Saraswati Vidyamandir Logo
Book Free Orientation
Home Our Courses CBSE Resources
NCERT Solutions
Science — Class 9 Science — Class 10 Maths — Class 9 Maths — Class 10 About Us Contact Us Book Free Orientation
1

Real Numbers

Class 10 Maths  ·  NCERT Solutions  ·  10 Questions

All Chapters
10 Questions & Answers
Chapter 1: Real Numbers — NCERT Solutions (rationalised syllabus 2026–27). This chapter covers the Fundamental Theorem of Arithmetic (every composite number can be expressed as a product of primes in a unique way), its use in finding HCF and LCM by prime factorisation, and proofs of irrationality by contradiction. Key result: for two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.

Exercise 1.1

Q1

Express each number as a product of its prime factors:

(i) 140   (ii) 156   (iii) 3825   (iv) 5005   (v) 7429

Answer

We divide each number successively by prime numbers, starting from the smallest:

(i) 140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7

\(140 = 2^{2} \times 5 \times 7\)

(ii) 156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13

\(156 = 2^{2} \times 3 \times 13\)

(iii) 3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17

\(3825 = 3^{2} \times 5^{2} \times 17\)

(iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13

\(5005 = 5 \times 7 \times 11 \times 13\)

(v) 7429 = 17 × 437 = 17 × 19 × 23

\(7429 = 17 \times 19 \times 23\)
Q2

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

Answer

(i) 26 and 91:   26 = 2 × 13 and 91 = 7 × 13

HCF = product of the smallest power of each common prime factor = 13
LCM = product of the greatest power of each prime factor = 2 × 7 × 13 = 182

\(\text{HCF} \times \text{LCM} = 13 \times 182 = 2366 = 26 \times 91\ \checkmark\)

(ii) 510 and 92:   510 = 2 × 3 × 5 × 17 and 92 = 2² × 23

HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460

\(\text{HCF} \times \text{LCM} = 2 \times 23460 = 46920 = 510 \times 92\ \checkmark\)

(iii) 336 and 54:   336 = 2⁴ × 3 × 7 and 54 = 2 × 3³

HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024

\(\text{HCF} \times \text{LCM} = 6 \times 3024 = 18144 = 336 \times 54\ \checkmark\)
Q3

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

Answer

(i) 12, 15 and 21:   12 = 2² × 3,   15 = 3 × 5,   21 = 3 × 7

The only prime common to all three is 3, so HCF = 3.

\(\text{LCM} = 2^{2} \times 3 \times 5 \times 7 = 420\)

(ii) 17, 23 and 29:   each number is prime, so they share no common factor other than 1.

HCF = 1

\(\text{LCM} = 17 \times 23 \times 29 = 11339\)

(iii) 8, 9 and 25:   8 = 2³,   9 = 3²,   25 = 5²

The three numbers are pairwise co-prime, so HCF = 1.

\(\text{LCM} = 2^{3} \times 3^{2} \times 5^{2} = 1800\)
Q4

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

Answer

For any two positive integers a and b:

\(\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b\)

Therefore:

\(\text{LCM}(306, 657) = \dfrac{306 \times 657}{9} = \dfrac{201042}{9} = 22338\)

LCM (306, 657) = 22338

Q5

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

Answer

A number ends with the digit 0 only if it is divisible by 10, i.e. its prime factorisation must contain both 2 and 5.

\(6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}\)

The prime factorisation of 6n contains only the primes 2 and 3 — the prime 5 does not occur anywhere.

By the Fundamental Theorem of Arithmetic, the prime factorisation of every composite number is unique. So no other prime (in particular, 5) can ever appear in the factorisation of 6n.

Hence 6n can never end with the digit 0 for any natural number n.

Q6

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer

A composite number has at least one factor other than 1 and itself. We take out the common factor in each expression:

First number:

\(7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times 78 = 1014\)

Since 13 and 78 are factors of 1014 (besides 1 and 1014), it is composite.

