CBSE Class 10 Mathematics Sample Paper 2025 Standard Set 1 with Solutions

Practicing this complete CBSE Class 10 Mathematics Sample Paper with step-by-step solutions is essential for mastering the latest board examination pattern, understanding the blueprint, and scoring 95+ marks. Designed by expert CBSE educators in strict alignment with the latest NCERT curriculum and CBSE Class 10 Mathematics (Standard - Subject Code 041) guidelines, this paper covers every critical chapter from Real Numbers to Statistics and Probability.

CBSE Class 10 Mathematics Paper Pattern and Marking Scheme

The CBSE Class 10 Mathematics board exam comprises an 80-mark theory paper with a duration of 3 hours, along with 20 marks allocated for internal assessment. The theory question paper contains 38 compulsory questions divided into five distinct sections:

Section	Question Numbers	Question Type	Marks per Question	Total Marks
Section A	Q1 – Q20	Multiple Choice Questions (including 2 Assertion-Reason)	1 Mark	20 Marks
Section B	Q21 – Q25	Very Short Answer Type (VSA)	2 Marks	10 Marks
Section C	Q26 – Q31	Short Answer Type (SA)	3 Marks	18 Marks
Section D	Q32 – Q35	Long Answer Type (LA)	5 Marks	20 Marks
Section E	Q36 – Q38	Case-Based / Source-Based Integrated Units	4 Marks	12 Marks
Total	38 Questions	Entire Syllabus	—	80 Marks

There is no overall choice in the question paper; however, internal choices are provided in 2 questions of Section B, 2 questions of Section C, 2 questions of Section D, and in the 2-mark sub-questions of 2 questions of Section E.

Complete CBSE Class 10 Mathematics Sample Paper with Step-by-Step Solutions

Section A: Multiple Choice Questions (1 Mark Each)

Q1. If two positive integers a and b are written as a = x³y² and b = xy³, where x, y are prime numbers, then HCF(a, b) is:
(a) xy
(b) xy²
(c) x³y³
(d) x²y²

Solution:
Given a = x³ × y² and b = x¹ × y³.
To find HCF(a, b), take the product of the smallest power of each common prime factor involved in the numbers.
Smallest power of x = x¹
Smallest power of y = y²
Therefore, HCF(a, b) = x × y² = xy².
Correct Option: (b)

Q2. If α and β are the zeroes of the quadratic polynomial p(x) = x² − 5x + 6, then the value of α⁻¹ + β⁻¹ is:
(a) 5/6
(b) −5/6
(c) 6/5
(d) −6/5

Solution:
For the polynomial p(x) = x² − 5x + 6, a = 1, b = −5, c = 6.
Sum of zeroes: α + β = −b/a = −(−5)/1 = 5
Product of zeroes: αβ = c/a = 6/1 = 6
Now, α⁻¹ + β⁻¹ = (1/α) + (1/β) = (α + β) / (αβ) = 5/6.
Correct Option: (a)

Q3. The pair of linear equations 2x + 3y = 7 and 4x + 6y = 15 represents two lines which are:
(a) Intersecting at exactly one point
(b) Coincident
(c) Parallel
(d) Perpendicular

Solution:
Comparing with standard forms a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = 3/6 = 1/2
c₁/c₂ = 7/15
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the system represents an inconsistent pair with parallel lines having no common solution.
Correct Option: (c)

Q4. If the common difference of an Arithmetic Progression is 5, then what is the value of a₁₈ − a₁₃?
(a) 5
(b) 20
(c) 25
(d) 30

Solution:
For an AP with first term a and common difference d = 5:
a₁₈ = a + 17d
a₁₃ = a + 12d
a₁₈ − a₁₃ = (a + 17d) − (a + 12d) = 5d = 5(5) = 25.
Correct Option: (c)

Q5. The distance of the point P(−6, 8) from the origin is:
(a) 8 units
(b) 2√7 units
(c) 10 units
(d) 6 units

Solution:
Distance of any point (x, y) from the origin (0, 0) is given by √(x² + y²).
Distance OP = √[(−6)² + (8)²] = √(36 + 64) = √100 = 10 units.
Correct Option: (c)

Q6. If ΔABC ~ ΔPQR with BC/QR = 1/3, then Area(ΔPRQ) / Area(ΔBCA) is equal to:
(a) 9
(b) 3
(c) 1/3
(d) 1/9

Solution:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Area(ΔPRQ) / Area(ΔBCA) = (QR / BC)² = (3 / 1)² = 9.
Correct Option: (a)

