CBSE Class 10 Maths Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 10 Mathematics paper built to the current board blueprint: 38 questions across five sections for 80 marks in 3 hours, with
TL;DR: This is a full-length CBSE Class 10 Mathematics paper built to the current board blueprint: 38 questions across five sections for 80 marks in 3 hours,…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 10 Mathematics paper built to the current board blueprint: 38 questions across five sections for 80 marks in 3 hours, with Section A objective, Sections B to D descriptive and Section E case-based. Every question is followed by a fully worked solution, not just the final answer, so you can check your method as well as your result.
Board: CBSE · Class: Class 10 · Subject: Mathematics · Total: 80 marks · Time: 3 hours
Section A — Objective (1 mark each) — 20 marks
Questions 1–18 are multiple choice. Questions 19 and 20 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. The HCF of 96 and 404 is (1 mark)
Answer: 4. 404 = 96×4 + 20; 96 = 20×4 + 16; 20 = 16×1 + 4; 16 = 4×4 + 0. The last non-zero remainder is 4.
Q2. The LCM of two numbers is 182 and their HCF is 13. If one number is 26, the other is (1 mark)
Answer: 91. Product of the numbers = HCF × LCM = 13 × 182 = 2366, so the other number = 2366 ÷ 26 = 91.
Q3. If the sum of the roots of 2x² − (k+1)x + (k−2) = 0 is 5/2, then k equals (1 mark)
Answer: k = 4. Sum of roots = (k+1)/2 = 5/2, so k + 1 = 5 and k = 4.
Q4. The quadratic equation x² + 4x + 5 = 0 has (1 mark)
Answer: No real roots. D = 16 − 20 = −4 < 0, so the roots are not real.
Q5. The 15th term of an AP whose first term is 5 and common difference is 3 is (1 mark)
Answer: 47. aₙ = a + (n−1)d = 5 + 14×3 = 5 + 42 = 47.
Q6. How many two-digit numbers are divisible by 7? (1 mark)
Answer: 13. They form the AP 14, 21, …, 98: 98 = 14 + (n−1)7 gives n = 13.
Q7. If the pair of equations 2x + 3y = 7 and 4x + 6y = k has infinitely many solutions, then k is (1 mark)
Answer: 14. For infinitely many solutions a₁/a₂ = b₁/b₂ = c₁/c₂, i.e. 2/4 = 3/6 = 7/k, so k = 14.
Q8. If sin A = 3/5, then cos A equals (1 mark)
Answer: 4/5. cos²A = 1 − sin²A = 1 − 9/25 = 16/25, and A is acute, so cos A = 4/5.
Q9. The value of (sin 30° · cos 60° + cos 30° · sin 60°) is (1 mark)
Answer: 1. It equals sin(30° + 60°) = sin 90° = 1. Directly: ½·½ + (√3/2)(√3/2) = ¼ + ¾ = 1.
Q10. A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2.5 m from the wall, the length of the ladder is (1 mark)
Answer: 5 m. cos 60° = base/hypotenuse, so ½ = 2.5/L and L = 5 m.
Q11. Two similar triangles have areas in the ratio 4 : 9. Their corresponding sides are in the ratio (1 mark)
Answer: 2 : 3. The ratio of areas of similar triangles equals the square of the ratio of corresponding sides, and √(4/9) = 2/3.
Q12. In a right triangle ABC right-angled at B, if AB = 6 cm and BC = 8 cm, then AC is (1 mark)
Answer: 10 cm. AC² = 6² + 8² = 36 + 64 = 100, so AC = 10 cm.
Q13. The distance of the point (−6, 8) from the origin is (1 mark)
Answer: 10 units. √((−6)² + 8²) = √(36 + 64) = √100 = 10.
Q14. The midpoint of the segment joining (2, −3) and (−6, 7) is (1 mark)
Answer: (−2, 2). Midpoint = ((2 + (−6))/2, (−3 + 7)/2) = (−4/2, 4/2) = (−2, 2).
Q15. The length of a tangent drawn from a point 13 cm from the centre of a circle of radius 5 cm is (1 mark)
Answer: 12 cm. The tangent is perpendicular to the radius, so length = √(13² − 5²) = √(169 − 25) = √144 = 12 cm.
Q16. The area of a sector of a circle of radius 7 cm with central angle 60° is (1 mark)
Answer: 77/3 cm² ≈ 25.67 cm². Area = (θ/360)πr² = (60/360)(22/7)(49) = (1/6)(154) = 77/3 cm².
Q17. If the mean of 5, 7, 9, x and 11 is 9, then x is (1 mark)
Answer: 13. The sum must be 5 × 9 = 45, and 5 + 7 + 9 + 11 = 32, so x = 13.
Q18. A die is thrown once. The probability of getting a prime number is (1 mark)
Answer: 1/2. The primes on a die are 2, 3 and 5 — three outcomes out of six, so 3/6 = 1/2.
