# Class 10 Math: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

## Class 10: Class 10 Math: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

The document Class 10 Math: CBSE Sample Question Paper- Term I (2021-22)- 2 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 product of two numbers is 1800 and their LCM is 200, then the HCF of the numbers is:
(a) 9

(b) 90
(c) 900
(d) 360000

HCF  = Product of two numbers/LCM
= 1800/200
= 9.

Q.2: If α and β are the zeroes of the quadratic polynomial f(x) = kx+ 4x + 4, then the value of α+β/αβ is:
(a) 2/3
(b) 1
(c) – 1
(d) – 2

It is given that,
f(x) = kx2 + 4x + 4
∴ α + β = − 4/k

and αβ = 4/k

∴ α+β/αβ = -4/k/4/k
= -1

Q.3: The value of k for which the system of equations kx + 10y = k – 5 and 10x + ky = k will have infinite many solutions is:
(a) – 3
(b) ± 10
(c) ± 6
(d) 4

Here,

The system will have infinite solutions if

Then, k/10 = 10/k
⇒ k2 = 100
⇒ k = ± 10

Q.4: A box contains 80 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 30, is:
(a) 1/8
(b) 9/80
(c) 11/80
(d) 3/20

Total number outcomes = 80
i.e., n(S) = 80
Prime number less than 30 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
i.e., n(E) = 10
∴ Required probability,
P(E) = n(E)/n(S)
= 10/80
= 1/8

Q.5: The area of circle that can be inscribed in a square of side 2.8 cm, is:
(a) 6.60 cm2
(b) 6.08 cm2
(c) 6.76 cm2
(d) 6.16 cm2

Since, side of a square = diameter of the circle
∴ Radius of the circle (r) = 2.8/2

= 1.4 cm
Now, area of the circle = πr2
= 22/7 x 1.4 x 1.4
= 6.16 cm2.

Q.6: The largest number which divides 377,286 and 421 leaving remainders 9, 10 and 7 respectively, is:
(a) 23
(b) 46
(c) 69
(d) 92

Required number
= 377 – 9 i.e., 368
286 – 10 i.e., 276 and 421 – 7 i.e., 414
368 = 24 × 23
276 = 22 × 3 × 23
and 414 = 2 × 32 × 23
Hence, HCF = 2 × 23
= 46.

Q.7: If 2 sin θ – √3 = 0, then the value of 4 cos3 θ – 3 cos θ is:
(a) – 1
(b) 1
(c) 2
(d) 4

2 sin θ − 3 = 0
⇒ sin θ = 3/2
⇒ sin θ = sin 60°
⇒ θ = 60°
Now, 4 cos3 θ – 3cos θ
= 4(cos 60°)3 –  3 cos 60°
= 4(cos 60°)3 – 3 cos 60°
= -1

Q.8: In DABC, DE || BC, then the value of x is:

(a) 1 unit
(b) 2 unit
(c) 3 unit
(d) 4 unit

Since, DE || BC, then
⇒ x/x-2 = x + 2/x - 1
⇒ x(x - 1) = x2 - 4
⇒ x2 - x = x2 - 4
⇒ x = 4 Unit

Q.9: The number 0.131313 ______ is the form of p/q will be equal:
(a) 4/11
(b) 2/11

(c) 13/99
(d) 2/33

0.131313 ........ =
So,  = 13/99

Q.10: The pair of linear equations 3x + 6 = 10y and 2x – 15y + 3 = 0 is:
(a) inconsistent
(b) consistent
(c) parallel
(d) none of these

Here,

∴ The given system of linear equations is consistent.

Q.11: Riya was drawing a rectangle on her worksheet, then length of diagonal of a rectangle having length 12 m and breadth 5 cm is:
(a) 13 m
(b) 12 cm
(c) 5 m
(d) 10 cm

Let ABCD be a rectangle, then, AB = 12 m and BC = 5 cm in right angled DABC,
AC2 = AB2 + BC2
(By using Pythagoras Theorem)
= (12)2 + (5)2
= 144 + 25
= 169
∴ AC = √169 = 13 m.

