CBSE Class 10 Maths Sample Paper 2027 with Solutions PDF — Complete Guide

In short: CBSE Class 10 Maths Sample Paper 2027 follows the latest pattern — 80 marks, 5 sections (A-E), 38 questions, 3 hours. Section A has 20 MCQs (1 mark each), Section B has 5 questions (2 marks each), Section C has 6 questions (3 marks each), Section D has 4 questions (5 marks each), and Section E has 3 case-based questions (4 marks each). This guide includes complete solutions, marking scheme breakdown, and strategies to maximize your score.

CBSE Class 10 Maths Sample Paper 2027 — Paper Pattern

Exam Structure

Section-wise Breakdown

Chapter-wise Weightage 2027

High-Weightage Chapters (Focus Areas)

Section A — MCQs (20 × 1 = 20 Marks)

Sample Questions with Solutions

Q1. If HCF(306, 657) = 9, then LCM(306, 657) is:

• (a) 22338
• (b) 23238
• (c) 23__(d) 22__(Correct: a)

Solution:

HCF × LCM = Product of numbers
9 × LCM = 306 × 657
LCM = (306 × 657) / 9
LCM = 201042 / 9 = 22338

Answer: (a) 22338

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Q2. The zeroes of the polynomial x² – 3x – 4 are:

• (a) 4, –1
• (b) –4, 1
• (c) 4, 1
• (d) –4, –1

Solution:

x² – 3x – 4 = 0
x² – 4x + x – 4 = 0
x(x – 4) + 1(x – 4) = 0
(x + 1)(x – 4) = 0
x = –1 or x = 4

Answer: (a) 4, –1

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Q3. If the distance between points (2, –2) and (–1, x) is 5, one value of x is:

• (a) 2
• (b) –2
• (c) 1
• (d) –1

Solution:

Distance = √[(x₂–x₁)² + (y₂–y₁)²]
5 = √[(–1–2)² + (x+2)²]
25 = 9 + (x+2)²
(x+2)² = 16
x + 2 = ±4
x = 2 or x = –6

Answer: (a) 2

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Q4. If tan θ = 5/12, then (sin θ + cos θ) / (sin θ – cos θ) equals:

• (a) –17/7
• (b) 17/7
• (c) 7/17
• (d) –7/17

Solution:

tan θ = 5/12 means opposite = 5, adjacent = 12
Hypotenuse = √(25 + 144) = √169 = 13
sin θ = 5/13, cos θ = 12/13

(sin θ + cos θ)/(sin θ – cos θ)
= (5/13 + 12/13)/(5/13 – 12/13)
= (17/13)/(–7/13)
= –17/7

Answer: (a) –17/7

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Q5. The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs is:

• (a) 7
• (b) 14
• (c) 21
• (d) 28

Solution:

Number of bad eggs = Probability × Total eggs
= 0.035 × 400
= 14

Answer: (b) 14

Section B — Very Short Answer (5 × 2 = 10 Marks)

Sample Questions with Solutions

Q21. Find the value of k for which the system of equations kx + 3y = k – 3 and 12x + ky = k has no solution.

Solution: For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

k/12 = 3/k ≠ (k–3)/k

From k/12 = 3/k: k² = 36 k = ±6

Check k = 6: 6/12 = 3/6 = 1/2, (6–3)/6 = 1/2 (Equal, so infinite solutions) Check k = –6: –6/12 = 3/–6 = –1/2, (–6–3)/–6 = 3/2 (Not equal ✓)

Answer: k = –6

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Q22. Find the sum of first 20 terms of an AP whose nth term is given by aₙ = 3 – 4n.

Solution: a₁ = 3 – 4(1) = –1 a₂₀ = 3 – 4(20) = 3 – 80 = –77

Sum = n/2 × (first term + last term) S₂₀ = 20/2 × (–1 + (–77)) S₂₀ = 10 × (–78) S₂₀ = –780

Answer: S₂₀ = –780

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Q23. In the figure, DE ∥ BC. If AD = 2.4 cm, DB = 3.6 cm, and AC = 5 cm, find AE.

Solution: By Basic Proportionality Theorem (Thales Theorem): AD/DB = AE/EC

Let AE = x, then EC = 5 – x

2.4/3.6 = x/(5 – x) 2.4(5 – x) = 3.6x 12 – 2.4x = 3.6x 12 = 6x x = 2

Answer: AE = 2 cm

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Q24. Find the value of sin 60° cos 30° + sin 30° cos 60°.