Q19. Assertion (A): The number 6ⁿ can end with the digit 0 for some natural number n. Reason (R): Any number ending in 0 has both 2 and 5 as prime factors. (1 mark)
Answer: (d) A is false, R is true. 6ⁿ = 2ⁿ·3ⁿ has no factor 5, so it can never end in 0; the reason itself is a correct statement.
Q20. Assertion (A): The points (1, 1), (2, 3) and (3, 5) are collinear. Reason (R): Three points are collinear if the area of the triangle they form is zero. (1 mark)
Answer: (a) Both true and R explains A. Area = ½|1(3−5) + 2(5−1) + 3(1−3)| = ½|−2 + 8 − 6| = 0, so the points are collinear by exactly the criterion in R.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. Prove that 3 + 2√5 is irrational, given that √5 is irrational. (2 marks)
Answer: Suppose 3 + 2√5 is rational, say equal to r. Then √5 = (r − 3)/2. The right side is a difference and quotient of rationals, hence rational, so √5 would be rational — contradicting the given. Therefore 3 + 2√5 is irrational.
Q2. Find the zeroes of the polynomial x² − 2x − 8 and verify the relation between the zeroes and the coefficients. (2 marks)
Answer: x² − 2x − 8 = (x − 4)(x + 2), so the zeroes are 4 and −2. Sum = 4 + (−2) = 2 = −(−2)/1 = −b/a. Product = 4 × (−2) = −8 = −8/1 = c/a. Both relations hold.
Q3. Solve for x and y: x + y = 14 and x − y = 4. (2 marks)
Answer: Adding the two equations gives 2x = 18, so x = 9. Substituting into x + y = 14 gives y = 5. Hence x = 9, y = 5.
Q4. In triangle ABC, the point D lies on AB and the point E lies on AC such that DE is parallel to BC. If AD = 3 cm, DB = 5 cm and AE = 4.5 cm, find EC. (2 marks)
Answer: By the Basic Proportionality Theorem, AD/DB = AE/EC, so 3/5 = 4.5/EC. Cross-multiplying, 3·EC = 22.5 and EC = 7.5 cm.
Q5. A bag contains 5 red, 8 white and 7 black balls. One ball is drawn at random. Find the probability that it is not black. (2 marks)
Answer: Total balls = 5 + 8 + 7 = 20. Balls that are not black = 5 + 8 = 13. P(not black) = 13/20.
Section C — Short Answer (3 marks each) — 18 marks
Answer all 6 questions. Each carries 3 marks.
Q1. Find the HCF and LCM of 510 and 92 by prime factorisation, and verify that HCF × LCM = product of the numbers. (3 marks)
Answer: 510 = 2 × 3 × 5 × 17 and 92 = 2² × 23. The only common prime factor is 2, so HCF = 2. LCM = 2² × 3 × 5 × 17 × 23 = 23460. Check: HCF × LCM = 2 × 23460 = 46920, and 510 × 92 = 46920. Verified.
Q2. Solve the quadratic equation 2x² − 7x + 3 = 0 by the quadratic formula. (3 marks)
Answer: Here a = 2, b = −7, c = 3, so D = 49 − 24 = 25 and √D = 5. x = (7 ± 5)/4, giving x = 12/4 = 3 and x = 2/4 = 1/2. The roots are 3 and 1/2.
Q3. The sum of the first 12 terms of an AP is 246 and its first term is 4. Find the common difference and the 12th term. (3 marks)
Answer: Sₙ = n/2[2a + (n−1)d] gives 246 = 6[8 + 11d], so 8 + 11d = 41, hence 11d = 33 and d = 3. The 12th term is a + 11d = 4 + 33 = 37.
Q4. Prove that (1 + cot²θ)(1 − cos θ)(1 + cos θ) = 1. (3 marks)
Answer: (1 − cos θ)(1 + cos θ) = 1 − cos²θ = sin²θ. Also 1 + cot²θ = cosec²θ = 1/sin²θ. Multiplying, (1/sin²θ)(sin²θ) = 1, as required.
Q5. Find the area of the triangle whose vertices are (1, −1), (−4, 6) and (−3, −5). (3 marks)
Answer: Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| = ½|1(6−(−5)) + (−4)((−5)−(−1)) + (−3)((−1)−6)| = ½|11 + 16 + 21| = ½(48) = 24 square units.
Q6. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use π = 3.14) (3 marks)
Answer: Area of the sector = (90/360)πr² = ¼ × 3.14 × 100 = 78.5 cm². Area of the triangle formed by the two radii = ½ × 10 × 10 = 50 cm². Minor segment = 78.5 − 50 = 28.5 cm².
Section D — Long Answer (5 marks each) — 20 marks
Answer all 4 questions. Each carries 5 marks.
