CBSE Class 10 Mathematics Question Paper 2025 with Solutions

The CBSE Class 10 Mathematics question paper is the ultimate preparation resource for students aiming to score 95+ in their board examinations. Solving the official CBSE Class 10 Mathematics question paper 2025 with solutions gives you direct insight into the revised question pattern, topic-wise weightage, step-wise marking scheme, and the exact difficulty level set by the Central Board of Secondary Education. Whether you are preparing for Mathematics Standard (Subject Code 041) or Mathematics Basic (Subject Code 241), working through this verified previous year question paper CBSE ensures thorough conceptual clarity, enhances problem-solving speed, and eliminates exam-day anxiety.

Question Paper Pattern and Structure

The CBSE Class 10 Mathematics Board Examination 2025 was conducted for a total of 80 marks with an exam duration of 3 hours (180 minutes), along with an additional 15 minutes of mandatory reading time. The passing mark for the theory paper is 33% (26 marks out of 80), alongside internal assessments (20 marks).

The 2025 question paper comprised 38 compulsory questions divided across five structured sections (A to E), incorporating an internal choice of approximately 33% across 2-mark, 3-mark, and 5-mark categories:

Section	Question Type	Number of Questions	Marks per Question	Total Marks
Section A	Multiple Choice Questions (MCQs) & Assertion-Reasoning	20 (Q1 to Q20)	1 Mark	20 Marks
Section B	Short Answer Type I (SA-I)	5 (Q21 to Q25)	2 Marks	10 Marks
Section C	Short Answer Type II (SA-II)	6 (Q26 to Q31)	3 Marks	18 Marks
Section D	Long Answer Type (LA)	4 (Q32 to Q35)	5 Marks	20 Marks
Section E	Case-Based / Source-Integrated Units	3 (Q36 to Q38)	4 Marks	12 Marks
Total		38 Questions	-	80 Marks

Unit-Wise Marks Distribution

Unit No.	Unit Name	Chapters Included	Weightage
I	Number Systems	Real Numbers	06 Marks
II	Algebra	Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions	20 Marks
III	Coordinate Geometry	Coordinate Geometry (Distance & Section Formulae)	06 Marks
IV	Geometry	Triangles, Circles	15 Marks
V	Trigonometry	Introduction to Trigonometry, Trigonometric Identities, Some Applications of Trigonometry	12 Marks
VI	Mensuration	Areas Related to Circles, Surface Areas and Volumes	10 Marks
VII	Statistics & Probability	Statistics, Probability	11 Marks
Grand Total			80 Marks

CBSE Class 10 Mathematics 2025 Solved Questions and Step-by-Step Solutions

Below are representative questions directly from the 2025 Class 10 Mathematics board examination, complete with detailed mathematical workings, step allocations, and final answers aligned with the official answer key.

Section A: 1-Mark Questions (MCQs & Assertion-Reason)

Question 1 (Real Numbers)

If two positive integers a and b are written as a = x³y² and b = xy³, where x and y are prime numbers, then HCF(a, b) is:

• (A) xy
• (B) xy²
• (C) x³y³
• (D) x²y²

Solution:

1. The highest common factor (HCF) of two numbers is the product of the smallest powers of each common prime factor involved.
2. Common prime factor x has the lowest exponent 1 (from x¹).
3. Common prime factor y has the lowest exponent 2 (from y²).
4. Therefore, HCF(a, b) = xy².

Correct Option: (B) xy²

Question 2 (Coordinate Geometry)

The point on the x-axis which is equidistant from points A(-1, 0) and B(5, 0) is:

• (A) (2, 0)
• (B) (0, 2)
• (C) (3, 0)
• (D) (-2, 0)

Solution:

1. Let the required point on the x-axis be P(x, 0).
2. Since P is equidistant from A and B, PA² = PB².
3. (x - (-1))² + (0 - 0)² = (x - 5)² + (0 - 0)²
4. (x + 1)² = (x - 5)²
5. x² + 2x + 1 = x² - 10x + 25
6. 12x = 24 ⇒ x = 2.

Correct Option: (A) (2, 0)

Question 3 (Statistics)

For a given frequency distribution, if the Mean is 27 and the Median is 30, then using the empirical relationship, the Mode is:

• (A) 33
• (B) 36
• (C) 30
• (D) 24

Solution:

1. The empirical relationship between mean, median, and mode is: Mode = 3 × Median - 2 × Mean.
2. Substituting the given values: Mode = 3(30) - 2(27)
3. Mode = 90 - 54 = 36.

Correct Option: (B) 36

Question 4 (Assertion-Reason)

Assertion (A): The HCF of two numbers is 5 and their product is 150, then their LCM is 30.
Reason (R): For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.

• (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
• (B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
• (C) (A) is true but (R) is false.
• (D) (A) is false but (R) is true.

Solution:

1. By formula: HCF × LCM = a × b. Thus, Reason (R) is true.
2. Substituting values: 5 × LCM = 150 ⇒ LCM = 150 / 5 = 30. Thus, Assertion (A) is true.
3. Reason (R) correctly explains Assertion (A).

