NCERT Math Solution Class 10 | Download Free PDF

Class 10 is a very important year in every student’s school journey, and Mathematics plays a very imp role in shaping your overall career outcome. The syllabus can seem a bit tough with topics starting from algebra to geometry and trigonometry. However, having the right books and notes can make a huge difference in how you tackle and master these topics. One such resource that can greatly affect your preparation is the NCERT Math Solutions for Class 10.

NCERT solutions for Class 10 Math provide detailed solutions of NCERT in-text questions, examples, and end-of-the-chapter questions that are always useful and all important. We at Educart have provided free, easy-to-download PDFs with detailed explanations for the NCERT Class 10 Maths Textbook (all chapters) as per the latest CBSE Class 10 Maths Syllabus.

In this blog, we’ll explore the benefits of using NCERT Math Solutions, how they can help you understand complex problems and ways to use them effectively to ace your board exams.

Full Math Solutions NCERT Class 10 PDF

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Some topics have been removed from NCERT books but are still INCLUDED in CBSE Math 2025 syllabus. (Refer to the table below)

Chapter-Wise Breakdown of NCERT Math Solutions for Class 10

The NCERT Math Solutions for Class 10 are structured chapter-wise to help you tackle each topic methodically. Here’s a quick breakdown of some key chapters:

Chapter 1: Real Numbers

In this chapter, you’ll learn about the Euclidean division algorithm and the Fundamental Theorem of Arithmetic. Key concepts include:

• Euclid’s Division Lemma: A method to compute the highest common factor (HCF) of two numbers.
• Prime Factorisation: breaking down numbers into their prime factors.
• Rational and Irrational Numbers: Understand the properties of rational numbers (expressible as fractions) and irrational numbers (which cannot be expressed as a simple fraction).
• Decimal Expansion: Learn how rational numbers have terminating or repeating decimals, while irrational numbers have non-terminating, non-repeating decimal expansions.

Examples:

Prove that √3 is an irrational number.

Solutions: Let 3 be a rational number in the form of p/q.

where q ≠ 0, p, q ∈ Z, and p, q are co-primes, i.e., GCD (p, q) = 1.

√3 = (p/q)

√3 q = p

Squaring on both sides

(√3q)2 = p2

3q2 = p2

q2 = p2/3 ------- (1)

As per equation 1, 3 is a factor of p. (Theorem: if a is a prime number and if a divides p2, then a divides p, where a is a positive integer.)

Here 3 is the prime number that divides p2, then 3 divides p, hence 3 is a factor of p.

If 3 is a factor of p, then 

p=3c (where c is the constant)

Substituting the value of p in eq 1

q2 = (3c)2/3

9c2/3 = q2

c2= q2/3 ------- (2)

Hence 3 is a factor of q (from 2).

eq 1 shows that 3 is a factor of p, and eq 2 shows that 3 is a factor of q. 

This contradicts the assumption that p and q are coprime numbers, so √3 isn’t a rational number. Thus, √3 is an irrational number. 

The sum of two numbers is 18 and the sum of their reciprocals is 9/40. Find the numbers.

Solutions: Let the first number be x and the second number be (18-x)

As per the question

(1/x) + ( 1/(18-x) = ¼

((18-x+x) /x(18-x))= ¼

4(18-x+x) = x(18-x) (-x+x = 0)

72 = 18x - x²

x²- 18x+ 72 = 0

x²- 12x-6x+ 72 = 0

x(x-12)-6(x-12) =0

(x-6) (x-12) =0

x =6 or x=12

If x=6 then

First number = 6 

Second number= (18-6) = 12

If x=12 then

First number = 12

Second number= (18-12) = 6

Chapter 2: Polynomials

This chapter covers polynomials in one variable and focuses on the relationship between zeros and coefficients. Key topics include:

• Zeros of a Polynomial: The value of x that makes the polynomial equal to zero.
• Relationship Between Zeros and Coefficients: For quadratic, cubic, and linear polynomials, explore how the sum and product of zeros relate to the coefficients.
• Division Algorithm for Polynomials: Learn how to divide one polynomial by another and express the remainder.

If 𝛼 and 𝛽 are zeroes of a polynomial 6𝑥² -5x+1 then form a quadratic polynomial whose zeroes are 𝛼² and 𝛽².

