CBSE Class 10 Maths Question Paper 2025, Check Difficulty Level, Download Set Wise PDF

CBSE Class 10th Maths Question Paper 2025: The Central Board of Secondary Education (CBSE) has concluded the CBSE Class 10th Maths Exam 2025 for Standard and Basic paper at 1.30 pm on March 10, 2025. Almost 24 lakhs registered students appeared for the CBSE Maths Paper 2025 Class 10 today. Educators and exam experts are expecting that the difficulty level of the Class 10 Maths Paper can be easy to slight moderate level. Students can soon access the CBSE Class 10 Maths Question Paper 2025 for all sets here.

CBSE Class 10 Maths Question Paper 2025

The CBSE Class 10 Maths Basic and Standard Exam Paper 2025 is held for 3 hours. As an important subject of Class 10, Mathematics contributes significantly a big part to overall CBSE board exam results 2025. The solutions to all sets of CBSE Class 10 Maths Question Paper 2025 will provide clarity on the correct responses. You will be able to download the CBSE Class 10th Maths Paper PDF from the direct links to be added below shortly. Along with that, students will receive the complete CBSE Board Maths Paper Analysis 2025 by experts on this Adda247 page, so stay tuned.

CBSE Class 10th Maths Paper 2025

The CBSE Class 10th Maths Paper 2025 is worth total 80 marks. The Class 10 Maths Board exam is administered in offline mode, so students will receive the question paper in hard copy. The Maths questions is asked from the main 7 units. Algebra holds the most of the marks in the paper, where coordinate geometry and number system presents the least marks. CBSE Class 10th Standard Maths papers 2025 have 20 multiple-choice questions (1 mark each), 5 questions (2 marks each), 6 questions (3 marks each), 4 questions (5 marks each), and 3 case-based questions (4 marks each).

CBSE Board 10th Maths Paper 2025 Chapter-wise Weightage

Students can check the CBSE Class 10 maths question paper 2025 chapter-wise weightage below.

• NUMBER SYSTEMS chapter is worth of 06 marks.
• ALGEBRA chapter is worth of 20 marks.
• COORDINATE GEOMETRY chapter is worth of 06 marks.
• GEOMETRY chapter is worth of 15 marks.
• TRIGONOMETRY chapter is worth of 12 marks.
• MENSURATION chapter is worth of 10 marks in the the CBSE Class 10 Maths question paper 2025.
• STATISTICS & PROBABILITY chapter is worth of 11 marks.

Class 10th Maths Board Paper 2025 Analysis

Our  exam experts will analyse Class 10 Maths Board Papers 2025 and soon let you know best attempts and complete solution to the paper.

• Generally the Maths Class 10th paper contains two sorts of questions: Basic and Standard.
• Class 10 maths papers have 20 Multiple Choice Questions (MCQs) worth 1 mark,
• 5 Short Answer-I (SA-I) questions worth 2 points,
• 6 Short Answer-II (SA-II) questions worth 3 marks,
• 4 Long Answer (LA) questions worth 5 marks.

Students who will take the CBSE Board Exam in 2026 should refer to the CBSE 10 Maths Paper 2025 to obtain a thorough understanding of the various types of questions.

CBSE Class 10 Maths Question Paper 2025 PDF Download All Sets

To assist students in analysing their performance, we will present the CBSE Class 10 Maths question Paper with Solutions for all possible sets. This expert-created maths paper solution is great resource for students to know the correct answers to the questions. Next year aspirants can improve their preparation and score higher on the real test. This website will soon contains links to download the 10th Maths question papers.

Class 10 Maths Questions with Solutions

Question: Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) 140
Using the division of a number by prime numbers method, we can get the product of prime factors of 140.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7

(ii) 156
Using the division of a number by prime numbers method, we can get the product of prime factors of 156.

Hence, 156 = 2 × 2 × 13 × 3 = 22 × 13 × 3

(iii) 3825
Using the division of a number by prime numbers method, we can get the product of prime factors of 3825.

Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17

(iv) 5005
Using the division of a number by prime numbers method, we can get the product of prime factors of 5005.

Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429
Using the division of a number by prime numbers method, we can get the product of prime factors of 7429.

Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23

Question: Check whether 6n can end with the digit 0 for any natural number n.

Solution:

If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.

Prime factorization of 6n = (2 × 3)n

Therefore, the prime factorization of 6n doesn’t contain the prime number 5.

Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.

Question: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36

(l + b) = 36 ……….(1)
Given, the length is 4 m more than its width.

Let width = x

And length = x + 4

Substituting this in eq(1), we get;

x + x + 4 = 36
2x + 4 = 36
2x = 32
x = 16

Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

Question:

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14
x – y = 4

(ii) 3x – y = 3
9x – 3y = 9

Solution:

(i) Given,
x + y = 14 and x – y = 4 are the two equations.
From 1st equation, we get,
x = 14 – y
Now, put the value of x in second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the value of x;
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.

(ii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From 1st equation, we get,
x = (3 + y)/3
Now, substitute the value of x in the given second equation to get,
9[(3 + y)/3] – 3y = 9

⇒ 3(3+y) – 3y = 9
⇒ 9 + 3y – 3y = 9
⇒ 9 = 9
Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

Question: Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Solution:

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

Then, AP = BP

AP2 = BP2

Using distance formula,

(x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2

x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25

x – y = 2

Hence, the relation between x and y is x – y = 2.

Question:  Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

Solution:

Given,

A(2, 3)= (x1, y1)

B(4, k) = (x2, y2)

C(6, -3) = (x3, y3)

If the given points are collinear, the area of the triangle formed by them will be 0.

½ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

½ [2(k + 3) + 4(-3 -3) + 6(3 – k)] = 0

½ [2k + 6 – 24 + 18 – 6k] = 0

½ (-4k) = 0

4k = 0

k = 0

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