Correct Option: (A)

Section B: 2-Mark Short Answer Questions

Question 5 (Real Numbers)

Prove that 3 + 2√5 is an irrational number, given that √5 is irrational.

Step-by-Step Solution:

1. Assumption: Let us assume to the contrary that 3 + 2√5 is rational. [Mark: ½]
2. Therefore, we can find co-prime integers a and b (where b ≠ 0) such that:
   3 + 2√5 = a / b [Mark: ½]
3. Rearranging the terms:
   2√5 = (a / b) - 3 = (a - 3b) / b
   √5 = (a - 3b) / (2b) [Mark: ½]
4. Since a and b are integers, (a - 3b) / (2b) is a rational number. This implies that √5 is rational.
5. However, this contradicts the given fact that √5 is irrational.
6. Hence, our assumption is incorrect, and 3 + 2√5 is irrational. [Mark: ½]

Question 6 (Arithmetic Progressions)

Find the 20^th term from the last term of the AP: 3, 8, 13, ..., 253.

Step-by-Step Solution:

1. Given AP: 3, 8, 13, ..., 253. Common difference d = 8 - 3 = 5.
2. Writing the AP in reverse order: 253, 248, 243, ..., 3. [Mark: ½]
3. For the reversed AP:
   • First term (a') = 253
   • Common difference (d') = -5
   [Mark: ½]
4. Using formula for n^th term: a_n = a' + (n - 1)d' [Mark: ½]
5. a₂₀ = 253 + (20 - 1)(-5)
6. a₂₀ = 253 + 19(-5) = 253 - 95 = 158. [Mark: ½]

Section C: 3-Mark Short Answer Questions

Question 7 (Trigonometry)

Prove that: (sin θ - 2 sin³ θ) / (2 cos³ θ - cos θ) = tan θ

Step-by-Step Solution:

1. Left Hand Side (LHS):
   LHS = (sin θ - 2 sin³ θ) / (2 cos³ θ - cos θ) [Mark: ½]
2. Factoring out sin θ from numerator and cos θ from denominator:
   LHS = [sin θ (1 - 2 sin² θ)] / [cos θ (2 cos² θ - 1)] [Mark: 1]
3. Expressing sin² θ in terms of cos² θ using identity sin² θ = 1 - cos² θ:
   Numerator factor = 1 - 2(1 - cos² θ) = 1 - 2 + 2 cos² θ = 2 cos² θ - 1 [Mark: 1]
4. Substituting back into expression:
   LHS = [sin θ (2 cos² θ - 1)] / [cos θ (2 cos² θ - 1)]
   LHS = sin θ / cos θ = tan θ = RHS.
5. Hence Proved. [Mark: ½]

Question 8 (Circles)

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Step-by-Step Solution:

1. Given: A circle with center O. A point P lies outside the circle. PQ and PR are two tangents drawn from P touching the circle at Q and R respectively. [Mark: ½]
2. To Prove: PQ = PR.
3. Construction: Join OP, OQ, and OR. [Mark: ½]
4. Proof:
   • In ΔOQP and ΔORP:
   • ∠OQP = ∠ORP = 90° (Tangent at any point of a circle is perpendicular to the radius through the point of contact). [Mark: ½]
   • OQ = OR (Radii of the same circle). [Mark: ½]
   • OP = OP (Common hypotenuse). [Mark: ½]
   • ∴ ΔOQP ≅ ΔORP (By RHS congruence criterion).
   • ⇒ PQ = PR (By Corresponding Parts of Congruent Triangles - CPCT). [Mark: ½]

Section D: 5-Mark Long Answer Questions

Question 9 (Applications of Trigonometry)

From the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use √3 = 1.732)

Step-by-Step Solution:

1. Diagram and Definition:
   Let AB be the lighthouse of height h = 75 m. Let C and D be the positions of the two ships on the sea level, such that ∠ADB = 30° and ∠ACB = 45°. [Mark: 1]
2. In right-angled triangle ΔABC:
   tan 45° = AB / BC
   1 = 75 / BC ⇒ BC = 75 m [Mark: 1]
3. In right-angled triangle ΔABD:
   tan 30° = AB / BD
   1 / √3 = 75 / BD ⇒ BD = 75√3 m [Mark: 1 ½]
4. The distance between the two ships is CD = BD - BC:
   CD = 75√3 - 75 = 75(√3 - 1) m [Mark: 1]
5. Substituting √3 = 1.732:
   CD = 75(1.732 - 1) = 75 × 0.732 = 54.9 m. [Mark: ½]

Final Answer: The distance between the two ships is 54.9 meters.

