CBSE Class 10 Maths Standard — MCQ Practice 2025-26

20 MCQs from the CBSE Maths Standard sample question paper and 30 MCQs from the official case study question bank. Click any option for instant feedback and explanation.

50 Total Questions 20 from Maths Standard SQP 30 from Question Bank Source: CBSE Official

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📄 Maths Standard Sample Question Paper 2025-26 — Section A (Q1–Q20)1 mark each · MCQ + Assertion-Reason

If a = 2² × 3^x, b = 2² × 3 × 5 and c = 2² × 3 × 7, and LCM(a, b, c) = 3780, then x equals:

✅ Explanation

LCM = 2² × 3^x × 5 × 7 = 140 × 3^x = 3780 → 3^x = 27 = 3³ → x = 3.

The shortest distance (in units) of the point (2, 3) from the y-axis is:

✅ Explanation

The shortest distance from any point to the y-axis equals the absolute value of its x-coordinate. Here x = 2, so distance = 2 units.

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are not parallel, then the value of k must be:

✅ Explanation

Two lines a₁x+b₁y=c₁ and a₂x+b₂y=c₂ are NOT parallel if a₁/a₂ ≠ b₁/b₂. Here 3/2 ≠ 2k/5 → k ≠ 15/4.

A quadrilateral ABCD is drawn to circumscribe a circle. If BC = 7 cm, CD = 4 cm and AD = 3 cm, then the length of AB is:

✅ Explanation

For a quadrilateral circumscribing a circle: AB + CD = AD + BC. So AB + 4 = 3 + 7 → AB = 6 cm.

If secθ + tanθ = x, then secθ − tanθ will be:

✅ Explanation

Using the identity sec²θ − tan²θ = 1: (secθ + tanθ)(secθ − tanθ) = 1. Since secθ + tanθ = x, we get secθ − tanθ = 1/x.

Which one of the following is not a quadratic equation?
(A) (x+2)² = 2(x+3) (B) x² + 3x = (−1)(1−3x²) (C) x³ − x² + 2x + 1 = (x+1)³ (D) (x+2)(x+1) = x² + 2x + 3

✅ Explanation

Expanding option D: x² + 3x + 2 = x² + 2x + 3 → x − 1 = 0. This is a linear equation, not quadratic.

Five congruent circles of radius 1 cm each are arranged like Olympic rings. Adjacent circles intersect such that the chord joining their intersection points is also 1 cm. The total area of all the dotted (lens-shaped) regions is:

✅ Explanation

Since chord = radius = 1 cm, the triangle formed at each intersection is equilateral with θ = 60°. There are 8 such segments (4 intersections × 2 segments each). Each segment area = π/6 − √3/4. Total = 8[π/6 − √3/4] cm².

A pair of dice is tossed. The probability of not getting a sum of 8 is:

✅ Explanation

Ways to get sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes. P(sum=8) = 5/36. P(not sum 8) = 1 − 5/36 = 31/36.

If 2 sin 5x = √3 and 0° ≤ x ≤ 90°, then x is equal to:

✅ Explanation

sin 5x = √3/2 → 5x = 60° → x = 12°.

Q10

In △ABC ~ △PQR, where AB = 3 cm, BC = 4 cm, AC = 5 cm and PR = 10 cm. The perimeter of △PQR (in cm) is:

✅ Explanation

Scale factor = PR/AC = 10/5 = 2. Perimeter △ABC = 3+4+5 = 12 cm. Perimeter △PQR = 2 × 12 = 24 cm.

Q11

The area of the base of a right circular cone is 51 cm² and its volume is 85 cm³. The height of the cone is:

✅ Explanation

V = (1/3) × base area × h → 85 = (1/3) × 51 × h → h = (85 × 3)/51 = 255/51 = 5 cm.

Q12

If the zeroes of the quadratic polynomial ax² + bx + c (a, c ≠ 0) are equal, then:

✅ Explanation

For equal roots, discriminant = 0: b² = 4ac → ac = b²/4 ≥ 0. Therefore c and a must have the same sign.

Q13

The area (in cm²) of a sector of a circle of radius 21 cm cut off by an arc of length 22 cm is:

✅ Explanation

Area of sector = (1/2) × arc length × radius = (1/2) × 22 × 21 = 231 cm².

Q14

If △ABC ~ △DEF, AB = 6 cm, DE = 9 cm, EF = 6 cm and FD = 12 cm, then the perimeter of △ABC is:

✅ Explanation

Scale factor: AB/DE = 6/9 = 2/3. Perimeter △DEF = 9+6+12 = 27 cm. Perimeter △ABC = (2/3) × 27 = 18 cm.

Q15

If the probability of a letter chosen at random from the word "MATHEMATICS" being a vowel is 2/(2x+1), then x is equal to:

✅ Explanation

MATHEMATICS has 11 letters; vowels: A, E, A, I = 4. P(vowel) = 4/11. Setting 4/11 = 2/(2x+1) → 4(2x+1) = 22 → 8x = 18 → x = 9/4.