Solutions: P(x) = 6𝑥² -5x+1

𝛼 + 𝛽 = -b/a

=-(-5)/6

= 5/6

𝛼 𝛽 = c/a

=1/6

quadratic polynomial whose zeroes are 𝛼² and 𝛽²

Required polynomial = P(x)

= 𝑥² -(sum of zeroes)x+product of zeroes

= 𝑥² -(𝛼 + 𝛽 )x+(𝛼 𝛽)

= x ² − [ ( α + β ) ² − 2 α β ] x + ( α β ) ²

= x ² − [ ( 5/6) ² − 2 x 1/6 ] x + (1/6 ) ²

=  x ² − [ ( 25/36) - 1/3 ] x + (1/36 ) 

=  x ² − [ ( 25 - 12)/36 ] x + (1/36 ) 

=  x ² − [ ( 13)/36 ] x + (1/36 ) 

=  36x ² − [ ( 13)/36 ] x + (1/36 ) 

⇒ Required polynomial =  36x ² − 13x +1

The graph of a quadratic polynomial p(x) passes through the points (-6,0), (0, -30), (4,-20) and (6,0). The zeroes of the polynomial are

Solutions:

y = ax² + bx + c

where a > 0

You can see from the table that –6 and 6 are zeroes of the quadratic polynomial. Also, note from the above graph that –6, 6, and 4 are the x-coordinates of the points. 

y = ax² + 6x + c

y = ax² - 6x + c

⇒ zeroes of the polynomial are =  -6 and 6

Chapter 3: Pair of Linear Equations in Two Variables

This chapter helps you solve simultaneous linear equations using different methods. Key methods include:

• Graphical Method: Plotting lines to find the point of intersection, which gives the solution.
• Substitution Method: Solve one equation for one variable and substitute it into the other equation.
• Elimination Method: Eliminate one variable by adding or subtracting equations.
• Cross-Multiplication Method: A method used for solving two-variable equations directly.

Solve the following system of linear equations graphically:

x+2y = 3, 2x-3y+8 = 0

Solutions:

Graph of  x+2y = 3

‍⇒ y = (3-x)/2  ………eq1

Putting x=1, we get y= 1

Putting x=-3, we get y= 3

Putting x=-5, we get y= 4

Table for equation  x+2y = 3

Graph

Graph of  2x-3y+8 = 0

⇒ y = (2x+8)/3  ………eq1

Putting x=-1, we get y= 2

Putting x=2, we get y= 4

Putting x=5, we get y= 6

Table for equation 2x-3y+8 = 0

Graph

Plotting the graph

The value of k for which the system of equations 3x-ky= 7 and 6x+ 10y =3 is inconsistent, is

Solutions: Substituting the value in the system of equations

‍a₁x + b₁y = c₁ | 3x-ky= 7a₂x + b₂y = c₂   | 6x+ 10y =3

‍There is only 1 case for an inconsistent solution where the graphs of the pair of equations are parallel.

a₁/a₂ = b₁/b₂ ≠ c₁/c₂‍

3/6 = -k/10 ≠ 7/3

3/6 = -k/10 and -k/10 ≠ 7/3

-6k = 30 and -3k ≠ 70

k = -5 and k  ≠ -70/3

∴ k = -5 

Chapter 4: Quadratic Equations

This chapter teaches various methods to solve quadratic equations. Students can make notes for CBSE Class 10 Maths Formulas. Key topics include:

• Factorisation Method: breaking down the quadratic expression into two factors.
• Quadratic Formula: The formula to solve quadratic equations x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac.
• Nature of Roots: Explore how the discriminant (b2−4ac)(b^2 - 4ac)(b2−4ac) determines whether the roots are real, distinct, or imaginary.

Examples:

A quadratic polynomial having zeroes -√5/2 and √5/2 is

A) 𝑥²− 5√2 x +1

B) 8𝑥² - 20 

C) 15𝑥²- 6 

D) 𝑥² - 2√5 x -1

Solutions: α = -√5/2 and β = √5/2

x² - (Sum of the zeroes) x + (product of zeroes) = 0

x² - (α +β ) x + (αβ) = 0

Substituting the values in the equation

x² - (-√5/2 + √5/2) x + ( -√5/2 * √5/2) = 0

x² - (0) x + ( -5/2) = 0

x² - 5/2 = 0

2x² - 5 = 0

To verify

The sum of the zeroes = α +β

= -√5/2 + √5/2

= 0

α = -b/a

= -0/2

= 0

Product of the zeroes = αβ

=  -√5/2 * √5/2

= -5/2

β = c/a

= -5/2

A quadratic polynomial having zeroes -√5/2 and √5/2 is x² - 5/2 = 0

As per the options

A) 𝑥²− 5√2 x +1

x = √2(5 ± √23)/2

B) 8𝑥² - 20 

x = ± √5/2

C) 15𝑥²- 6 

x = ± √3/5

D) 𝑥² - 2√5 x -1

Roots - √5(2 ± √23)/2

A quadratic polynomial having zeroes -√5/2 and √5/2 is Option B) 8𝑥² - 20 

The roots are irrational and distinct

Chapter 5: Arithmetic Progressions (AP)