Q.12: While playing with circular toy. Mohit, started measuring it, if he measured perimeter of a sector of a circle of radius 4 cm is 20 cm, then the arc ACB is:

(a) 8 cm
(b) 28 cm
(c) 12 cm
(d) 8.5 cm

Here OACB is a sector of circle
and OA= OB = 4 cm
∴ Perimeter = 20 cm (given)
OA+ OB +arc ACB = 20
⇒ 4 + 4 + arc ACB = 20
⇒ arc ACB = (20 – 8) cm
⇒ 12 cm

Q.13: The value of 4 sin2 30° + cosec30° + 3 cos 0° + 4 sin 90° – tan2 60° is:
(a) 9

(b) 10
(c) 11
(d) 12

4 sin2 30° + cosec2 30° + 3 cos 0° + 4 sin 90°  – tan2 60°.

= 4(1/2)2 + (2)2 + 3 x 1 + 4 x 1 - (√3)2

= 4 x 1/4 + 4 + 7 - 3
= 1 + 4 + 4

= 9.

Q.14: The decimal expansions of 17/54 × 23 is:
(a) 0.034
(b) 0.3040
(c) 0.0034
(d) 0.3440

= 34/10000
= 0.0034

Q.15: In the given figure, ΔABC ~ ΔADE. If AB = 8 cm, DE = 4.5 cm and BC = 9 cm, then AE is:
(a) 4.5 cm
(b) 4 cm
(c) 3.8 cm
(d) 4.8 cm

⇒ AB/AE = BC/DE
⇒ 8/AE = 9/4.5

⇒ AE = 8 x 4.5/9
= 4 cm

Q.16: A letter is chosen at random from the letters of the word 'MATHEMATICS' then the probability that the chosen letter is a vowel; is:
(a) 4/11
(b) 7/11
(c) 1
(d) 0

Total no. of letters in the given word = 11
i.e., n(S) = 11
and no. of vowels in the given word = 4
i.e., n(E) = 4
∴ P(choosing a vowel) = 4/11.

Q.17: If three vertices of a triangle are (2, 4), (5, 7) and (3, 7), then its centroid is:
(a) (3, 6)
(b) (10/3, 6)
(c) ( 11/3, 17/3)
(d) ( 10/3, 6)

Centroid of a triangle

Q.18: If x = a sin θ and y = b cos θ, then the value of b2x2 + a2y2 is:
(a) ab
(b) 1
(c) a2b2
(d) a2 + b
2

We have,
x = α sin θ
and y = b cos θ
Then, b2x2 + a2y2 = b2 × a2 sin2 q + a2 × b2 cos2 θ
= a2b2 (sin2 θ + cos2 θ)
= a2b2 × 1
[∵ sin2 θ + cos2 θ = 1]
= a2b2.

Q.19: If the numbers are 2, 2, 2, 3, 3, 5, 5, 7, 7, 7, 7, 9, 9, 9 given and one number is selected at random, then the probability (getting 7) is:
(a) 3/14
(b) 1/7
(c) 2/7
(d) 3/7

Total outcome,
n(S) = 14
Favourable outcomes,
n(E) = 4
∴ (getting 7) = n(E)/n(S)
= 4/14
= 2/7

Q.20: If the points A(5, – 3), B(– 6, – 3), C(8, 4) and D(k, 5) are the vertices of a parallelogram, taken in order, then the value of k is:
(a) 17
(b) 18
(c) 19
(d) 10

In parallelogram, Mid point of AC = Mid Point of BD

⇒ k – 6 =  13
⇒ k =  19.

Section - B

Q.21: If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are
(a) 2
(b) 3
(c) 4
(d) 5

Since HCF = 81, two numbers can be taken as 81x and 81y,
ATQ 81x + 81y = 1215
Or x + y = 15
which gives four co prime pairs
(1,14), (2,13), (4,11), (7, 8)

Q.22. Given below is the graph representing two linear equations by lines AB and CD respectively. What is the area of the triangle formed by these two lines and the line x = 0?