Q16

The points A(9, 0), B(9, −6), C(−9, 0) and D(−9, 6) are the vertices of a:

✅ Explanation

Midpoint of diagonal AC = (0,0); midpoint of diagonal BD = (0,0). Since diagonals bisect each other, ABCD is a parallelogram. Sides AB = 6 ≠ BC = √360, so it's not a rectangle.

Q17

The median of a set of 9 distinct observations is 20.5. If each observation is increased by 2, the median of the new set:

✅ Explanation

When every observation is increased by the same constant, the median also increases by that constant. New median = 20.5 + 2 = 22.5 (increased by 2).

Q18

The length of a tangent drawn from a point at a distance of 41 cm from the centre of a circle of radius 9 cm is:

✅ Explanation

Using Pythagoras (tangent ⊥ radius): tangent² = 41² − 9² = 1681 − 81 = 1600. Tangent = 40 cm.

Q19

Assertion (A): The number 5ⁿ cannot end with the digit 0, where n is a natural number.
Reason (R): A number ends with 0 if its prime factorisation contains both 2 and 5.

✅ Explanation

5ⁿ has only 5 as its prime factor — it contains no factor of 2. So it can never be a multiple of 10 and cannot end with 0. R correctly explains A. Both true, R is the correct explanation.

Q20

Assertion (A): If cos A + cos²A = 1, then sin²A + sin⁴A = 1.
Reason (R): sin²A + cos²A = 1.

✅ Explanation

From A: cos A = 1 − cos²A = sin²A. Substituting into the assertion: sin²A + (sin²A)² = sin²A + sin⁴A. Also sin²A + cos²A = 1 → sin²A + sin⁴A = 1 ✓. R is the correct explanation. Both true, R is correct explanation.

📚 CBSE Case Study Question Bank — Mixed Topics (Q21–Q50)Real Numbers · Polynomials · Linear Equations · Quadratic Eq. · AP · Coordinate Geometry

Q21

A class library is to be set up for sections A (32 students) and B (36 students). What is the minimum number of books so that they can be distributed equally among students of either section?

✅ Explanation

The minimum number = LCM(32, 36). 32 = 2⁵, 36 = 2² × 3². LCM = 2⁵ × 3² = 32 × 9 = 288.

Q22

The product of two positive integers equals the product of their HCF and LCM. Using this, HCF(32, 36) is:

✅ Explanation

32 = 2⁵, 36 = 2² × 3². HCF = 2² = 4. Verify: HCF × LCM = 4 × 288 = 1152 = 32 × 36 ✓.

Q23

36 can be expressed as a product of its prime factors as:

✅ Explanation

36 = 4 × 9 = 2² × 3². Verify: 4 × 9 = 36 ✓.

Q24

The number 7 × 11 × 13 × 15 + 15 is:

✅ Explanation

7 × 11 × 13 × 15 + 15 = 15(7 × 11 × 13 + 1) = 15 × (1001 + 1) = 15 × 1002. Since it has factors 15 and 1002 (other than 1 and itself), it is a composite number.

Q25

If p and q are positive integers such that p = ab² and q = a²b, where a and b are prime numbers, then LCM(p, q) is:

✅ Explanation

For LCM, take the highest power of each prime: LCM = a² × b² = a²b².

Q26

A seminar has 60 Hindi, 84 English and 108 Maths teachers. To seat them in rooms with the same number and same subject, the maximum number of teachers per room is:

✅ Explanation

Maximum per room = HCF(60, 84, 108). 60 = 2²×3×5, 84 = 2²×3×7, 108 = 2²×3³. HCF = 2² × 3 = 12.

Q27

Using the data from the above question, the minimum number of rooms required for the seminar is:

✅ Explanation

Rooms = 60/12 + 84/12 + 108/12 = 5 + 7 + 9 = 21 rooms.

Q28

The LCM of 60, 84 and 108 is:

✅ Explanation

60 = 2²×3×5, 84 = 2²×3×7, 108 = 2²×3³. LCM = 2²×3³×5×7 = 4×27×5×7 = 3780.

Q29

The product of HCF and LCM of 60, 84 and 108 is: (HCF = 12, LCM = 3780)

✅ Explanation

HCF(60,84,108) = 12, LCM = 3780. Product = 12 × 3780 = 45360.

Q30

108 can be expressed as a product of its primes as:

✅ Explanation

108 = 4 × 27 = 2² × 3³. So 108 = 2² × 3³. Verify: 4 × 27 = 108 ✓.

Q31

In the standard form of a quadratic polynomial, ax² + bx + c, the coefficients a, b and c are:

✅ Explanation

In ax²+bx+c, a must be non-zero (otherwise it is not a quadratic polynomial), while b and c can be any real numbers.

Q32

If the roots of the quadratic polynomial are equal, where discriminant D = b² − 4ac, then:

✅ Explanation

For a quadratic to have equal (repeated) real roots: b² − 4ac = 0, i.e., D = 0. D > 0 gives two distinct real roots; D < 0 gives no real roots.