Solution: sin 60° = √3/2, cos 30° = √3/2 sin 30° = 1/2, cos 60° = 1/2

sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 4/4 = 1

Answer: 1

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Q25. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls.

Solution: Let blue balls = x Total balls = 5 + x

P(red) = 5/(5+x) P(blue) = x/(5+x)

Given: P(blue) = 2 × P(red) x/(5+x) = 2 × 5/(5+x) x = 10

Answer: 10 blue balls

Section C — Short Answer (6 × 3 = 18 Marks)

Sample Questions with Solutions

Q26. Prove that √5 is irrational.

Solution: Assume √5 is rational. Then √5 = p/q where p, q are co-prime integers and q ≠ 0.

Squaring both sides: 5 = p²/q² p² = 5q²

This means p² is divisible by 5. Therefore, p is divisible by 5. (If p² is divisible by prime, p is divisible by that prime)

Let p = 5m for some integer m. (5m)² = 5q² 25m² = 5q² q² = 5m²

This means q² is divisible by 5. Therefore, q is divisible by 5.

But p and q both divisible by 5 contradicts that they are co-prime. Hence, our assumption is wrong.

Therefore, √5 is irrational. ∎

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Q27. Find the zeroes of the polynomial p(x) = 6x² – 3 – 7x and verify the relationship between zeroes and coefficients.

Solution: p(x) = 6x² – 7x – 3 = 0

Using factorization: 6x² – 9x + 2x – 3 = 0 3x(2x – 3) + 1(2x – 3) = 0 (3x + 1)(2x – 3) = 0

x = –1/3 or x = 3/2

Zeroes: α = –1/3, β = 3/2

Verification:

• Sum of zeroes = α + β = –1/3 + 3/2 = (–2 + 9)/6 = 7/6

• From polynomial: –b/a = –(–7)/6 = 7/6 ✓

• Product of zeroes = αβ = (–1/3)(3/2) = –1/2

• From polynomial: c/a = –3/6 = –1/2 ✓

Verified.

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Q28. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits are reversed. Find the number.

Solution: Let tens digit = x, units digit = y Number = 10x + y

Given:

1. x + y = 9 ... (i)
2. 10x + y + 27 = 10y + x ... (ii)

From (ii): 9x – 9y = –27 x – y = –3 ... (iii)

Adding (i) and (iii): 2x = 6 x = 3

From (i): y = 9 – 3 = 6

Answer: The number is 36

Verification: 36 + 27 = 63 (digits reversed) ✓

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Q29. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution: Given: A circle with center O, tangent XY at point P.

To Prove: OP ⊥ XY

Proof: Take any point Q on XY other than P. Q lies outside the circle (tangent touches at only one point). Therefore, OQ > OP (OP is radius, OQ is distance to external point)

This is true for every point Q on XY except P. Therefore, OP is the shortest distance from O to line XY.

The shortest distance from a point to a line is the perpendicular distance.

Hence, OP ⊥ XY. ∎

Section D — Long Answer (4 × 5 = 20 Marks)

Sample Questions with Solutions

Q32. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution: Let speed of train = x km/h Time taken = 360/x hours

New speed = (x + 5) km/h New time = 360/(x + 5) hours

Given: Original time – New time = 1 hour 360/x – 360/(x + 5) = 1

360(x + 5) – 360x = x(x + 5) 360x + 1800 – 360x = x² + 5x 1800 = x² + 5x x² + 5x – 1800 = 0

Using quadratic formula: x = (–5 ± √(25 + 7200))/2 x = (–5 ± √7225)/2 x = (–5 ± 85)/2

x = 80/2 = 40 or x = –90/2 = –45 (rejected, speed can't be negative)

Answer: Speed of train = 40 km/h

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Q33. State and prove Basic Proportionality Theorem (Thales Theorem).

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides those sides in the same ratio.

Given: In △ABC, DE ∥ BC where D is on AB and E is on AC.

To Prove: AD/DB = AE/EC

Construction: Join BE and CD. Draw EM ⊥ AB and DN ⊥ AC.