In this chapter, you’ll learn how sequences of numbers progress linearly. Key topics include:

• Finding the nth term of an AP: Using the formula an=a+(n−1)da_n = a + (n - 1)dan​=a+(n−1)d, where aaa is the first term and ddd is a common difference.
• Sum of n Terms of an AP: Using the formula Sn=n2×[2a+(n−1)d]S_n = \frac{n}{2} \times [2a + (n - 1)d] Sn​=2n​×[2a+(n−1)d].

Examples:

Ms. Sheela visited a store near her house and found that the glass jars were arranged one above the other in a specific pattern. On the top layer, there are 3 jars. In the next layer, there are 6 jars. In the 3rd layer from the top there are 9 jars and so on till the 8th layer.

Based on the above situation answer the following questions.

(i) Write an A.P. whose terms represent the number of jars in different layers starting from the top. Also, find the common difference.

(ii) Is it possible to arrange 34 jars in a layer if this pattern is continued? Justify your answer.

(iii) (A) If there are ‘n’ number of rows in a layer then find the expression for finding the total number of jars in terms of n. Hence find 𝑆₈.

(B) The shopkeeper added 3 jars in each layer. How many jars are there in the 5th layer from the top?

Solutions:

(i) The jars are in AP of 

3,6,9,12…upto 8 terms

First term = a

a=3

Common difference= d = a₁ - a

d= 6-3

d= 3

(ii) Given AP = 3,6,9,12…upto 8 terms

Let a_n = 34

a+ (n-1)d = a_n

3 + (n-1)3 = 34

3 +3n -3 = 34

3n = 34

n= 34/3

n= 11.33

The number of terms can’t be in decimals. Hence, 34 isn’t the term in the AP.

(iii) (A) Given AP = 3,6,9,12…upto 8 terms

S_n = n/2 [2a+(n-1)d]

S_n = n/2 [2(3)+(n-1)3]

S_n = n/2 [6 + 3n - 3]

S_n = n/2 [ 3n + 3]

Putting n=8

S₈ = 8/2 [ 3(8) + 3]

S₈ = 4 [ 24 +3]

S₈ = 4 [ 27]

S₈ = 108

(iii) (B) Given AP = 3,6,9,12…upto 8 terms

After adding 3 jars in each layer

New AP = 3+3,6+3,9+3,12+3…upto 8 terms

New AP = 6,9,12,15…upto 8 terms

Now a= 6

d = a₁ -a

d= 9 -6

d= 3

For 5 terms

a+ (n-1)d = a_n

6+ (5-1)3 = a₅

6+ (4)3 = a₅

18 = a₅

There are 18 jars in the 5th layer.

If the nth term of an A.P. is 7n-4 then the common difference of the A.P. is

Solutions:

a_n= 7n-4

Put n=1

a = 7(1)-4

a= 3

Put n=2

a₁ = 7(2)-4

a= 10

Common difference d= 

d= a₁ -a

d= 10-3

d= 7

Chapter 6: Triangles

This chapter focuses on the concept of the similarity of triangles. Key topics include:

• Basic Proportionality Theorem (Thales Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
• Criteria for Similarity: Explore different criteria like AA (angle-angle), SAS (side-angle-side), and SSS (side-side-side) for determining the similarity between triangles.
• Pythagoras Theorem: Prove that in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

Chapter 7: Coordinate Geometry

This chapter introduces geometric interpretation using a coordinate plane. Key topics include:

• Distance Formula: Calculate the distance between two points (x1,y1)(x_1, y_1)(x1,y1) and (x2,y2)(x_2, y_2)(x2,y2) using the formula (x2−x1)2+(y2−y1). 2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}(x2​−x1​)2+(y2​−y1​)2.
• Section Formula: Find the coordinates of a point dividing a line segment in a given ratio.
• Area of a Triangle: Use the coordinates of a triangle’s vertices to calculate its area.

