MCQ's On Mathematics With Detailed Solutions

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Class 10 Maths MCQs provide a full and step-by-step explanation of every MCQ problem for CBSE Class 10th Maths. Class 10 Maths MCQs are given for all the chapters (1 to 15) with detailed solutions. The MCQs are based on the NCERT Curriculum and are as per the latest CBSE Syllabus (2021 – 2022). Given below are objective-type questions that cover the important concepts in the Class 10 Maths Chapters such as Real Numbers, Polynomials, Quadratic equations, Arithmetic Progression, Surface Areas And Volumes, Areas Related To Circles, Trigonometry, Probability and others.

The students studying for the Class 10th board examinations can benefit greatly from the Class 10 Maths MCQs as practising these questions will help the students to do a quick revision for all the chapters.

Ques 1. n² – 1 is divisible by 8, if n is

an integer
a number that isn’t a prime number
a number that isn't even.
an integer that is even.

Click here for the answer

Ans. (c) an odd integer

Explanation: An odd integer with the form (2Q + 1), where Q is a natural number, is known.

so, n² -1 = (2Q + 1)² -1

= 4Q² + 4Q + 1 -1

= 4Q² + 4Q

Substituting Q = 1, 2,…

When Q = 1,

4Q² + 4Q = 4(1)² + 4(1) = 4 + 4 = 8 , it is divisible by 8.

When Q = 2,

4Q² + 4Q = 4(2)² + 4(2) =16 + 8 = 24, it is also divisible by 8 .

When Q = 3,

4Q² + 4Q = 4(3)² + 4(3) = 36 + 12 = 48 , divisible by 8

4Q2 + 4Q is divisible by 8 for all natural numbers, it is concluded.

As a result, for any odd values of n, n2 -1 is divisible by 8.

Ques 2. If one of the cubic polynomial's zeroes is x³+ ax²+ bx + c, then the product of the other two zeroes is:

b-a-1
b-a+1
a-b+1
a-b-1

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Ans. (b) b-a+1

Explanation: Since one zero equals -1,

P(x) = x³+ax²+bx+c

P(-1) = (-1)³+a(-1)²+b(-1)+c

0 = -1+a-b+c

c=1-a+b

-constant term/coefficient of x3= Product of zeroes, αβγ

(-1)βγ = -c/1

c=βγ

βγ = b-a+1

Ques 3. –3 and 4 are the zeroes of which quadratic polynomial:

x² – x + 12
x² + x + 12
(x²/2) – (x/2) – 6
2x² + 2x – 24

Click here for the answer

Ans. (c) (x²/2) – (x/2) – 6

Explanation: Let's say the specified zeros are α = -3 and β = 4.

Sum of zeroes, α + β= -3 + 4 = 1

αβ = -3 4 = -12, Product of Zeroes

As a result, the quadratic polynomial = x2 – (sum of zeroes)x + (product of zeroes)

= x² – (α + β)x + (αβ)

= x² – (1)x + (-12)

= x² – x – 12

Dividing by 2,

= (x²/2) – (x/2) – 6

Read More: Polynomials in One Variable

Ques 4. The x² + 99x + 127 quadratic polynomial's zeroes are

they are both positive
they are both negative
they are one positive and one negative
they are both equal

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Ans. (b) both negative

Explanation: x² + 99x + 127 is the given quadratic polynomial.

When we compare it to the conventional form, we get:

a = 1, b = 99, and c = 127 are the values of a, b, and c.

a, b, and c are all greater than 0.

We know that if all of the coefficients in a quadratic polynomial have the same sign, the polynomial's zeroes will be negative.

As a result, the above quadratic polynomial's zeroes are negative.

Ques 5. If the quotient of a polynomial p(x) divided by a polynomial g(x) is zero, the relationship between the degrees of p(x) and g(x) is.

degree of g(x) > degree of p(x)
degree of p(x) = degree of g(x)
degree of p(x) > degree of g(x)
nothing can be asserted about p(x) and g (x)

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Ans. (a) degree of g(x) > degree of p(x)

Explanation: We know that, p(x)= g(x) × q(x) + r(x)

Given that, q(x) = 0

In the case when q(x) = 0, then r(x) = 0

As a result, when we divide p(x) by g, we get (x),

The value of p(x) should then be zero.

If r(x) = 0, then the degree of p(x) < degree of g(x).

Read More: Degree of a Polynomial

Ques 6. When 1 is subtracted from the numerator, a fraction becomes 1/3, and when 8 is added to the denominator, it becomes 1/4. The resultant fraction is:

3/12
4/12
5/12
7/12

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Ans. (c) 5/12

Explanation: Let's call the fraction x/y.

