NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers

These Class 10 Maths Chapter 1 solutions cover Real Numbers from the NCERT textbook (Reprint 2026–27). Every question of Exercise 1.1 and Exercise 1.2 is solved step by step — the Fundamental Theorem of Arithmetic, prime factorisation, HCF & LCM, and proofs of irrationality — with each numerical answer cross-checked against the book’s answer appendix.

Class: 10 Subject: Mathematics Chapter: 1 – Real Numbers Exercises: 1.1, 1.2 Session: 2026–27 Board: CBSE (NCERT)
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Chapter 1 Overview

Chapter 1 of Class 10 Mathematics, Real Numbers, builds on the real numbers you met in Class IX. It opens with the Fundamental Theorem of Arithmetic — the statement that every composite number can be factorised as a product of primes in a unique way. This single idea is then used in two powerful ways: first to compute the HCF and LCM of numbers by prime factorisation, and second to prove that numbers like √2, √3 and √5 are irrational. Along the way you revisit the relation HCF(a, b) × LCM(a, b) = a × b, and learn the technique of proof by contradiction. The Class 10 Maths Chapter 1 solutions below work through every question of Exercise 1.1 and Exercise 1.2 step by step.

Key Concepts & Definitions

Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Prime factorisation: writing a number as a product of powers of primes, e.g. 32760 = 23 × 32 × 5 × 7 × 13.

HCF (Highest Common Factor): the product of the smallest power of each common prime factor of the numbers.

LCM (Least Common Multiple): the product of the greatest power of each prime factor involved in the numbers.

Rational number: a number that can be written as p⁄q where p and q are integers and q ≠ 0.

Irrational number: a real number that cannot be written in the form p⁄q; e.g. √2, √3, √5, π.

Coprime numbers: two numbers whose only common factor is 1, e.g. 8 and 9.

Proof by contradiction: assume the opposite of what you want to prove, then show it leads to an impossibility, so the original statement must be true.

Important Formulas & Results (Chapter 1)

HCF × LCM rule (two numbers): HCF(a, b) × LCM(a, b) = a × b.

Finding LCM from HCF: LCM(a, b) = (a × b) ⁄ HCF(a, b).

HCF by prime factorisation: product of the smallest power of each common prime.

LCM by prime factorisation: product of the greatest power of every prime appearing.

Key irrationality lemma: If a prime p divides a2, then p divides a (a a positive integer).

Note: For three numbers, HCF(p, q, r) × LCM(p, q, r) is generally not equal to p × q × r.

Exercise 1.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given in the appendix.

1. Express each number as a product of its prime factors: (i) 140   (ii) 156   (iii) 3825   (iv) 5005   (v) 7429

SOLUTION (i) 140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7 = 22 × 5 × 7. (ii) 156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13 = 22 × 3 × 13. (iii) 3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17. (iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13. (v) 7429 = 17 × 437 = 17 × 19 × 23 = 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   (ii) 510 and 92   (iii) 336 and 54

SOLUTION (i) 26 = 2 × 13,   91 = 7 × 13. HCF = 13; LCM = 2 × 7 × 13 = 182. Check: 13 × 182 = 2366 = 26 × 91. ✓ (HCF = 13, LCM = 182) (ii) 510 = 2 × 3 × 5 × 17,   92 = 22 × 23. HCF = 2; LCM = 22 × 3 × 5 × 17 × 23 = 23460. Check: 2 × 23460 = 46920 = 510 × 92. ✓ (HCF = 2, LCM = 23460) (iii) 336 = 24 × 3 × 7,   54 = 2 × 33. HCF = 2 × 3 = 6; LCM = 24 × 33 × 7 = 3024. Check: 6 × 3024 = 18144 = 336 × 54. ✓ (HCF = 6, LCM = 3024)

3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21   (ii) 17, 23 and 29   (iii) 8, 9 and 25

SOLUTION (i) 12 = 22 × 3,   15 = 3 × 5,   21 = 3 × 7. HCF = 3 (only common prime); LCM = 22 × 3 × 5 × 7 = 420. (HCF = 3, LCM = 420) (ii) 17, 23 and 29 are all primes, with no common factor. HCF = 1; LCM = 17 × 23 × 29 = 11339. (HCF = 1, LCM = 11339) (iii) 8 = 23,   9 = 32,   25 = 52. No common prime, so HCF = 1; LCM = 23 × 32 × 52 = 1800. (HCF = 1, LCM = 1800)

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

SOLUTION Using HCF × LCM = product of the two numbers: LCM(306, 657) = (306 × 657) ⁄ HCF(306, 657) = 201042 ⁄ 9 = 22338.