(a) 3sq. units
(b) 4sq. units
(c) 6sq. units
(d) 8sq. units

Required Area is area of triangle
ACD   = 1/2× 6×2

= 6 sq units

Q.23: If tan α + cot α = 2, then tan20α + cot20α =
(a) 0
(b) 2
(c) 20
(d) 220

tan α + cot α = 2
gives α = 45°.
So tan α = cot α = 1
tan20α + cot20α = 120 + 120 = 1+1 = 2

Q.24: If 217x + 131y = 913, 131x + 217y = 827, then x + y is
(a) 5
(b) 6
(c) 7
(d) 8

Adding the two given equations we get:
348x + 348y = 1740.
So x + y = 5

Q.25: The LCM of two prime numbers p and q (p > q) is 221. Find the value of 3p – q.
(a) 4
(b) 28
(c) 38
(d) 48

LCM of two prime numbers = product of the numbers
221 = 13 × 17.
So p = 17 & q = 13
∴ 3p – q = 51-13 = 38

Q.26: A card is drawn from a well shuffled deck of cards. What is the probability that the card drawn is neither a king nor a queen?
(a) 11/13
(b) 12/13
(c) 11/26
(d) 11/52

Probability that the card drawn is neither a king nor a queen
= 52 - 8/52
= 44/52 = 11/13

Q.27: Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is
(a) 5/36
(b) 11/36
(c) 12/36
(d) 23/36

Outcomes when 5 will come up at least once are(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6) Probability that 5 will come up at least once = 11/36

Q.28: If 1+ sin2α = 3 sinα cosα, then values of cot α are
(a) –1, 1
(b) 0,1
(c) 1, 2
(d) –1, –1

1+ sin2α = 3 sinα cos α
sin2α + cos2α + sin2α = 3 sinα cos α
2 sin2α – 3sinα cos α + cos2α = 0
(2sinα – cos α)(sinα– cosα) =0
∴ cotα = 2 or cotα = 1

Q.29: The vertices of a parallelogram in order are A(1, 2), B(4, y), C(x, 6) and D(3, 5). Then (x, y) is
(a) (6, 3)
(b) (3, 6)
(c) (5, 6)
(d) (1, 4)

Since ABCD is a parallelogram, diagonals AC and BD bisect each other,
∴ mid point of AC= mid point of BD

Comparing the co-ordinates, we get,

∴ (x, y) = (6, 3)

Q.30: In the given figure, ∠ACB = ∠CDA, AC = 8cm, AD = 3cm, then BD is

(a) 22/3 cm
(b) 26/3 cm
(c) 55/3 cm
(d) 64/3 cm

ΔACD ~ Δ ABC(AA)

8/AB = 3/8
This gives AB = 64/3 cm.
So BD = AB – AD = 64/3 – 3 = 55/3cm.

Q.31: The equation of the perpendicular bisector of line segment joining points A(4,5) and B(-2,3) is 1
(a) 2x – y +7 = 0
(b) 3x + 2y – 7 = 0
(c) 3x – y – 7 = 0
(d) 3x + y – 7 = 0

Any point (x, y) of perpendicular bisector will be equidistant from A & B.

Solving we get, –12x – 4y + 28 = 0
or 3x + y – 7 = 0

Q.32: In the given figure, D is the mid-point of BC, then the value of cot y°/cot x° is

(a) 2
(b) 1/2
(c) 1/3
(d) 1/4

Q.33: The smallest number by which 1/13 should be multiplied so that its decimal expansion terminates after two decimal places is
(a) 13/100
(b) 13/10
(c) 10/13
(d) 100/13

The smallest number by which 1/13 should  be multiplied so that its decimal expansion terminates after two decimal points is 13/100 as

13/100

Q.34: Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of largest square FDGB that can be inscribed in the triangle ABE is

(a) 32/3cm
(b) 16/3cm
(c) 8/3cm
(d) 4/3cm

ΔABE is a right triangle & FDGB is a square of side x cm ΔAFD ~Δ DGE(AA)

128 = 24x or x = 16/3cm

Q.35: Point P divides the line segment joining R(–1, 3) and S(9, 8) in ratio k :1. If P lies on the line x – y + 2 = 0, then value of k is
(a) 2/3
(b) 1/2
(c) 1/3
(d) 1/4

Since P divides the line segment joining R(-1, 3) and S(9,8) in ratio k:1
∴ coordinates of P are
Since P lies on the line x – y + 2 = 0, then

9k –1 –8k–3 +2k+2 = 0
which gives k = 2/3

Q.36: In the figure given below, ABCD is a square of side 14 cm with E, F, G and H as the mid points of sides AB, BC, CD and DA respectively. The area of the shaded portion is

