CBSE Class 10 Maths Basic — MCQ Practice 2025-26

20 MCQs from the CBSE Maths Basic 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 Basic SQP 30 from Question Bank Source: CBSE Official

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

The exponent of 3 in the prime factorisation of 2025 is:

✅ Explanation

2025 = 45² = (9 × 5)² = 3⁴ × 5². The exponent of 3 is 4.

If 2024x + 2025y = 1 and 2025x + 2024y = −1, then x − y =

✅ Explanation

Subtracting the first equation from the second: (2025x − 2024x) + (2024y − 2025y) = −1 − 1 → x − y = −2.

The number of polynomials having −2 and 5 as its zeroes is:

✅ Explanation

Any polynomial of the form k(x+2)(x−5) where k ≠ 0 has −2 and 5 as zeroes. Since k can be any non-zero real number, there are infinitely many such polynomials.

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

✅ Explanation

Expanding C: x²+x−2 = x²−2x−3 → 3x+1 = 0. This is linear, not quadratic. Options A, B, D all reduce to quadratic equations.

The value of x for which 2x, (x + 10) and (3x + 2) are three consecutive terms of an AP is:

✅ Explanation

For an AP: 2(x+10) = 2x + (3x+2) → 2x+20 = 5x+2 → 18 = 3x → x = 6. Verify: 12, 16, 20 ✓ (common difference 4).

If 1 + 2 + 3 + 4 + … + 50 = 25k, then k =

✅ Explanation

Sum = n(n+1)/2 = 50 × 51/2 = 1275 = 25k → k = 1275/25 = 51.

The distance between the points (cos 30°, sin 30°) and (cos 60°, −sin 60°) is:

✅ Explanation

Points: (√3/2, 1/2) and (1/2, −√3/2). Distance = √[(√3/2−1/2)² + (1/2+√3/2)²] = √[(√3−1)²/4 + (1+√3)²/4] = √[(3−2√3+1+1+2√3+3)/4] = √[8/4] = √2 units.

The co-ordinates of the point which is the mirror image of the point (−3, 5) about the x-axis are:

✅ Explanation

Reflection about the x-axis: the x-coordinate stays the same; the y-coordinate changes sign. Mirror image of (−3, 5) = (−3, −5).

In △ABC and △DEF, AB/EF = AC/DE. They will be similar when:

✅ Explanation

Given AB/EF = AC/DE (sides about ∠A in △ABC correspond to sides about ∠E in △DEF). By SAS similarity criterion, the included angle must be equal: ∠A = ∠E.

Q10

△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

A circle is inscribed in a triangle with sides 25 cm, 24 cm and 7 cm. The radius r of the circle is: (Hint: 7, 24, 25 is a Pythagorean triple)

✅ Explanation

For a right triangle (7, 24, 25), inradius r = (a + b − c)/2 = (7 + 24 − 25)/2 = 6/2 = 3 cm. Alternatively, area = (1/2)×7×24 = 84 cm² = r×s where s = (7+24+25)/2 = 28, so r = 84/28 = 3 cm.

Q12

Which one of the following is not equal to unity (1)?

✅ Explanation

cot²x − cosec²x = cot²x − (1 + cot²x) = −1, not 1. The identity is cosec²x − cot²x = 1, so cot²x − cosec²x = −1.

Q13

For the frequency distribution below, the upper limit of the median class is:
Classes: 0−5, 5−10, 10−15, 15−20, 20−25 with frequencies 11, 12, 13, 9, 11.

✅ Explanation

Total n = 56, n/2 = 28. Cumulative frequencies: 0−5: 11; 5−10: 23; 10−15: 36. The cumulative frequency first exceeds 28 in class 10−15. Upper limit of median class 10−15 is 15.

Q14

The empirical relationship between the three measures of central tendency is a(Median) = Mode + b(Mean). The value of (2b + 3a) is:

✅ Explanation

The empirical formula is 3 Median = Mode + 2 Mean. So a = 3, b = 2. Therefore 2b + 3a = 2(2) + 3(3) = 4 + 9 = 13.

Q15

From an external point Q, the length of the tangent to a circle is 12 cm, and Q is 13 cm from the centre. The radius of the circle (in cm) is:

✅ Explanation

Tangent ⊥ radius: r² = 13² − 12² = 169 − 144 = 25. r = 5 cm.

Q16

In the figure, PA is a tangent from external point P to a circle with centre O and diameter AB. If ∠POB = 115°, then ∠APO is:

✅ Explanation

PA is tangent → ∠PAO = 90°. In triangle APO: ∠APO = ∠POB − ∠PAO = 115° − 90° = 25°.

Q17

The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is:

✅ Explanation

Circumference = 2πr, so r₁/r₂ = 3/4. Areas ratio = πr₁²/πr₂² = (r₁/r₂)² = (3/4)² = 9 : 16.

Q18

An event is most unlikely to happen. Its probability is closest to:

✅ Explanation

Probability ranges from 0 (impossible) to 1 (certain). "Most unlikely" means probability closest to 0. Among the options, 0.0001 is the smallest and closest to 0.

Q19

Assertion (A): The line joining the midpoints of two sides of a triangle is parallel to the third side.
Reason (R): If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.

✅ Explanation

Both are true: A states the Midpoint Theorem; R states the converse of Basic Proportionality Theorem (BPT). Since joining midpoints divides both sides in ratio 1:1 (same ratio), R directly justifies A. Both true, R is correct explanation.

Q20

Assertion (A): Two coins are tossed simultaneously. Possible outcomes are two heads, one head & one tail, two tails. Hence P(two heads) = 1/3.
Reason (R): Probabilities of equally likely outcomes are always equal.

✅ Explanation

The Assertion is FALSE: HH, TT each occur 1 way, but HT can occur 2 ways (HT or TH). So P(HH) = 1/4, not 1/3. The three listed events are NOT equally likely. The Reason is TRUE. Hence D.

📚 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.