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

SOLUTION A number ends with 0 only if it is divisible by 10 = 2 × 5, i.e. its prime factorisation must contain both 2 and 5. Now 6n = (2 × 3)n = 2n × 3n. The only primes here are 2 and 3; the prime 5 never appears. By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), there is no 5 in 6n, so 6n is never divisible by 10. ∴ 6n can never 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 A composite number has at least one factor other than 1 and itself. Take out the common factor in each expression. First number: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78. Since it is 13 × 78 (a product of two numbers greater than 1), it is composite. (Value = 1014.) Second number: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009. Being 5 × 1009, it has a factor 5 other than 1 and itself, so it is composite. (Value = 5045.)

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 They meet again at the starting point after a time that is a common multiple of both their round times. The first such time is the LCM of 18 and 12. 18 = 2 × 32,   12 = 22 × 3. LCM = 22 × 32 = 36. ∴ they will meet again at the starting point after 36 minutes.

Exercise 1.2 Solutions

These are proof questions; each is proved by the method of contradiction using the result “if a prime p divides a2, then p divides a.”

1. Prove that √5 is irrational.

SOLUTION Assume, to the contrary, that √5 is rational. Then we can write √5 = a⁄b, where a and b are coprime integers and b ≠ 0. So b√5 = a. Squaring both sides: 5b2 = a2. Hence 5 divides a2, and since 5 is prime, 5 divides a. Write a = 5c for some integer c. Substituting: 5b2 = (5c)2 = 25c2, so b2 = 5c2. Hence 5 divides b2, and so 5 divides b. Thus 5 divides both a and b, so they have a common factor 5. This contradicts the assumption that a and b are coprime. The contradiction arose from assuming √5 is rational. ∴ √5 is irrational. ■

2. Prove that 3 + 2√5 is irrational.

SOLUTION Assume, to the contrary, that 3 + 2√5 is rational. Then we can write 3 + 2√5 = a⁄b, where a and b are integers and b ≠ 0. Rearranging: 2√5 = a⁄b − 3 = (a − 3b)⁄b, so √5 = (a − 3b)⁄(2b). Since a, b and 3 are integers, (a − 3b)⁄(2b) is a rational number, which would make √5 rational. But √5 is irrational (proved in Q1). This is a contradiction. The contradiction arose from assuming 3 + 2√5 is rational. ∴ 3 + 2√5 is irrational. ■

3. Prove that the following are irrationals: (i) 1⁄√2   (ii) 7√5   (iii) 6 + √2

SOLUTION (i) 1⁄√2. Assume 1⁄√2 is rational, say 1⁄√2 = a⁄b (a, b integers, b ≠ 0). Then √2 = b⁄a, which is rational since a and b are integers. But √2 is irrational — contradiction. ∴ 1⁄√2 is irrational. (ii) 7√5. Assume 7√5 is rational, say 7√5 = a⁄b. Then √5 = a⁄(7b), which is rational. But √5 is irrational — contradiction. ∴ 7√5 is irrational. (iii) 6 + √2. Assume 6 + √2 is rational, say 6 + √2 = a⁄b. Then √2 = a⁄b − 6 = (a − 6b)⁄b, which is rational. But √2 is irrational — contradiction. ∴ 6 + √2 is irrational.

Common Mistakes to Avoid

Watch out for these

  • For HCF, taking the greatest power of common primes — HCF uses the smallest power; LCM uses the greatest.
  • Applying HCF × LCM = product to three or more numbers — this relation only holds for two numbers.
  • In the “ends with 0” question, forgetting that ending in 0 requires the prime 5 (and 2), not just an even number.
  • In irrationality proofs, forgetting to first state “a and b are coprime” — the whole contradiction depends on it.
  • Writing “5 divides a2 so 5 divides a” without justification — cite the result that for a prime p, p | a2 ⇒ p | a.
  • Treating numbers like 7 × 11 × 13 + 13 as prime — always check for a common factor that can be taken out.

