NCERT Solutions For Class 10 Maths Chapter 1 – 2025-26 Free PDF
Class 10 Maths Chapter 1 Solutions
Welcome to our comprehensive guide to NCERT Solutions For Class 10 Maths Chapter 1 – Real Numbers. If you’re looking for step-by-step solutions of all exercises of update 2025-26 session, conceptual explanations, and a free downloadable PDF, you’ve landed at the right place.
We at Vidyayan Academy provide complete NCERT solutions for Class 10 Maths Chapter 1 in an easy-to-understand format — perfect for exam preparation, school revision, or self-study.
All Exercises of NCERT Solutions For Class 10 Maths Chapter 1- Download pdf:
We provide comprehensive and accurate solutions for NCERT Class 10 Maths Chapter 1 – Real Numbers, designed to help students for session 2025-26 to understand each concept clearly. Our step-by-step explanations cover all questions from Exercise 1.1 and 1.2, including topics like prime factorization, HCF & LCM, and proofs of irrational numbers. These solutions follow the latest syllabus and are ideal for both regular study and quick revision before exams. You can easily download the PDF solutions and boost your preparation with confidence.
Exercise 1.1 – Prime Factorisation, HCF & LCM
This exercise focuses on:
- Expressing numbers as product of prime factors
- Finding HCF and LCM using the prime factorisation method
- Using the identity:
HCF×LCM=Product of the numbers - Simple reasoning-based questions on divisibility (e.g., whether a number can end with 0)
- Real-life application-based questions like finding when two events will occur together using LCM.
Exercise 1.2 – Proving Irrational Numbers
This exercise deals with:
- Proof-based questions to show that:
- √2, √3, √5 are irrational
- Using the proof by contradiction method
- Applying Fundamental Theorem of Arithmetic in logical proofs
👉 Every exercise is solved line-by-line in our Class 10 Maths Chapter 1 Solutions PDF file.
Chapter 1: Real Numbers
NCERT Solution of Class 10 Maths Exercise 1.1
1. Express each number as product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Solution:
(i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
Solution:
26 = 2 × 13
91 = 7 × 13
LCM = 2 × 7 × 13 = 182
HCF = 13
Product of the two numbers = 26 × 91 = 2366
HCF × LCM = 13 × 182 = 2366
Hence, product of two numbers = HCF × LCM
(ii) 510 and 92
Solution:
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of the two numbers = 510 × 92 = 46920
HCF × LCM = 2 × 23460 = 46920
Hence, product of two numbers = HCF × LCM
(iii) 336 and 54
Solution:
336 = 2 × 2 × 2 × 2 × 3 × 7 = 24 × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 33
HCF = 2 × 3 = 6
LCM = 24 × 33 × 7 = 3024
Product of the two numbers= 336 × 54 = 18144
HCF × LCM = 6 × 3024 = 18144
Hence, product of two numbers = HCF × LCM
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution:
12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3 and LCM = 22 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
Solution:
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
Solution:
8 = 2 × 2 × 2 = 23
9 = 3 × 3 = 32
25 = 5 × 5 = 52
HCF = 1
LCM = 23 × 32 × 52 = 8 × 9 × 25 = 1800
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
We know that,
LCM × HCF = Product of two numbers
Therefore, LCM = Product of two numbers/HCF
So, LCM = (306 × 657)/9 = 22338
Hence, LCM (306, 657) = 22338
5. Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5.
Prime factorisation of 6ⁿ = (2 × 3)ⁿ
It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6ⁿ will not be divisible by 5.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Numbers are of two types – prime and composite.
Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that:
7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
Now consider the next number:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 ×1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path.
As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia.
The total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 × 3 × 3 and
12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
NCERT Solution of Class 10 Maths Exercise 1.2
1. Prove that √5 is irrational.
Solution:
Let √5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b. Let a and b have a common factor other than 1. Then we can divide them by the common factor and assume that a and b are co-prime.
a = √5b
⇒ a² = 5b²
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer.
⇒ (5k)² = 5b²
⇒ 25k² = 5b²
⇒ 5k² = b²
This means that b² is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
This is a contradiction to the fact that a and b are co-prime.
Hence, √5 cannot be expressed as p/q or it can be said that √5 is irrational.