Chapter 8: Introduction to Trigonometry

This chapter covers the basics of trigonometry, a branch of mathematics that deals with the angles and sides of triangles. Key topics include:

• Trigonometric Ratios: Learn the ratios like sine, cosine, and tangent, which relate the angles of a right triangle to the lengths of its sides.
• Trigonometric Identities: Key identities like sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1sin2θ+cos2θ=1 are introduced.
• Complementary Angles: Understand how the trigonometric ratios of complementary angles relate to each other.

Chapter 9: Applications of Trigonometry

This chapter applies trigonometric concepts to real-world scenarios. Key applications include:

• Height and Distance: Learn to calculate the height of an object or the distance between two points using trigonometric ratios, angles of elevation, and depression.

Chapter 10: Circles

In this chapter, you’ll explore the properties of circles, especially tangents. Key topics include:

• Tangent to a Circle: A line that touches a circle at exactly one point.
• Properties of Tangents: Learn that the radius drawn to the point of contact is perpendicular to the tangent.

Chapter 11: Areas Related to Circles

This chapter focuses on finding areas related to circles, such as sectors and segments. Key topics include:

• Area of a Sector: Using the formula θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2360∘θ​×πr2, where θ\thetaθ is the angle of the sector.
• Area of a Segment: The area of the segment formed by a chord and the corresponding arc.

Chapter 12: Surface Areas and Volumes

This chapter covers three-dimensional shapes and their surface areas and volumes. Key topics include:

• Surface Area of Solids: Learn to calculate the surface areas of cylinders, cones, spheres, and hemispheres.
• The volume of Solids: Learn the formulas to calculate volumes of the same shapes, useful for real-life applications.

Chapter 13: Statistics

In this chapter, you’ll explore data collection and analysis. Key topics include:

• Mean, median, and mode: Calculate these measures of central tendency from grouped data.
• Cumulative Frequency and Graphical Representation: Learn how to represent data using cumulative frequency curves (ogives).

Examples:

Find the mean and median of the following data:

Solutions: 

Maximum frequency = 25

Modal class = 110-115

Hence

l= 110

h=5

f=25

f1=22

f2=20

F=42

Mean = Σx_if_i/ Σf_i

= 12,105/120

= 100.875

The mean of the given data is 100.875

Median= l + (N/2 -C.F.)/f x h

Median= 110 + (120/2 -42.)/25 x 5

Median= 110 + (60 -42.)/25 x 5

Median= 110 + (18)/25 x 5

Median= 110 + 3.6

Median= 113.60

The median of the given data is 113.60

The monthly expenditure on milk in 200 families of a Housing Society is given below

Find the value of x and also find the mean expenditure

Solutions:

Total number of families= f_i= 200

∴ 24 + 40+33+x+30+22+16+7 = 200

x= 200-172

x= 28

Mean = Σx_if_i/ Σf_i

= 5,32,000/200

= ₹2662.5

Median class = N/2

=200/2

=100th term

The 100th observation is in 2500-3000 (median class)

l= 2500; N=200; f=28; C.F.=97; h=500

Median= l + (N/2 -C.F.)/f x h

= 2500 + (200/2 -97)/28 x 500

= 2500 + (100 -97)28 x500

=2500 + 3/7 x125

=2500 +53.571

=2553.571

Chapter 14: Probability

This chapter introduces the basics of probability and how it applies to everyday situations. Key topics include:

• Probability of Events: Calculate the probability of a simple event using P(E)=Number of favourable outcomes Total number of outcomes P(E) = \frac {\text{Number of favourable outcomes}}{\text{Total number of outcomes}} P(E)=Total number of outcomes Number of favourable outcomes.
• Complementary Events: Understand that the probability of the complement of an event is 1−P(E)1 - P(E)1−P(E).

Examples:

Two dice are rolled together bearing numbers 4, 6, 7, 9, 11, 12. Find the probability that the product of numbers obtained is an odd number

Solutions:

Number on each dice= 4, 6, 7, 9, 11, 12

Odd numbers= 7,9,11

Total possibilities when two dices are rolled= 6x6

= 36

Favourable outcomes when the number is odd= 3x3  (∵ product of odd numbers is always odd)

=9

P(favorable outcomes)= Favourable outcomes/Total outcomes

=9/36

=1/4

How many positive three-digit integers have the hundredth digit 8 and unit’s digit 5? Find the probability of selecting one such number out of all three-digit numbers.

Solutions: Total three-digit integers= 900

Number with the hundredth digit 8 and unit’s digit 5= 805, 815,825,...,895

Favourable outcomes= 10

P(favorable outcomes)= Favourable outcomes/Total outcomes

= 10/900

=1/9

Why Choose NCERT Math Solutions for Class 10?