Q7. If sin θ + sin² θ = 1, then the value of cos² θ + cos⁴ θ is:
(a) −1
(b) 0
(c) 1
(d) 2

Solution:
Given: sin θ + sin² θ = 1 ⇒ sin θ = 1 − sin² θ = cos² θ.
Squaring both sides gives sin² θ = cos⁴ θ.
Substitute into the expression: cos² θ + cos⁴ θ = sin θ + sin² θ = 1.
Correct Option: (c)

Q8. A card is drawn at random from a well-shuffled pack of 52 playing cards. The probability of getting a red face card is:
(a) 3/13
(b) 3/26
(c) 1/26
(d) 3/52

Solution:
Total number of possible outcomes = 52.
Total number of red cards = 26 (Hearts and Diamonds).
Face cards per suit = 3 (Jack, Queen, King).
Total red face cards = 2 × 3 = 6.
Probability = Number of favorable outcomes / Total outcomes = 6/52 = 3/26.
Correct Option: (b)

Q9 (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:
By property: HCF × LCM = Product of numbers ⇒ 5 × LCM = 150 ⇒ LCM = 150/5 = 30.
Both Assertion and Reason are true, and Reason is the exact formula used to verify the Assertion.
Correct Option: (a)

Q10 (Assertion & Reason).
Assertion (A): The point (0, 4) lies on the y-axis.
Reason (R): The x-coordinate of any point lying on the y-axis is zero.
(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:
Any point with x = 0 has the coordinate format (0, y) and strictly lies on the y-axis. Hence, both statements are true and Reason directly explains the Assertion.
Correct Option: (a)

Section B: Very Short Answer Questions (2 Marks Each)

Q11. Prove that √5 is an irrational number.

Solution:
Let us assume to the contrary that √5 is a rational number.
Therefore, √5 = a/b, where a and b are co-prime integers and b ≠ 0.
Rearranging, we get a = √5 b.
Squaring both sides:
a² = 5b² …… (Equation 1)
This means 5 divides a². By Theorem (Fundamental Theorem of Arithmetic), 5 also divides a.
Let a = 5c for some integer c.
Substituting a = 5c in Equation 1:
(5c)² = 5b² ⇒ 25c² = 5b² ⇒ b² = 5c².
This implies that 5 divides b², and therefore 5 divides b.
Thus, 5 is a common factor of both a and b.
This contradicts the fact that a and b are co-prime.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
Hence, √5 is irrational. (Proved)

Q12. Find the coordinates of the point P which divides the line segment joining the points A(4, −3) and B(8, 5) in the ratio 3 : 1 internally.

Solution:
Let the coordinates of point P be (x, y).
Here, (x₁, y₁) = (4, −3), (x₂, y₂) = (8, 5), and m₁ : m₂ = 3 : 1.
Applying the section formula:
x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
x = [3(8) + 1(4)] / (3 + 1) = (24 + 4) / 4 = 28 / 4 = 7
y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
y = [3(5) + 1(−3)] / (3 + 1) = (15 − 3) / 4 = 12 / 4 = 3
Therefore, the required coordinates of point P are (7, 3).

Q13. Evaluate: [5 cos² 60° + 4 sec² 30° − tan² 45°] / [sin² 30° + cos² 30°]

Solution:
Using standard trigonometric ratios:
cos 60° = 1/2, sec 30° = 2/√3, tan 45° = 1, sin 30° = 1/2, cos 30° = √3/2.
Denominator: sin² 30° + cos² 30° = (1/2)² + (√3/2)² = 1/4 + 3/4 = 1 (or by identity sin² θ + cos² θ = 1).
Numerator:
5(1/2)² + 4(2/√3)² − (1)² = 5(1/4) + 4(4/3) − 1
= 5/4 + 16/3 − 1
Taking LCM of 4 and 3 (LCM = 12):
= [15 + 64 − 12] / 12 = 67/12.
Hence, the value of the expression is 67/12.

Section C: Short Answer Questions (3 Marks Each)

Q14. Find the zeroes of the quadratic polynomial f(x) = 6x² − 3 − 7x and verify the relationship between the zeroes and its coefficients.

Solution:
Rewriting the polynomial in standard form ax² + bx + c:
f(x) = 6x² − 7x − 3
To find zeroes, set f(x) = 0:
6x² − 9x + 2x − 3 = 0
3x(2x − 3) + 1(2x − 3) = 0
(2x − 3)(3x + 1) = 0
⇒ 2x − 3 = 0 or 3x + 1 = 0
⇒ x = 3/2 or x = −1/3.
Let α = 3/2 and β = −1/3.
Verification:
Here, a = 6, b = −7, c = −3.
1. Sum of zeroes: α + β = 3/2 + (−1/3) = (9 − 2)/6 = 7/6
Formula value: −b/a = −(−7)/6 = 7/6
Hence, Sum of zeroes = −b/a (Verified).
2. Product of zeroes: αβ = (3/2) × (−1/3) = −3/6 = −1/2
Formula value: c/a = −3/6 = −1/2
Hence, Product of zeroes = c/a (Verified).