As a result, in response to the question,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4 => 4x –y =8 ………………..(2)

On Subtracting equation (1) from equation (2), we obtain

x = 5 ………………………………………….(3)

We may get the following result by plugging this number into equation (2):

4×5 – y = 8

y= 12

Therefore, the fraction is 5/12.

Ques 7. Ritu can row 20 kilometres downstream in 2 hours and 4 kilometres upstream in 2 hours. Her rowing speed on calm water and the current speed is:

6 km/hr and 3kph
7 kph and 4 kph
6 kph and 4 kph
10 kph and 6 kph

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Ans. (c) 6km/hr and 4km/hr

Explanation: Let's say Ritu's speed in calm water is x km/hr.

Stream Speed = y km/hr

Now, Ritu's pace during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

When both equations are added together, we obtain

2x=12

x = 6

When we plug in the value of x into eq.1, we obtain

y = 4

Therefore,

Ritu's speed in calm water is 6 km/hr.

Stream speed = 4 km/hr

Ques 8. The value of c for which the equations cx–y = 2 and 6x–2y = 3 have an unlimited number of solutions is

3
-3
-12
no value

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Ans. (d) no value

Explanation: cx – y = 2 and 6x – 2y = 3 are the given equations.

When compared to the conventional form,

a1 = c, b1 = -1, c1 = -2

c2= -3, a2 = 6, b2 = -2

a1/a2 = c/6

b1/b2 = -1/-2 = 1/2

c1/c2 = -2/-3 = 2/3

There must be a condition for there to be an infinite number of solutions.

a1/a2 = b1/b2 = c1/c2

c/6 = 1/2 = 2/3

Therefore,

c = 3 and c = 4

Here, c has different values.

As a result, the pair of equations will have an endless number of solutions for any value of c.

Ques 9. The reciprocals of Rehman's ages three years ago and five years from now add up to 1/3. Rehman is currently of the following age:

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Ans. (a) 7

Explanation: Let x be Rehman's current age.

3 years ago, His age was x – 3 years ago.

After five years, his age equals x + 5.

Given that the sum of the reciprocals of Rehman's ages three years ago and after five years is equal to 1/3,

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x2 + 2x – 15

⇒ x2 – 4x – 21 = 0

⇒ x2 – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

Because we know that age cannot be negative, the solution is 7.

Ques 10. Which of the equations below doesn’t qualify the criteria of a quadratic equation?

(x + 2)² = 2(x + 3)
x2 + 3x = (–1) (1 – 3x)²
(x + 2) (x – 1) = x2 – 2x – 3
x³– x² + 2x + 1 = (x + 1)³

Click here for the answer

Ans. (c) (x + 2) (x – 1) = x2 – 2x – 3

Explanation: We know that a quadratic equation has a degree of two.

By double-checking the selections,

(a) (x + 2)2 = 2(x + 3)

x2 + 4x + 4 = 2x + 6

x2 + 2x – 2 = 0

Thus, This is an example of a quadratic equation.

(b) x2 + 3x = (–1) (1 – 3x)2

x2 + 3x = -1(1 + 9x2 – 6x)

x2 + 3x + 1 + 9x2 – 6x = 0

10x2 – 3x + 1 = 0

Thus, This is an example of a quadratic equation.

x2 + x – 2 = x2 – 2x – 3

x2 + x – 2 – x2 + 2x + 3 = 0

3x + 1 = 0

This is not an example of a quadratic equation.

Ques 11. 11th term of the A.P. -3, -1/2, 2 …. is

28
22
-38
-48

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Ans. (b) 22

Explanation: A.P. = -3, -1/2, 2 …

First-term a = – 3

Common difference, d = a2 − a1 = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

Nth term;

an = a+(n−1)d

a11 = 3+(11-1)(5/2)

a11 = 3+(10)(5/2)

a11 = -3+25

a11 = 22

Read More: Arithmetic Sequence

Ques 12. The following words are missing from AP: __, 13, __, 3:

11 and 9
17 and 9
18 and 8
18 and 9

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Ans. (c) 18 and 8

Explanation: a2 = 13 and

a4 = 3

The nth term of an AP;

an = a+(n−1) d

a2 = a +(2-1)d

13 = a+d ………………. (i)

a4 = a+(4-1)d

3 = a+3d ………….. (ii)

When we subtract equation I from equation (ii), we obtain

– 10 = 2d

d = – 5

Fill in the value of d in equation 1 now.

13 = a+(-5)

a = 18 (first term)

a3 = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8 (third term).

Ques 13. ∠B = ∠E, ∠F = ∠C and AB = 3 DE in triangles ABC and DEF. The two triangles are

congruent but not similar
similar but not congruent
neither similar nor congruent
similar as well as congruent

Click here for the answer

Ans. (b) similar but not congruent

Explanation: In ΔABC and ΔDEF,

∠B = ∠E, ∠F = ∠C and AB = 3 DE

By AA similarity criterion,

ΔABC ~ ΔDEF

AB = 3DE

⇒ AB/DE = 3

⇒ AB/DE = BC/EF = AC/DF = 3

Triangles must have a side ratio of 1 to be congruent.