Question 10 (Surface Areas and Volumes)

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. (Take π = 3.14)

Step-by-Step Solution:

1. Given Parameters:
   • Radius of hemisphere and cone (r) = Diameter / 2 = 8 / 2 = 4 cm.
   • Height of cone (h) = 4 cm.
   [Mark: 1]
2. Formulas:
   • Volume of hemisphere = (2/3)πr³
   • Volume of cone = (1/3)πr²h
   [Mark: 1]
3. Total Volume of Toy (V):
   V = (1/3)πr²h + (2/3)πr³
   V = (1/3)πr²(h + 2r) [Mark: 1]
4. Substituting numerical values:
   V = (1/3) × 3.14 × (4)² × [4 + 2(4)]
   V = (1/3) × 3.14 × 16 × (4 + 8)
   V = (1/3) × 3.14 × 16 × 12 [Mark: 1]
5. Simplifying:
   V = 3.14 × 16 × 4 = 3.14 × 64 = 200.96 cm³. [Mark: 1]

Final Answer: The total volume of the solid toy is 200.96 cm³.

Section E: 4-Mark Case-Based Study

Question 11 (Case Study: Coordinate Geometry in Architecture)

A sports complex designer is mapping out a new running track on a Cartesian plane where 1 grid unit represents 10 meters. The main pavilion is situated at point A(2, 3), the medical emergency station is at point B(6, 7), and the water refreshment booth is located at point C(8, 9).

1. Find the direct straight-line distance between the main pavilion (A) and the medical station (B). [1 Mark]
2. Check whether points A, B, and C lie in a straight line (collinear). [1 Mark]
3. Find the coordinates of point M which divides the line segment joining A(2, 3) and B(6, 7) internally in the ratio 1:3. [2 Marks]

Solution:

1. Distance AB:
   Using distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
   AB = √[(6 - 2)² + (7 - 3)²] = √[4² + 4²] = √[16 + 16] = √32 = 4√2 units.
   Actual distance = 4√2 × 10 = 40√2 meters (or approx. 56.57 m). [1 Mark]
2. Collinearity Check:
   Slope of AB = (7 - 3) / (6 - 2) = 4 / 4 = 1.
   Slope of BC = (9 - 7) / (8 - 6) = 2 / 2 = 1.
   Since Slope(AB) = Slope(BC) and B is a common point, points A, B, and C are collinear (lie on the same straight line). [1 Mark]
3. Section Formula Coordinates:
   Using section formula for ratio m:n = 1:3 with A(2, 3) and B(6, 7):
   x = (mx₂ + nx₁) / (m + n) = (1 × 6 + 3 × 2) / (1 + 3) = (6 + 6) / 4 = 12 / 4 = 3.
   y = (my₂ + ny₁) / (m + n) = (1 × 7 + 3 × 3) / (1 + 3) = (7 + 9) / 4 = 16 / 4 = 4.
   Therefore, the coordinates of point M are (3, 4). [2 Marks]

Difficulty Analysis of CBSE Class 10 Mathematics 2025

A granular difficulty analysis reveals that the 2025 Mathematics paper maintained a balanced blend of standard textbook questions and higher-order thinking skill (HOTS) questions:

Section	Difficulty Rating	Key Observations
Section A (MCQs)	Moderate	Direct formula checks alongside 3 conceptual questions in Coordinate Geometry and Trigonometry. Assertion-Reasoning required careful reading of definitions.
Section B (2 Marks)	Easy to Moderate	Direct NCERT textbook derivations (Irrationality proof, basic AP terms, and circle chord properties).
Section C (3 Marks)	Moderate	Required rigorous algebraic manipulation for trigonometric proofs and calculation precision in grouped mean/median calculations.
Section D (5 Marks)	Moderate to Challenging	Theorems (Basic Proportionality Theorem) were predictable, while multi-step heights & distances and combination surface volume questions demanded high geometric visualization.
Section E (Case-Based)	Moderate	Practical, real-world context scenarios testing core concepts of Arithmetic Progressions and Coordinate Geometry.

CBSE Step-Wise Marking Scheme Insights

CBSE evaluators use a rigorous step-marking system where every stage of mathematical derivation carries dedicated weightage. Understanding this prevents loss of avoidable marks:

• Formula Declaration (½ Mark): Always state the general formula before substituting values (e.g., writing d = √[(x₂ - x₁)² + (y₂ - y₁)²] or Volume of Cone = (1/3)πr²h).
• Geometric Reasoning (½ to 1 Mark): In circle and triangle proofs, missing standard reasons (such as "Radii of the same circle" or "Tangents from an external point are equal") results in direct deduction even if the final result is correct.
• Units and Dimensional Notation (½ Mark): Always conclude mensuration and heights & distances answers with appropriate SI units (cm, m, cm², m³).
• Final Concluding Statement: Always underline or box your final numerical answer for high visibility during examiner evaluation.

How to Use This Question Paper for Exam Prep

1. Timed 3-Hour Simulation: Print the question paper and attempt it under strict exam hall conditions (10:30 AM to 1:30 PM) without checking solutions in between.
2. Master the 15-Minute Reading Period: Use the initial 15 minutes exclusively to read Sections D and E, selecting which internal choice questions you can execute with 100% precision.
3. Self-Audit with the Solution Key: Grade your answer script strictly against the step-marking rubric provided above to identify calculation errors or skipped working steps.
4. Target Weak Conceptual Nodes: If you lost marks in trigonometric identities or surface areas, revisit NCERT exemplar problems for targeted remediation.

Download More Practice Papers and CBSE Class 10 Solutions

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