Q15. Prove that the lengths of tangents drawn from an external point to a circle are equal.

Solution:
Given: A circle with center O, an external point P, and two tangents PQ and PR touching the circle at points Q and R respectively.
To Prove: PQ = PR.
Construction: Join OP, OQ, and OR.
Proof:
We know that a tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OQP = 90° and ∠ORP = 90°.
In right triangles ΔOQP and ΔORP:
1. ∠OQP = ∠ORP = 90° (Radius perpendicular to tangent)
2. OP = OP (Common hypotenuse)
3. OQ = OR (Radii of the same circle)
By RHS (Right angle-Hypotenuse-Side) congruence criterion:
ΔOQP ≅ ΔORP.
Therefore, by CPCT (Corresponding Parts of Congruent Triangles):
PQ = PR. (Hence Proved)

Q16. The sum of the 4th and 8th terms of an AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:
Let a be the first term and d be the common difference of the AP.
The n-th term is a_n = a + (n − 1)d.
Condition 1: a₄ + a₈ = 24
(a + 3d) + (a + 7d) = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 …… (Equation 1)
Condition 2: a₆ + a₁₀ = 44
(a + 5d) + (a + 9d) = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22 …… (Equation 2)
Subtracting Equation 1 from Equation 2:
(a + 7d) − (a + 5d) = 22 − 12
2d = 10 ⇒ d = 5.
Substitute d = 5 into Equation 1:
a + 5(5) = 12 ⇒ a + 25 = 12 ⇒ a = −13.
First term: a₁ = −13
Second term: a₂ = a + d = −13 + 5 = −8
Third term: a₃ = a + 2d = −13 + 10 = −3
Therefore, the first three terms of the AP are −13, −8, −3.

Section D: Long Answer Questions (5 Marks Each)

Q17. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:
Let the speed of the stream be x km/h.
Given speed of the motor boat in still water = 18 km/h.
Speed of the boat upstream = (18 − x) km/h.
Speed of the boat downstream = (18 + x) km/h.
Distance = 24 km.
Time taken upstream (T₁) = Distance / Speed = 24 / (18 − x) hours.
Time taken downstream (T₂) = Distance / Speed = 24 / (18 + x) hours.
According to the question, T₁ − T₂ = 1 hour.
[24 / (18 − x)] − [24 / (18 + x)] = 1
24 × [ (18 + x − (18 − x)) / ((18 − x)(18 + x)) ] = 1
24 × [ (18 + x − 18 + x) / (324 − x²) ] = 1
24(2x) = 324 − x²
48x = 324 − x²
Rearranging into standard quadratic equation form:
x² + 48x − 324 = 0.
Solving by splitting the middle term (since 54 × (−6) = −324 and 54 − 6 = 48):
x² + 54x − 6x − 324 = 0
x(x + 54) − 6(x + 54) = 0
(x + 54)(x − 6) = 0
⇒ x = 6 or x = −54.
Since speed cannot be negative, we reject x = −54.
Therefore, the speed of the stream is 6 km/h.

Q18. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the total height of the tower.

Solution:
Let AB be the building of height AB = 7 m.
Let CD be the cable tower of height H meters.
Let the horizontal distance between the building and the tower be BD = d meters.
From top of building A, draw a horizontal line AE perpendicular to CD such that AE = BD = d, and ED = AB = 7 m.
Let CE = h. Thus, total height CD = CE + ED = h + 7.
In right ΔABD (looking down at the foot of the tower D with angle of depression ∠EAD = 45° ⇒ ∠ADB = 45°):
tan 45° = AB / BD
1 = 7 / d ⇒ d = 7 m.
Therefore, AE = BD = 7 m.
In right ΔAEC (looking up at the top of the tower with angle of elevation ∠CAE = 60°):
tan 60° = CE / AE
√3 = h / 7 ⇒ h = 7√3 m.
Total height of the cable tower CD = CE + ED = 7√3 + 7 = 7(√3 + 1) m.
If √3 ≈ 1.732:
Height = 7(1.732 + 1) = 7(2.732) = 19.124 m.
Hence, the height of the cable tower is 7(√3 + 1) m (or 19.12 m).