Q33

If α and 1/α are the zeroes of the quadratic polynomial 2x² − x + 8k, then k is:

✅ Explanation

Product of zeroes = α × (1/α) = 1 = c/a = 8k/2 = 4k. Solving: 4k = 1 → k = 1/4.

Q34

The graph of x² + 1 = 0:

✅ Explanation

x² + 1 = 0 → x² = −1. There are no real roots (square of a real number is never negative). The parabola y = x² + 1 lies entirely above the x-axis — it neither touches nor intersects the x-axis.

Q35

If the sum of the roots is −p and the product of the roots is −(1/p), then the quadratic polynomial is:

✅ Explanation

A quadratic with sum S and product P of roots is k[x² − Sx + P]. Here S = −p, P = −1/p. So the polynomial is k[x² − (−p)x + (−1/p)] = k[x² + px − 1/p] = k(x² + px − 1/p).

Q36

In City A, a journey of 10 km costs Rs 75 and a journey of 15 km costs Rs 110. If the fixed charge is Rs x and per-km charge is Rs y, the pair of linear equations is:

✅ Explanation

Fixed charge x + (per km charge y) × (km) = amount. For 10 km: x+10y=75; for 15 km: x+15y=110. Correct option: x+10y=75, x+15y=110.

Q37

Using City A's charges (x+10y=75, x+15y=110, giving x=5, y=7), the amount for a 50 km journey is:

✅ Explanation

Solving: subtracting gives 5y=35 → y=7; x=75−70=5. For 50 km: 5 + 50×7 = 5+350 = Rs 355.

Q38

In City B, 8 km costs Rs 91 and 14 km costs Rs 145. What will a person have to pay for travelling 30 km?

✅ Explanation

Setup: x+8y=91, x+14y=145. Subtracting: 6y=54 → y=9; x=91−72=19. For 30 km: 19+30×9 = 19+270 = Rs 289.

Q39

Raj's car travels at x km/h. Ajay's car travels 5 km/h faster. The distance covered by Ajay's car in two hours is:

✅ Explanation

Ajay's speed = (x+5) km/h. Distance in 2 hours = 2 × (x+5) = 2(x+5) km.

Q40

Raj takes 4 hours more than Ajay to travel 400 km. The quadratic equation describing Raj's speed x (km/h) is:

✅ Explanation

400/x − 400/(x+5) = 4. Simplifying: 400(x+5)−400x = 4x(x+5) → 2000 = 4x²+20x → x²+5x−500=0.

Q41

Solving x² + 5x − 500 = 0, Raj's speed is:

✅ Explanation

x²+5x−500=0 → (x+25)(x−20)=0. Since speed > 0, x = 20 km/h (rejecting x=−25).

Q42

If Raj's speed is 20 km/h, Ajay's speed is 25 km/h. How long did Ajay take to travel 400 km?

✅ Explanation

Time = Distance ÷ Speed = 400 ÷ 25 = 16 hours.

Q43

Veer currently runs 200 m in 51 seconds. With each day of practice, he takes 2 seconds less. Which AP represents the time taken after each day?

✅ Explanation

Time decreases by 2 each day: 51, 49, 47, … → common difference d = −2. This is an AP: 51, 49, 47, …

Q44

Veer wants to run 200 m in 31 seconds. The minimum number of days of practice needed is:

✅ Explanation

aₙ = 51 + (n−1)(−2) ≤ 31 → 51−2n+2 ≤ 31 → 53−2n ≤ 31 → 2n ≥ 22 → n ≥ 11. Minimum = 11 days. Check: a₁₁ = 51−20 = 31 ✓.

Q45

Which of the following numbers is NOT a term in the AP 51, 49, 47, …?

✅ Explanation

The AP has first term 51 and d=−2, so all terms are odd (51, 49, 47, …). 30 is even and cannot be a term in this AP. (Verify: 51−2(n−1)=30 → n=11.5, not an integer.)

Q46

If the n^th term of an AP is given by aₙ = 2n + 3, the common difference of the AP is:

✅ Explanation

d = aₙ₊₁ − aₙ = [2(n+1)+3] − [2n+3] = 2(n+1)+3−2n−3 = 2.

Q47

Find the value of x for which 2x, x + 10 and 3x + 2 are three consecutive terms of an AP:

✅ Explanation

For an AP: 2(x+10) = 2x + (3x+2) → 2x+20 = 5x+2 → 3x = 18 → x = 6. Check: 12, 16, 20 ✓.

Q48

Niharika runs 1/4^th the distance AD (where AD = 100 m) along the 2^nd line and posts a green flag. The position of the green flag is:

✅ Explanation

The 2nd line is at x = 2. Distance = 1/4 × 100 = 25 m along AD. Position = (2, 25).

Q49

Preet runs 1/5^th the distance AD along the 8^th line and posts a red flag. The position of the red flag is:

✅ Explanation

The 8th line is at x = 8. Distance = 1/5 × 100 = 20 m along AD. Position = (8, 20).

Q50

Green flag at (2, 25) and red flag at (8, 20). The distance between both flags is:

✅ Explanation

Distance = √[(8−2)² + (20−25)²] = √[6² + (−5)²] = √[36 + 25] = √61 units.