Q1. A motor boat whose speed in still water is 18 km/h takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. (5 marks)
Answer: Let the stream speed be x km/h. Upstream time = 24/(18 − x), downstream time = 24/(18 + x), and the difference is 1 hour: 24/(18 − x) − 24/(18 + x) = 1. This gives 24(18 + x − 18 + x) = (18 − x)(18 + x), i.e. 48x = 324 − x². So x² + 48x − 324 = 0, which factors as (x + 54)(x − 6) = 0. Rejecting the negative root, the stream flows at 6 km/h.
Q2. Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio (Basic Proportionality Theorem). (5 marks)
Answer: In triangle ABC let DE ∥ BC meet AB at D and AC at E. Join BE and CD, and drop perpendiculars EM ⊥ AB and DN ⊥ AC. Now ar(ADE)/ar(BDE) = (½·AD·EM)/(½·DB·EM) = AD/DB, since both triangles share the same height EM. Similarly ar(ADE)/ar(DEC) = AE/EC. But triangles BDE and DEC stand on the same base DE and lie between the same parallels DE and BC, so ar(BDE) = ar(DEC). Hence AD/DB = AE/EC.
Q3. The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 30 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower and the original distance from its foot. (5 marks)
Answer: Let the height be h and the nearer distance be x. From the nearer point, tan 60° = h/x, so h = x√3. From the farther point, tan 30° = h/(x + 30), so h = (x + 30)/√3. Equating, x√3 = (x + 30)/√3, hence 3x = x + 30, so x = 15 m and h = 15√3 ≈ 25.98 m. The original distance was x + 30 = 45 m.
Q4. A solid is in the shape of a cone standing on a hemisphere, both having the same radius 7 cm, and the total height of the solid is 21 cm. Find the volume of the solid. (Use π = 22/7) (5 marks)
Answer: The hemisphere contributes a height equal to its radius, 7 cm, so the cone's height is 21 − 7 = 14 cm. Volume of the hemisphere = (2/3)πr³ = (2/3)(22/7)(343) = 718.67 cm³. Volume of the cone = (1/3)πr²h = (1/3)(22/7)(49)(14) = 718.67 cm³. Total volume ≈ 1437.33 cm³.
Section E — Case-based questions (4 marks each) — 12 marks
Answer all 3 case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — A school is organising a sports day. The seats in the stadium are arranged in rows: the first row has 20 seats, the second 24, the third 28, and so on, each row having 4 more seats than the one before. (i) Which progression do the seat counts form, and what are its first term and common difference? (ii) How many seats are in the 15th row? (iii) How many seats are there in the first 15 rows altogether? (4 marks)
Answer: (i) The counts 20, 24, 28, … form an AP with first term a = 20 and common difference d = 4. (ii) a₁₅ = a + 14d = 20 + 56 = 76 seats. (iii) S₁₅ = 15/2 (2×20 + 14×4) = 15/2 (40 + 56) = 15/2 × 96 = 720 seats.
Q2. CASE STUDY 2 — A student stands 20 m from the base of a vertical flagpole. The angle of elevation of the top of the pole from the student's eye, which is 1.5 m above the ground, is 45°. (i) Sketch the situation and name the right triangle involved. (ii) Find the height of the pole above eye level. (iii) Find the total height of the flagpole. (4 marks)
Answer: (i) The horizontal line from the eye to the pole, the vertical part of the pole above eye level, and the line of sight form a right triangle, right-angled at the pole. (ii) tan 45° = h/20, and tan 45° = 1, so h = 20 m above eye level. (iii) Total height = 20 + 1.5 = 21.5 m.
Q3. CASE STUDY 3 — A survey recorded the daily pocket money (in ₹) of 40 students: 0–10 → 4 students, 10–20 → 8, 20–30 → 12, 30–40 → 10, 40–50 → 6. (i) Which class is the modal class? (ii) Find the mode. (iii) Find the mean pocket money using the direct method. (4 marks)
Answer: (i) The highest frequency is 12, so the modal class is 20–30. (ii) Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 20 + [(12 − 8)/(24 − 8 − 10)] × 10 = 20 + (4/6)(10) = 26.67 (about ₹26.7). (iii) Using midpoints 5, 15, 25, 35, 45: Σfᵢxᵢ = 4(5) + 8(15) + 12(25) + 10(35) + 6(45) = 20 + 120 + 300 + 350 + 270 = 1060. Mean = 1060/40 = ₹26.50.
FAQs
What is the total duration of the Class 10 Maths board exam?
3 hours is the standard duration for the CBSE Class 10 Maths board exam.
How many marks is the board exam out of?
The board exam is out of 80 marks, with internal assessment contributing 20 marks.
Are there optional questions in the exam?
Yes, there are internal choices within some questions, but all sections must be attempted.
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