2. Prove that is 3 + 2√5 irrational.
Solution:
Let 3 + 2√5 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
3 + 2√5 = a/b
⇒ 2√5 = a/b – 3
⇒ √5 = 1/2(a/b – 3)
Since a and b are integers, 1/2(a/b – 3) will also be rational.
Therefore, √5 is rational.
This contradicts the fact that √5 is irrational.
Hence, our assumption that 3 + 2√5 is rational is false.
Therefore, 3 + 2√5 is irrational.
3. Prove that the following are irrationals:
(i) 1/√2
Solution:
Let 1/√2 is rational.
Therefore, we can assume two co-prime integers a, b (b ≠ 0) such that
1/√2 = a/b
⇒ √2 = b/a
b/a is rational as a and b are integers.
Therefore, √2 is also rational, which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and 1/√2 is irrational.
(ii) 7√5
Solution:
Let 7√5 is rational.
Therefore, we assume find two co-prime integers a, b (b ≠ 0) such that
7√5 = a/b
⇒ √5 = a/7b
a/7b is rational as a and b are integers.
Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational.
Therefore, our assumption that 7√5 is rational is false.
Hence, 7√5 is irrational.
(iii) 6 + √2
Solution:
Let 6 + √2 be rational.
Therefore, we can assume two co-prime integers a, b (b ≠ 0) such that
6 + √2 = a/b
⇒ √2 = a/b – 6
Since a and b are integers, a/b – 6 is also rational and hence, √2 should be rational.
This contradicts the fact that √2 is irrational.
Therefore, our assumption is false and hence, 6 + √2 is irrational.
Exercise 1.1 – Prime Factorisation, HCF & LCM
Exercise 1.2 – Proving Irrational Numbers
What’s Inside NCERT Solutions For Class 10 Maths Chapter 1?
Chapter 1 – Real Numbers — is one of the most important chapters in Class 10. We’ve covered:
- Euclid’s Division Lemma and Algorithm
- Fundamental Theorem of Arithmetic
- Prime Factorization
- HCF and LCM
- Rational and Irrational Numbers
- Decimal Expansions
Every topic is explained clearly and matched with corresponding NCERT textbook exercises.
Class 10 Maths Chapter 1 Solutions – Real Numbers (Explained Conceptually)
Looking for Class 10 Maths Chapter 1 Solutions explained in detail without just exercise answers? This blog gives you complete conceptual clarity of Chapter 1 – Real Numbers based on the NCERT Class 10 Maths textbook. You’ll find all the key theorems, formulas, examples, and practical use-cases explained in simple language.
Let’s break down everything you need to master class 10 maths chapter 1 solutions for your exams.
What Is Chapter 1 – Real Numbers All About?
The chapter begins with the foundation of number theory and helps students understand how numbers are built and behave. Key topics in NCERT Solutions For Class 10 Maths Chapter 1 include:
- Euclid’s Division Lemma and Algorithm
- The Fundamental Theorem of Arithmetic
- Prime factorization
- HCF and LCM
- Properties of irrational numbers
- Rational numbers and decimal expansions
These are core concepts for understanding mathematics at a higher level.
1. Euclid’s Division Lemma & Algorithm
Euclid’s Division Lemma
This is the base of NCERT Solutions For Class 10 Maths Chapter 1. It states:
For any two positive integers a and b, there exist unique integers q and r such that
a = bq + r, where 0 ≤ r < b
This lemma helps in dividing two numbers and is used repeatedly in mathematical problem-solving.
Euclid’s Division Algorithm
It is the application of the lemma to find the HCF (Highest Common Factor) of two integers. In NCERT Solutions For Class 10 Maths Chapter 1, this algorithm is applied step-by-step to simplify HCF problems easily.
2. Fundamental Theorem of Arithmetic
This theorem is a centerpiece of NCERT Solutions For Class 10 Maths Chapter 1. It states:
“Every composite number can be factorized into primes, and the factorization is unique apart from the order.”
Examples:
- 20 = 2 × 2 × 5
- 32760 = 2³ × 3² × 5 × 7 × 13
This concept plays a major role in calculating HCF, LCM, and solving various number system problems.
3. Prime Factorization, HCF & LCM
One of the most frequent applications in class 10 maths chapter 1 solutions is using prime factorization to find HCF and LCM of two or more numbers.