Section E: Case Study-Based Integrated Questions (4 Marks Each)

Q19. Case Study on Coordinate Geometry (Sports Day Ground Track Layout)
To conduct Sports Day activities in a rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, so that AD = 100 m. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the 8th line and posts a red flag.

Based on the above information, answer the following questions:
(i) What are the coordinates of the green flag posted by Niharika? (1 Mark)
(ii) What are the coordinates of the red flag posted by Preet? (1 Mark)
(iii) Find the shortest distance between both flags. (2 Marks)
OR (iii) If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, at what position should she post her flag? (2 Marks)

Solution:
(i) Niharika runs on the 2nd line, so x₁ = 2.
Distance covered along AD = (1/4) × 100 m = 25 m, so y₁ = 25.
Therefore, coordinates of the green flag are (2, 25).

(ii) Preet runs on the 8th line, so x₂ = 8.
Distance covered along AD = (1/5) × 100 m = 20 m, so y₂ = 20.
Therefore, coordinates of the red flag are (8, 20).

(iii) Distance between Green Flag G(2, 25) and Red Flag R(8, 20):
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]
= √[(8 − 2)² + (20 − 25)²]
= √[(6)² + (−5)²]
= √[36 + 25] = √61 meters ≈ 7.81 meters.

OR (iii) Position of the blue flag (Mid-point of GR):
Using midpoint formula M = ((x₁ + x₂)/2, (y₁ + y₂)/2):
x = (2 + 8)/2 = 10/2 = 5
y = (25 + 20)/2 = 45/2 = 22.5
Rashmi must post the blue flag on the 5th line at a distance of 22.5 meters along AD, with coordinates (5, 22.5).

Section-Wise Difficulty Analysis and Chapter Weightage

Understanding the syllabus distribution allows candidates to allocate their 3-hour exam time strategically. Below is the CBSE Class 10 Mathematics standard blueprint weightage:

Unit No.	Unit Name	Chapters Included	Board Marks Weightage
I	Number Systems	Real Numbers	06 Marks
II	Algebra	Polynomials, Pair of Linear Equations, Quadratic Equations, Arithmetic Progressions	20 Marks
III	Coordinate Geometry	Coordinate Geometry	06 Marks
IV	Geometry	Triangles, Circles	15 Marks
V	Trigonometry	Introduction to Trigonometry, Trigonometric Identities, Heights and Distances	12 Marks
VI	Mensuration	Areas Related to Circles, Surface Areas and Volumes	10 Marks
VII	Statistics & Probability	Statistics, Probability	11 Marks
—	Total	14 NCERT Chapters	80 Marks

Cognitive Level Breakdown

• Remembering & Understanding (approx. 43-45 marks): Direct formula substitutions, definitions, fundamental theorem proofs, and standard textbook problems in Sections A and B.
• Applying (approx. 19-20 marks): Multi-step word problems involving speed-distance-time, heights and distances, and coordinate geometry section formulas.
• Analyzing, Evaluating & Creating (approx. 16-18 marks): Integrated case studies and geometric theorem applications requiring situational reasoning.

How to Practice Effectively with This Sample Paper

To maximize your score on exam day, implement this proven 3-phase revision technique using our sample paper:

1. Replicate Strict Exam Conditions: Sit in a quiet room with no textbooks, set a timer for 3 hours (plus 15 minutes initial reading time), and solve the paper in one continuous sitting on ruled answer sheets.
2. Follow Recommended Time Allocation:
   • Reading Time (15 mins): Read Section D and Section E questions carefully, deciding internal choices beforehand.
   • Section A (30 mins): Spend at most 1.5 minutes per MCQ.
   • Section B (20 mins): 4 minutes per 2-mark question.
   • Section C (35 mins): 5-6 minutes per 3-mark question.
   • Section D (45 mins): 10-11 minutes per 5-mark question with complete geometric constructions/sketches.
   • Section E (25 mins): 8 minutes per case study.
   • Revision & Unit Check (10 mins): Verify calculations, check unit tags (cm, m², km/h), and ensure question numbers match your script.
3. Audit with the Step-Wise Marking Scheme: CBSE awards marks for every correct step, formula, and diagram even if the final calculation has an arithmetic slip. Review where you gained or lost partial marks to identify conceptual gaps.

Download More CBSE Class 10 Sample Papers and Mock Tests

Ready to test your preparation with more high-yield model question papers? Visit Theorify QPTool (qptool.theorify.in) to access a comprehensive library of CBSE Class 10 Mathematics Standard and Basic sample papers, chapter-wise question banks, previous 10 years' solved board papers, and AI-driven instant rubric generators.