Practice MCQs & Assertion–Reason

1. The prime factorisation of 140 is:

(a) 2 × 72 × 5    (b) 22 × 5 × 7    (c) 22 × 52 × 7    (d) 2 × 5 × 7

2. The HCF of 26 and 91 is:

(a) 7    (b) 13    (c) 26    (d) 182

3. If HCF(a, b) = 9 and a × b = 201042, then LCM(a, b) is:

(a) 9    (b) 2234    (c) 22338    (d) 201042

4. For which value can 6n end with the digit 0 (n a natural number)?

(a) n = 1    (b) n = 5    (c) n = 10    (d) for no value of n

5. The LCM of 8, 9 and 25 is:

(a) 1800    (b) 900    (c) 200    (d) 3600

6. The number 7 × 11 × 13 + 13 is:

(a) a prime number    (b) a composite number    (c) an irrational number    (d) 1

7. According to the Fundamental Theorem of Arithmetic, the prime factorisation of a composite number is:

(a) never unique    (b) unique except for the order of factors    (c) unique only for even numbers    (d) always a single prime

8. Which of the following is an irrational number?

(a) 0.25    (b) 22⁄7    (c) √5    (d) −3

9. If a prime p divides a2 (a a positive integer), then:

(a) p divides a    (b) p divides 2a    (c) p2 divides a    (d) p does not divide a

10. The smallest number after which Sonia (18 min/round) and Ravi (12 min/round) meet again at the start is:

(a) 6 min    (b) 30 min    (c) 36 min    (d) 216 min

Answer key: 1-(b), 2-(b), 3-(c), 4-(d), 5-(a), 6-(b), 7-(b), 8-(c), 9-(a), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: 6n can never end with the digit 0 for any natural number n.

Reason: 6n = 2n × 3n contains no factor of 5, so it is not divisible by 10.

A-R 2. Assertion: The HCF of two numbers can be greater than their LCM.

Reason: For any two positive integers, HCF(a, b) × LCM(a, b) = a × b.

A-R 3. Assertion: √5 is an irrational number.

Reason: If a prime p divides a2, then p divides a, where a is a positive integer.

A-R 4. Assertion: The number 7 × 11 × 13 + 13 is composite.

Reason: 7 × 11 × 13 + 13 = 13 × 78, which has a factor 13 other than 1 and itself.

A-R 5. Assertion: The prime factorisation of every composite number is unique.

Reason: This uniqueness is stated by Euclid’s division algorithm.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(C).

Quick Revision Summary

  • Fundamental Theorem of Arithmetic: every composite number is a unique product of primes (apart from order).
  • HCF = product of smallest powers of common primes; LCM = product of greatest powers of all primes.
  • For two numbers, HCF × LCM = product of the numbers; so LCM = (a × b) ⁄ HCF.
  • A number ends in 0 only if both 2 and 5 are in its prime factorisation; 6n never does.
  • If a prime p divides a2, then p divides a — the key lemma for irrationality proofs.
  • √2, √3, √5 are irrational; proved by contradiction using coprime a, b.
  • The sum or product of a non-zero rational with an irrational is irrational (e.g. 3 + 2√5, 7√5).

How to score full marks in this chapter

Always start HCF/LCM questions by writing each number’s prime factorisation neatly, then read off the smallest powers (HCF) and greatest powers (LCM). In irrationality proofs, follow the fixed skeleton — assume rational, write as a⁄b with a, b coprime, square, deduce p divides a then p divides b, and conclude a contradiction. Quote the lemma “p | a2 ⇒ p | a” explicitly to secure the method marks.

Frequently Asked Questions

What is Class 10 Maths Chapter 1 Real Numbers about?

Chapter 1, Real Numbers, covers the Fundamental Theorem of Arithmetic, finding HCF and LCM by prime factorisation, the relation HCF × LCM = product of two numbers, and proofs that numbers like √2, √3 and √5 are irrational using the method of contradiction.

How many exercises are there in Class 10 Maths Chapter 1?

There are two exercises — Exercise 1.1 (prime factorisation, HCF and LCM, with 7 questions) and Exercise 1.2 (proving numbers irrational, with 3 questions) — all solved step by step on this page.

How do you prove √5 is irrational?

Assume √5 = a⁄b with a and b coprime. Squaring gives 5b2 = a2, so 5 divides a, say a = 5c. Then b2 = 5c2, so 5 also divides b. This makes 5 a common factor of a and b, contradicting that they are coprime — so √5 is irrational.

Are these Class 10 Maths Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Class 10 Mathematics textbook for the 2026–27 session, with every numerical answer verified against the book’s answer appendix.

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