The NCERT textbook forms the foundation of the CBSE syllabus and board exams. Every year, a majority of the questions in the board exams come directly from NCERT books, making it essential to thoroughly understand the material. The NCERT Math Solutions provide clear, step-by-step answers to every exercise, helping you gain in-depth knowledge of each concept.

Here’s why NCERT Math Solutions are indispensable for Class 10 students:

Clear and Detailed Explanations

The solutions break down every problem in a simple, understandable manner. Whether it’s solving quadratic equations, working through geometric proofs, or mastering coordinate geometry, these solutions ensure that every step is clear. This helps students tackle even the most complex problems with ease.

Alignment with CBSE Exam Pattern

NCERT Math Solutions follows the CBSE exam pattern, ensuring that you are well-prepared for the types of questions you will encounter in the exams. The solutions are designed to help you understand how to frame answers in a way that will earn you maximum marks.

Comprehensive Coverage

The solutions cover every chapter and topic from the NCERT Class 10 Math textbook, including key areas like algebra, trigonometry, geometry, probability, and statistics. This comprehensive coverage ensures that you are fully prepared for every aspect of the exam.

Improves problem-solving skills

Mathematics requires more than just theoretical knowledge; it demands strong problem-solving skills. By practicing with NCERT Math Solutions, you develop a deeper understanding of the concepts and techniques, helping you solve questions more efficiently and accurately.

Free and easily accessible

NCERT Math Solutions are available for free in PDF format, making them accessible to all students. You can download these solutions and refer to them anytime, anywhere, making your study sessions more flexible and efficient.

How to Use NCERT Math Solutions for Effective Preparation

Preparing for the Class 10 board exams can be daunting, especially when it comes to mathematics. The key to mastering this subject lies in understanding the core concepts and practicing efficiently. This is where NCERT Math Solutions come into play. Designed specifically to complement the NCERT textbook, these solutions break down complex problems into simple, easy-to-follow steps, allowing you to grasp even the toughest concepts. In this guide, we’ll explore how to use NCERT Math Solutions effectively to streamline your preparation, enhance your problem-solving skills, and boost your confidence for the exams.

1. Practice Regularly
   Mathematics is a subject that improves with consistent practice. Use the NCERT Math Solutions as a guide, but ensure that you are solving the problems on your own first before checking the solutions.
2. Understand the Concepts
   Don’t just memorise the steps—focus on understanding the underlying concepts. This will help you solve unfamiliar questions in the exam with confidence.
3. Review Mistakes
   When you encounter a mistake, use the solutions to understand where you went wrong. This process helps with in-detail learning and prevents future errors.
4. Time Yourself
   While practicing, time yourself to simulate exam conditions. This will help you improve your speed and accuracy for the actual board exam.

The NCERT Math Solutions for Class 10 is a must-have resource for any student aiming to do well or top in their exams. With great coverage, clear explanations, and alignment with the CBSE exam pattern, these solutions can significantly boost your preparation and help you score high marks. Whether you're solving complex problems or revising key concepts, NCERT Solutions provides the perfect support for a successful study strategy.

Download your free PDF copies of Class 10 NCERT Math Solutions and start preparing today!

FAQs (Frequently Asked Questions)

How can I effectively solve difficult problems in Class 10 math using NCERT solutions?

Solving difficult problems in Class 10 math becomes easier when you break them down step by step. The key is to start by thoroughly understanding the concept behind the question. Educart’s NCERT Math Solutions are designed in a way that simplifies complex problems by providing detailed explanations.

Are NCERT Math Solutions enough to score above 90 in the Class 10 board exams?

Yes, NCERT Math Solutions is sufficient to help you score above 90 in the Class 10 board exams if you use them effectively. However, to go beyond just solving textbook exercises, it’s recommended to solve past-year papers, and sample papers, and take mock tests regularly. 

What is the best way to revise Class 10 math before the board exams?

The best way to revise for Class 10 math is to focus on solving all exercises from the NCERT textbook, followed by solving sample papers and previous years’ question papers. 

How can I improve my speed and accuracy in solving NCERT math questions for Class 10?

Improving speed and accuracy requires consistent practice and a clear understanding of the concepts. First, ensure that you have mastered the basics by using Educart’s NCERT Math Solutions, which breaks down each concept and provides easy-to-understand explanations. Once you’re comfortable with the theory, practice solving questions within a set time limit to build speed.