General Formula:
- HCF = Product of smallest powers of common prime factors
- LCM = Product of highest powers of all prime factors
- HCF × LCM = Product of the numbers
Understanding this relationship is key to solving many problems in class 10 maths chapter 1 solutions.
4. Irrational Numbers and Their Proofs
What is an Irrational Number?
Any number that cannot be written in the form p/q (where p and q are integers and q ≠ 0) is irrational.
Examples: √2, √3, √5, π
In class 10 maths chapter 1 solutions, we are taught how to prove irrationality using proof by contradiction and the Fundamental Theorem of Arithmetic.
Key Proofs in Chapter:
- √2 is irrational
- √3 is irrational
- 2√5, 5 – √3 are irrational
Mastering these examples ensures better performance in exams and makes you confident in handling irrationality-based questions in class 10 maths chapter 1 solutions.
5. Rational Numbers and Decimal Expansions
Another important part of NCERT Solutions For Class 10 Maths Chapter 1 is determining whether a rational number has a terminating or non-terminating repeating decimal.
Rule:
If the denominator (after simplifying) has only 2 and/or 5 as prime factors, the decimal terminates.
Otherwise, it will be non-terminating but repeating.
Examples:
- 3/8 = 0.375 → Terminating
- 7/9 = 0.777… → Repeating
This concept is simple but frequently asked in both theory and MCQ sections. Therefore, it’s a must-learn part of class 10 maths chapter 1 solutions.
Summary of NCERT Solutions For Class 10 Maths Chapter 1
Here is a summary of the key points covered in Class 10 Maths Chapter 1 Solutions:
NCERT Solutions For Class 10 Maths Chapter 1 covers essential topics like Euclid’s Division Lemma, Fundamental Theorem of Arithmetic, Prime Factorization, Irrational Numbers, and Decimal Expansions to help students grasp foundational mathematical concepts. Mastery of Chapter 1 is crucial for board exams as it lays the groundwork for subsequent topics and frequently tested proofs. Understanding the content in Chapter 1 not only enhances exam performance but also paves the way for success in competitive tests and real-world problem-solving scenarios.
- Euclid’s Division Lemma – Used to divide integers and find HCF
- Fundamental Theorem of Arithmetic – All composite numbers have unique prime factorizations
- Prime Factorization – Helps calculate HCF & LCM
- Irrational Numbers – Proofs using contradiction and arithmetic theorem
- Decimal Expansions – Recognize terminating vs repeating decimals based on prime factors
Why Chapter 1 Is Important in Class 10 Maths
Understanding class 10 maths chapter 1 solutions helps build a strong foundation for:
- Algebra
- Number systems
- Polynomials
- Real-life problems involving divisibility, LCM, or HCF
- Mathematical proofs and logic
It’s also a frequently asked chapter in CBSE and school-level exams.
FAQs – Class 10 Maths Chapter 1 Solutions
Q1. What is Euclid’s Lemma in NCERT Solutions For Class 10 Maths Chapter 1?
It’s a formula used to divide two integers. It states: for any two positive integers a and b, there exist q and r such that a = bq + r, with 0 ≤ r < b.
Q2. How does the Fundamental Theorem of Arithmetic help?
It ensures every composite number has a unique prime factorization, which is useful in solving HCF, LCM, and irrational number proofs.
Q3. What is the difference between rational and irrational numbers in this chapter?
Rational numbers can be written as p/q and have predictable decimal expansions. Irrational numbers like √2 cannot be written in p/q form and their decimals never repeat or terminate.
Q4. Is Chapter 1 important for board exams?
Absolutely! Class 10 Maths Chapter 1 Solutions form the base for several other topics. Proofs from this chapter also appear frequently in exams.
Q5. How do I know if a decimal will terminate?
Check the prime factors of the denominator. If only 2 and/or 5 are present, the decimal terminates.
Final Note
NCERT Solutions For Class 10 Maths Chapter 1 is the first step toward scoring well in mathematics. This chapter blends logic, proof, arithmetic, and applications — making it fundamental not just for exams, but also for competitive tests and real-life problem-solving.
Whether you’re a student or educator, understanding these key ideas in class 10 maths chapter 1 solutions can build a strong mathematical foundation for future success.