(a) 44 cm²
(b) 49 cm²
(c) 98 cm²
(d) 49π/2 cm²

Shaded area = Area of semicircle + (Area of half square – Area of two quadrants) = Area of semicircle +(Area of half square – Area of semicircle) = Area of half square= 1/2 × 14 ×14 = 98cm²

Q.37: Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1 cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1 cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is

(a) 4(π/12 – √3/4) cm²
(b) (π/6 – √3/4) cm²
(c) 4(π/6 – √3/4) cm²
(d) 8(π/6 – √3/4) cm²

Let O be the center of the circle.
OA = OB = AB = 1cm.
So ΔDAB is an equilateral triangle and
∴ ∠AOB Required Area = 8x Area of one segment with r = 1cm, q = 60°

= 8(π/6 – √3 /4)cm²

Q.38: If 2 and 1/2 are the zeros of px2 + 5x + r, then
(a) p = r = 2
(b) p = r = – 2
(c) p = 2, r = – 2
(d) p = – 2, r = 2

Sum of zeroes = 2 + 1/2 = –5/p
i.e. 5/2 = –5/p. So p = –2
Product of zeroes = 2 x 1/2 = r/p
i.e. r/p = 1 or r = p = –2

Q.39: The circumference of a circle is 100 cm. The side of a square inscribed in the circle is
(A) 50√2 cm
(B) 100/π cm
(C) 50√2 /π cm
(D) 100√2/π cm

2πr = 100.
So, Diameter = 2r =100/π = diagonal of the square. side √2 = diagonal of square = 100√2 π
∴ side = 100√2 π = 50√2/π

Q.40: The number of solutions of 3x+y = 243 and 243x–y = 3 is
(A) 0
(B) 1
(C) 2
(D) infinite

3x+y = 243 = 35
So x + y = 5    (1)
243x–y = 3
(35)x–y = 31
So 5x–5y = 1
Since, so unique Solution.

Section - C

Q. 41 to 45 are based on Case Study - 1

Read the following text and answer the following questions on the basis of the same: The four angles of a cyclic quadrilateral ABCD are given in the following figure:

Q.41: The value of x is:
(a) 11
(b) 22
(c) 33
(d) 50

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.42: The value of (∠A + ∠D) is:
(a) 180°
(b) 190°
(c) 200°
(d) 150°

∠A + ∠D = (2x – 1)° + (4x – 7)°
= (6x – 8)°
= (6 × 33 – 8)°
[From question (i), x = 33]

Q.43: The value of y is:
(a) 25
(b) 30
(c) 40
(d) 50

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

Q.44: The measure of ∠D is:
(a) 120°
(b) 115°
(c) 125°
(d) 95°

∠D = (4x – 7)°
= (4 × 33 – 7)°
= 125°.

Q.45: The measure of (∠B + ∠C) is:
(a) 170°
(b) 150°
(c) 190°
(d) 180°

∠B + ∠C = (y + 5)° + (2y + 15)°
= (3y + 20)°
= (3 × 50° + 20°)
= 170°.

Q.46 to 50 are based on case Study - 2
One small size and another big size pizza was ordered by Mr. Rohan for a breakfast. Instead of eating this he sit to compare the slices of the pizza and then many questions came into his mind. The size of the slice of small and big pizza is shown in below figure.

Q.46: The slices of smaller pizza and bigger pizza are similar using:
(a) AA criterian
(b) SAS rule
(c) SSS rule
(d) slices are not similar.

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

Q.47: The scale factor is:
(a) 1
(b) 2
(c) 3
(d) 4

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

Q.48: If the small pizza has cut into 4 slices, then what will be the total area of the pizza ?
(a) 39 cm2
(b) 20√39 cm2
(c) 5√39 cm2
(d) 40 cm
2

By using Heron's formula,
area of DABC =
then

Q.49: The ratio of area of small slices to bigger slice is:
(a) 1 : 4
(b) 1 : 8
(c) 1 : 16
(d) 1 : 2

Q.50: If the bigger pizza has cut into 3 slices, then the total area of bigger pizza is:
(a) 60 cm2
(b) 20√39 cm2
(c) 539 cm2
(d) 60√39 cm
2

Now, total area of bigger pizza
= 3 × area of one slice
= 3 × 20√39 cm2
= 60√39 cm2

The document Class 10 Math: CBSE Sample Question Paper- Term I (2021-22)- 2 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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