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 MCQs with one correct answer. Q17–Q18 are case-based questions on a short source. Q19–Q20 are assertion–reason questions that use the board's own four answer options.

The difficulty mix is 8 Easy · 8 Medium · 4 Hard, balanced the way the board balances Section A. You get 25 minutes, about 75 seconds a question, which is the speed the real paper needs. 8 of the questions are solved in full below, each with the correct option and a worked explanation. Sign in free to attempt the other 12, timed and scored.

From 2026-27, CBSE Class 10 has two board exams a year. The first, around 15 February – 10 March 2027, is compulsory. An optional improvement exam in May 2027 lets you take up to three subjects again, and the better score counts. So you get two real attempts, and timed practice between February and May matters far more than it used to.

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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Questions students ask about this paper

Is this an official CBSE sample paper?

No, and that is on purpose. This is an original paper written by our subject team in the CBSE board pattern. It is not a copy of CBSE's official Sample Question Paper. So you get fresh questions you have not already seen on a dozen other sites, with instant scoring and worked solutions instead of 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. A board paper takes nothing away for a wrong answer, and leaving a question blank can only cost you.

How long should this paper take?

25 minutes for 20 questions, about 75 seconds each. The timer keeps running whether you watch it or not, and that is the point. Students lose board marks to poor timing at least as often as to gaps in what they know.

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 makes timed practice more useful. The better of your two scores counts, so a February attempt that goes badly is no longer final. But that only helps if you know exactly which topics cost you marks. A scored full-length paper tells you that. Re-reading the chapter does not.

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