Therefore, triangles are similar but not congruent.

Read More: Congruence Of Triangles

Ques 14. ΔABC ~ ΔDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm are provided. If that's the case, then the following is correct:

DE = 12 cm, ∠ F = 50 degrees
DE = 12 cm, ∠ F = 100 degrees
∠D = 100°, EF = 12 cm
∠D = 30°, EF = 12 cm

Click here for the answer

Ans. (b) DE = 12 cm, ∠F = 100°

Explanation: Given,

ΔABC ~ ΔDFE, ∠A =30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF= 7.5 cm

In triangle ABC,

∠A + ∠B + ∠C = 180°

∠B = 180° – 30° – 50° = 100°

The corresponding angles are equivalent because ΔABC ~ ΔDFE,

Thus, ∠D = ∠A = 30°

∠F = ∠B = 100°

∠E = ∠C = 50°

And

AB/DF = AC/DE

5/7.5 = 8/DE

DE = (8 × 7.5)/5 = 12 cm

Ques 15. A triangle with vertices (a, b + c), (b, c + a), and (c, a + b) has an area of :

(a + b + c)²
0
a + b + c
ab

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Ans. (b) 0

Explanation: Let the triangle's vertices be:

A = (x1, y1) = (a, b + c)

B = (x2, y2) = (b, c + a)

C = (x3, y3) = (c, a + b)

Area of triangle ABC = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

= (1/2)[a(c + a – a – b) + b(a + b – b – c) + c(b + c – c – a)]

= (1/2)[a(c – b) + b(a – c) + c(b – a)]

= (1/2)[ac – ab + ab – bc + bc – ac]

= (1/2)(0)

= 0

Read More: Similarity of Triangles

Ques 16. If the vertices of a parallelogram are A(6, 1), B(8, 2), C(9, 4) and D(p, 3) in that sequence, then the value of p is

4
-6
7
-2

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Ans. (c) 7

Explanation: The vertices of a parallelogram are A(6, 1), B(8, 2), C(9, 4) and D(p, 3).

A parallelogram's diagonals are known to bisect each other.

As a result, the coordinates of AC's midpoint are the same as the coordinates of BD's midpoint.

⇒ [(6 + 9)/2, (1 + 4)/2] = [(8 + p)/2, (2 + 3)/2]

⇒ (15/2, 5/2) = [(8 + p)/2, 5/2]

The x-coordinates are equalised by equating them.

(8 + p)/2 = 15/2

⇒ 8 + p = 15

⇒ p = 7

Ques 17. At what point does the perpendicular bisector of the line segment joining the points A(1,5) and B(4,6) cut the y-axis?

(0, 13)
(0, –13)
(0, 12)
(13, 0)

Click here for the answer

Ans. (a) (0, 13)

Explanation: The perpendicular bisector of line segment AB, as we know, is perpendicular at AB and goes across AB's midpoint.

Let P be the AB's midpoint.

Mid – point of AB = [(1 + 4)/2, (5 + 6)/2]

P = (5/2, 11/2)

The line AB’s slope = (6 – 5)/(4 – 1) = 1/3 is now calculated.

As a result, the bisector's slope = -1/slope of line AB = -1/(1/3) = -3.

Using the point-slope form, the equation for the line passing through the point P(5/2, 11/2) with slope -3 is:

y – (11/2) = -3[x – (5/2)]

(2y – 11)/2 = -3[(2x – 5)/2]

2y – 11 = -6x + 15

2y = -6x + 15 + 11

2y = -6x + 26

y = -3x + 13

The formula is y = mx + c.

The y-intercept is c = 13 in this case.

As a result, the y-axis is cut at A(1, 5) by the perpendicular bisector of the line segment linking the points A(1, 5) and B(4, 6). (0, 13).

Ques 18. The expression sin⁶θ + cos⁶θ + 3 sin²θ cos²θ has a value of:-

Click here for the answer

Ans. (d) 1

Explanation: We know that, sin²θ + cos²θ = 1

Taking cube on both sides,

(sin²θ + cos²θ)3 = 1

(sin²θ)3 + (cos²θ)3 + 3 sin²θ cos²θ (sin²θ + cos²θ) = 1

sin⁶θ + cos⁶θ + 3 sin²θ cos²θ = 1

Read More: Trigonometry Values

Ques 19. In a given location, it is recommended to construct a single circle park with an area equal to the total of the areas of two circular parks with diameters of 16 m and 12 m. The new park's radius would be

10 m
15 m
20 m
24 m

Click here for the answer

Ans. (a) 10 m

Explanation: Radii of two circular parks will be:

R1 = 16/2 = 8 m

R2 = 12/2 = 6 m

Let R be the new circular park's radius.

