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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.
Some topics have been removed from NCERT books but are still INCLUDED in CBSE Math 2025 syllabus. (Refer to the table below)
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:
In this chapter, you’ll learn about the Euclidean division algorithm and the Fundamental Theorem of Arithmetic. Key concepts include:
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.
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
This chapter covers polynomials in one variable and focuses on the relationship between zeros and coefficients. Key topics include:
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
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
This chapter helps you solve simultaneous linear equations using different methods. Key methods include:
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
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
This chapter teaches various methods to solve quadratic equations. Students can make notes for CBSE Class 10 Maths Formulas. Key topics include:
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
In this chapter, you’ll learn how sequences of numbers progress linearly. Key topics include:
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 an = 34
a+ (n-1)d = an
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
Sn = n/2 [2a+(n-1)d]
Sn = n/2 [2(3)+(n-1)3]
Sn = n/2 [6 + 3n - 3]
Sn = 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 = an
6+ (5-1)3 = a5
6+ (4)3 = a5
18 = a5
There are 18 jars in the 5th layer.
Solutions:
an= 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
This chapter focuses on the concept of the similarity of triangles. Key topics include:
This chapter introduces geometric interpretation using a coordinate plane. Key topics include:
This chapter covers the basics of trigonometry, a branch of mathematics that deals with the angles and sides of triangles. Key topics include:
This chapter applies trigonometric concepts to real-world scenarios. Key applications include:
In this chapter, you’ll explore the properties of circles, especially tangents. Key topics include:
This chapter focuses on finding areas related to circles, such as sectors and segments. Key topics include:
This chapter covers three-dimensional shapes and their surface areas and volumes. Key topics include:
In this chapter, you’ll explore data collection and analysis. Key topics include:
Solutions:
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Maximum frequency = 25
Modal class = 110-115
Hence
l= 110
h=5
f=25
f1=22
f2=20
F=42
Mean = Σxifi/ Σfi
= 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
Solutions:
Total number of families= fi= 200
∴ 24 + 40+33+x+30+22+16+7 = 200
x= 200-172
x= 28
%252520(1).avif)
Mean = Σxifi/ Σfi
= 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
This chapter introduces the basics of probability and how it applies to everyday situations. Key topics include:
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
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
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.
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.
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!
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.
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.
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.
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.
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