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CBSE Class 10 Mathematics Board Paper Set 1 — MCQs with Answers

Original paper · CBSE patternWritten by our subject team, not a reprint of an official sample paper.

Questions: 20
Time: 25 min
Marking: +1, no negative marking
Mix: 8 Easy · 8 Medium · 4 Hard

Board Paper Set 1 — CBSE Class 10 Mathematics is a full 20-question objective paper for Class 10 Mathematics, written in the CBSE board pattern and marked the way the board marks: +1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. It follows Section A of a real CBSE paper: Q1–Q16 are single-correct MCQs, Q17–Q18 are case-based questions built on a short source, and Q19–Q20 are assertion–reason items using the board's own four option strings.

Difficulty is 8 Easy · 8 Medium · 4 Hard, weighted the way the board weights Section A. You get 25 minutes — about 75 seconds a question, the pace the real paper demands. 8 of the questions are solved in full below, with the correct option and a worked explanation for each; the remaining 12 are timed and scored when you sign in free.

From 2026-27 CBSE Class 10 has two board exams a year: the first around 15 February – 10 March 2027 is compulsory, and an optional improvement exam in May 2027 lets you re-attempt up to three subjects with the better score counting. That gives you two genuine attempts — and makes timed practice between February and May worth far more than it used to be.

8 solved questions from this paper

Answer and worked explanation shown for each. The remaining 12 are timed and scored when you sign in free.

Q1Easy
The LCM of the numbers 12 and 18 is:
A.6
B.72
C.36✓ Correct
D.216
Answer: C. 36
Explanation: Writing each number as a product of primes, 12 = 2² × 3 and 18 = 2 × 3². The LCM takes the highest power of every prime that appears, giving 2² × 3² = 36. The value 216 is simply the product of the two numbers, which equals HCF × LCM, not the LCM.
Q2Easy
The sum and the product of the zeroes of the polynomial 2x² − 7x + 3 are respectively:
A.7/2 and 3/2✓ Correct
B.−7/2 and 3/2
C.7/2 and −3/2
D.7 and 3
Answer: A. 7/2 and 3/2
Explanation: For ax² + bx + c the sum of the zeroes is −b/a and the product is c/a. Here a = 2, b = −7 and c = 3, so the sum is 7/2 and the product is 3/2. Forgetting that the formula already carries a negative sign is what produces −7/2.
Q3Easy
The value of the discriminant of the quadratic equation 2x² − 4x + 3 = 0 is:
A.8
B.−8✓ Correct
C.40
D.−40
Answer: B. −8
Explanation: The discriminant is D = b² − 4ac. With a = 2, b = −4 and c = 3, D = (−4)² − 4 × 2 × 3 = 16 − 24 = −8. Note that squaring −4 gives +16, so the first term is never negative.
Q4Medium
If HCF(a, 96) = 8 and LCM(a, 96) = 480, then the value of a is:
A.40✓ Correct
B.60
C.32
D.48
Answer: A. 40
Explanation: For any two positive integers, HCF × LCM = product of the numbers, so a = (8 × 480) ÷ 96 = 40. Checking, 40 = 2³ × 5 and 96 = 2⁵ × 3 share exactly 2³ = 8, and their LCM is 2⁵ × 3 × 5 = 480.
Q5Medium
For which value of k does the pair of equations 2x + 3y = 7 and 6x + ky = 21 have infinitely many solutions?
A.3
B.6
C.1/9
D.9✓ Correct
Answer: D. 9
Explanation: A pair a₁x + b₁y = c₁ and a₂x + b₂y = c₂ has infinitely many solutions when a₁/a₂ = b₁/b₂ = c₁/c₂. Here 2/6 = 7/21 = 1/3, so 3/k must also equal 1/3, giving k = 9. Choosing k = 3 only makes the two y-coefficients equal to each other, which is not the required ratio.
Q6Medium
If one root of the equation x² + kx + 12 = 0 is 3, then the value of k is:
A.7
B.4
C.−7✓ Correct
D.−4
Answer: C. −7
Explanation: A root satisfies the equation, so substituting x = 3 gives 9 + 3k + 12 = 0. Then 3k = −21 and k = −7. Dropping the negative sign gives 7, which would instead make x = −3 a root.
Q7Hard
If sin θ + cos θ = √2, then the value of sin θ · cos θ is:
A.1
B.1/2✓ Correct
C.√2/2
D.0
Answer: B. 1/2
Explanation: Squaring both sides gives sin²θ + cos²θ + 2 sin θ cos θ = 2. Since sin²θ + cos²θ = 1, this leaves 2 sin θ cos θ = 1, so sin θ cos θ = 1/2. Forgetting to divide by 2 at the last step leaves 1.
Q8Hard
A metal workshop melts a solid steel sphere of radius 6 cm and recasts all of it into small spherical ball bearings of radius 2 cm each, with no metal lost. How many ball bearings are obtained?
A.3
B.9
C.27✓ Correct
D.18
Answer: C. 27
Explanation: Recasting conserves volume, so the number of bearings = (4/3)π(6)³ ÷ (4/3)π(2)³ = 216 ÷ 8 = 27. Volume grows with the cube of the radius, so the count is 3³ and not 3, which is merely the ratio of the two radii.

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More Class 10 Mathematics board papers

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Questions students ask about this paper

Is this an official CBSE sample paper?

No — and that is deliberate. This is an original paper written in the CBSE board pattern by our subject team, not a copy of CBSE's official Sample Question Paper. It means you get fresh questions you have not already seen on a dozen other sites, with instant scoring and worked solutions rather than a PDF.

How is this Class 10 Mathematics paper marked?

+1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. So attempt every question — there is no penalty for a wrong answer on a board paper, and leaving a question blank can only cost you.

How long should this paper take?

25 minutes for 20 questions, roughly 75 seconds each. The timer runs whether or not you are watching it, which is the point: board marks are lost to pacing at least as often as to gaps in knowledge.

Do I need to pay to attempt it?

No. Sign in free with Google and the full paper opens with a live timer, automatic scoring and a worked solution for every question — including the ones not shown on this page.

Does the May 2027 improvement exam change how I should practise?

It raises the value of timed practice. Because the better of your two scores counts, a February attempt that goes badly is no longer final — but only if you know precisely which topics cost you the marks. A scored full-length paper tells you that; re-reading the chapter does not.

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