Proof: Area of △ADE = (1/2) × AD × EM Area of △BDE = (1/2) × DB × EM

ar(△ADE)/ar(△BDE) = AD/DB ... (i)

Similarly, Area of △ADE = (1/2) × AE × DN Area of △CDE = (1/2) × EC × DN

ar(△ADE)/ar(△CDE) = AE/EC ... (ii)

Now, △BDE and △CDE are on the same base DE and between the same parallels DE and BC. Therefore, ar(△BDE) = ar(△CDE) ... (iii)

From (i), (ii), and (iii): AD/DB = AE/EC

Hence proved. ∎

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Q34. The median of the following data is 525. Find the values of x and y if the total frequency is 100.

Solution: Total frequency = 100 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100 76 + x + y = 100 x + y = 24 ... (i)

Median = 525, so median class is 500-600. n/2 = 50

Cumulative frequency before median class = 2 + 5 + x + 12 + 17 = 36 + x

Median = l + [(n/2 – cf)/f] × h 525 = 500 + [(50 – (36 + x))/20] × 100 25 = [(14 – x)/20] × 100 25 = 5(14 – x) 5 = 14 – x x = 9

From (i): y = 24 – 9 = 15

Answer: x = 9, y = 15

Section E — Case-Based Questions (3 × 4 = 12 Marks)

Case Study 1: Trigonometry in Real Life

A lighthouse is located on a cliff. A ship is sailing towards the cliff. The angle of depression from the top of the lighthouse to the ship changes as the ship approaches.

Q36. The height of the lighthouse above sea level is 100 m. When the ship is at point A, the angle of depression is 30°. When the ship reaches point B, the angle of depression is 60°.

(i) Find the distance of the ship from the base of the cliff when at point A.

Solution: tan 30° = 100/distance at A 1/√3 = 100/d₁ d₁ = 100√3 m ≈ 173.2 m

Answer: 100√3 m or 173.2 m

(ii) Find the distance of the ship from the base of the cliff when at point B.

Solution: tan 60° = 100/distance at B √3 = 100/d₂ d₂ = 100/√3 = 100√3/3 m ≈ 57.7 m

Answer: 100√3/3 m or 57.7 m

(iii) Find the distance traveled by the ship from A to B.

Solution: Distance AB = d₁ – d₂ = 100√3 – 100√3/3 = 100√3(1 – 1/3) = 100√3 × 2/3 = 200√3/3 m ≈ 115.5 m

Answer: 200√3/3 m or 115.5 m

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Case Study 2: Statistics in Sports

A school conducted a sports event. The following table shows the number of students who scored different ranges of points:

Q37.

(i) Find the modal class.

Solution: Modal class has highest frequency. Highest frequency = 29 (for class 60-80)

Answer: Modal class is 60-80

(ii) Find the mode.

Solution: Mode = l + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

l = 60, f₁ = 29, f₀ = 21, f₂ = 17, h = 20

Mode = 60 + [(29 – 21)/(2×29 – 21 – 17)] × 20 Mode = 60 + [8/(58 – 38)] × 20 Mode = 60 + [8/20] × 20 Mode = 60 + 8 = 68

Answer: Mode = 68 points

(iii) Find the mean using direct method.

Solution:

Mean = Σfᵢxᵢ / Σfᵢ = 5300/100 = 53

Answer: Mean = 53 points

Marking Scheme — How Examiners Award Marks

Section-wise Marking Guidelines

Common Mistakes That Cost Marks

How to Score 95%+ in CBSE Class 10 Maths

Time Management Strategy

Chapter-wise Preparation Priority

Last 30 Days Strategy

Download Official Sample Paper

CBSE releases official sample papers on their academic website:

Note: Always practice with the latest official sample paper released by CBSE.

Frequently Asked Questions

Is this sample paper based on the latest CBSE pattern?

Yes. This follows the 2024-25 pattern which continues for 2027 — 5 sections, 38 questions, 80 marks theory, case-based questions in Section E.

How many sample papers should I solve before the exam?

Solve at least 10-15 sample papers. Start with untimed practice, then move to timed practice (3 hours strict).

Are case-based questions difficult?

Case-based questions test application of concepts. They're not difficult if you understand the underlying concepts. Practice 20-30 case studies before the exam.

Should I attempt all questions?

Yes. There's no negative marking. Attempt every question. Even partial answers get partial marks.

How important are diagrams?

Very important. Diagrams carry 1-2 marks in geometry questions. Always draw neat, labeled diagrams.

What if I forget a formula during the exam?

Write what you remember and attempt the question. Partial marks are given for correct method even if the formula has minor errors.

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