Second number:

\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009 = 5045\)

Since 5 and 1009 are factors of 5045 (besides 1 and 5045), it is also composite.

Q7

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?

Answer

Both will be at the starting point together after a time that is a common multiple of their round times. The earliest such time is the LCM of 18 and 12.

\(18 = 2 \times 3^{2}, \qquad 12 = 2^{2} \times 3\)
\(\text{LCM}(18, 12) = 2^{2} \times 3^{2} = 36\)

Sonia and Ravi will meet again at the starting point after 36 minutes (by then Sonia completes 2 rounds and Ravi completes 3 rounds).

Exercise 1.2

Q8

Prove that \(\sqrt{5}\) is irrational.

Answer

Proof by contradiction: Assume, to the contrary, that \(\sqrt{5}\) is rational. Then we can write:

\(\sqrt{5} = \dfrac{a}{b}\)

where a and b are co-prime integers (no common factor other than 1) and b ≠ 0.

Rearranging: \(b\sqrt{5} = a\). Squaring both sides:

\(5b^{2} = a^{2}\)

So 5 divides a². Since 5 is prime, by the theorem “if a prime p divides a², then p divides a” — 5 divides a. Write a = 5c for some integer c. Substituting:

\(5b^{2} = (5c)^{2} = 25c^{2} \;\Rightarrow\; b^{2} = 5c^{2}\)

So 5 divides b², and hence 5 divides b.

Therefore a and b have 5 as a common factor — this contradicts the fact that a and b are co-prime.

This contradiction has arisen because of our wrong assumption that \(\sqrt{5}\) is rational. Hence \(\sqrt{5}\) is irrational. ∎

Q9

Prove that \(3 + 2\sqrt{5}\) is irrational.

Answer

Proof by contradiction: Assume, to the contrary, that \(3 + 2\sqrt{5}\) is rational. Then:

\(3 + 2\sqrt{5} = \dfrac{a}{b}\)

where a and b are co-prime integers and b ≠ 0. Rearranging:

\(2\sqrt{5} = \dfrac{a}{b} - 3 \;\Rightarrow\; \sqrt{5} = \dfrac{a - 3b}{2b}\)

Since a and b are integers, \(\dfrac{a - 3b}{2b}\) is a rational number. This would mean \(\sqrt{5}\) is rational.

But this contradicts the established fact that \(\sqrt{5}\) is irrational (proved in Q8).

Therefore our assumption is false, and \(3 + 2\sqrt{5}\) is irrational. ∎

Q10

Prove that the following are irrationals:

(i) \(\dfrac{1}{\sqrt{2}}\)    (ii) \(7\sqrt{5}\)    (iii) \(6 + \sqrt{2}\)

Answer

In each part we assume the number is rational and derive a contradiction.

(i) Assume \(\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}\) with a, b co-prime integers, b ≠ 0, a ≠ 0. Then:

\(\sqrt{2} = \dfrac{b}{a}\)

Since a and b are integers, \(\dfrac{b}{a}\) is rational — so \(\sqrt{2}\) would be rational, contradicting the fact that \(\sqrt{2}\) is irrational. Hence \(\dfrac{1}{\sqrt{2}}\) is irrational.

(ii) Assume \(7\sqrt{5} = \dfrac{a}{b}\) with a, b co-prime integers, b ≠ 0. Then:

\(\sqrt{5} = \dfrac{a}{7b}\)

Since a and b are integers, \(\dfrac{a}{7b}\) is rational — so \(\sqrt{5}\) would be rational, a contradiction. Hence \(7\sqrt{5}\) is irrational.

(iii) Assume \(6 + \sqrt{2} = \dfrac{a}{b}\) with a, b co-prime integers, b ≠ 0. Then:

\(\sqrt{2} = \dfrac{a}{b} - 6 = \dfrac{a - 6b}{b}\)

Since a and b are integers, \(\dfrac{a - 6b}{b}\) is rational — so \(\sqrt{2}\) would be rational, a contradiction. Hence \(6 + \sqrt{2}\) is irrational. ∎

All Chapters