If the areas of two circles with radii R1 and R2 are identical, then the area of a circle with radius R is also equal.

R² = R1² + R2²

= (8)² + (6)²

= 64 + 36

= 100

R = 10 m

Ques 20. A hollow cube with an interior edge of 22 cm is filled with spherical marbles with a diameter of 0.5 cm, with 1/8 space remaining empty. The amount of marbles that the cube can hold is thus

142296
142396
142496
142596

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Ans. (a) 142296

Explanation: Volume of cube = 223 = 10648 cm³

The volume of the empty cube = (1/8) 10648 = 1331 cm³

10648 1331 = 9317 cm3 = volume filled by spherical marbles

The spherical marble's radius is 0.5/2 = 0.25 cm = 1/4 cm.

1 spherical marble has a volume of (4/3) (22/7) (1/4)3 = 11/168 cm³.

n = 9317 (11/168) = 142296 number of spherical marbles

Read More: Cube Formula

Ques 21. A solid sphere is formed from a solid piece of iron in the shape of a cuboid with dimensions of 49 cm* 33 cm *24 cm. The sphere's diameter is

21 cm
23 cm
25 cm
19 cm

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Ans. (a) 21 cm

Explanation: For the given cuboid,

Length, l = 49 cm

Breadth, b = 33 cm

Height, h = 24 cm

Cube’s Volume = 49 × 33 × 24 cm³

Let r be the radius of the sphere.

Volume of sphere = 4/3 πr³

Volume of cuboid = volume of sphere molded

49 × 33 × 24 = 4/3 πr³

⇒ πr³ = 29106

⇒ r³ = 29106 × (22/7)

⇒ r³ = 9261

⇒ r³ = (21)³

⇒ r = 21 cm

As a result, the sphere's radius is 21 cm.

Ques 22. The bucket's two-round ends have sizes of 44 cm and 24 cm. The bucket stands 35 centimetres tall. The bucket has a capacity of

32.7 litres
33.7 litres
34.7 litres
31.7 litres

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Ans. (a) 32.7 litres

Explanation: Given,

The bucket's height is h = 35 cm.

One bucket's circular end has a diameter of 44 cm.

The radius R = 22 cm is then calculated.

Another end's diameter is 24 cm.

The radius r = 12 cm is then calculated.

We know that Volume of the bucket = (1/3)πh[R² + r² + Rr]

= (1/3) × (22/7) × 35 × [(22)² + (12)² + 22 × 12]

= (35/3) × (22/7) × (484 + 144 + 264)

= (5 × 22 × 892)/3

= 32706.6 cm³

= 32.7 litres.

Ques 23. The probability of a non-leap year with 53 Sundays being chosen at random is

1/7
2/7
3/7
5/7

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Ans. (a) 1/7

Explanation: Non-leap year = 365 days

52 weeks Plus 1 day Equals 365 days

The number of Sundays in a 52-week period equals 52.

Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday are all possible options for the last day.

The total number of possible outcomes: 7.

1 is the number of positive results.

As a result, the chance of receiving 53 Sundays is 1/7.

Related links:

Properties of Inverse Trigonometric Functions	Trigonometry Table	Some Applications of Trigonometry
Tangent Circle Formula	Inverse Process of Differentiation	Applications of Derivatives
Trigonometric Identities	Angle Between a Line and a Plane	Difference Between Codomain and Range
Volume of Sphere	Sin 30 Degrees	Sin Squared X Formula

MCQ's On Mathematics With Detailed Solutions

CBSE X Related Questions

1.
An arc of length $2.2\text{ cm}$ subtends an angle $\theta$ at the centre of the circle with radius $2.8\text{ cm}$. The value of $\theta$ is

2.
Two dice are rolled together. The probability of getting an outcome $(x, y)$ where $x \gt y$, is

3.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $3/5$, then find the number of yellow balls.

4.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

5.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

6.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

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MCQ's On Mathematics With Detailed Solutions

CBSE X Related Questions

1.An arc of length $2.2\text{ cm}$ subtends an angle $\theta$ at the centre of the circle with radius $2.8\text{ cm}$. The value of $\theta$ is

2.Two dice are rolled together. The probability of getting an outcome $(x, y)$ where $x \gt y$, is

3.A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $3/5$, then find the number of yellow balls.

4.A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

5.The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

6.In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
An arc of length $2.2\text{ cm}$ subtends an angle $\theta$ at the centre of the circle with radius $2.8\text{ cm}$. The value of $\theta$ is

2.
Two dice are rolled together. The probability of getting an outcome $(x, y)$ where $x \gt y$, is

3.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $3/5$, then find the number of yellow balls.

4.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

5.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

6.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.