n(E) = x

⇒ 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

Section - B

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) a5b
6

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

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°

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

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

Let p(x) = 2x3 – x– 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

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

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

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

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

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

= 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

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

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

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

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°

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

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

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

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

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

∵ tan θ = cot θ

Here tan 45° = cot 45°

= 1.

∴ θ = 45°

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

Section - C

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

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

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

∠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

∠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

∠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

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.

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

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

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

The document Class 10 Mathematics: CBSE Sample Question Paper- Term I (2021-22) - 3 Notes | Study CBSE Sample Papers For Class 10 - Class 10 is a part of the Class 10 Course CBSE Sample Papers For Class 10.
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