---
title: "NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers"
url: https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions
dateModified: 2026-10-07T14:41:11+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

This chapter's questions cover prime factorisation, finding HCF and LCM of integers, checking properties of numbers such as whether they can end with the digit zero or are composite, and proving the irrationality of various numbers.

Free PDF (7 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-1-real-numbers-a484766ea2.pdf

## EXERCISE 1.1

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Divide 140 by the smallest prime factor 2 to get 70.
2. Divide 70 by 2 to get 35, divide 35 by 5 to get 7, and 7 by 7 to get 1.
3. Express 140 as $2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7$.

Answer (i): $2^2 \times 5 \times 7$

**Part (ii)**

1. Divide 156 by the smallest prime factor 2 to get 78.
2. Divide 78 by 2 to get 39, divide 39 by 3 to get 13, and 13 by 13 to get 1.
3. Express 156 as $2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$.

Answer (ii): $2^2 \times 3 \times 13$

**Part (iii)**

1. Divide 3825 by 3 to get 1275, and again by 3 to get 425.
2. Divide 425 by 5 to get 85, divide 85 by 5 to get 17, and 17 by 17 to get 1.
3. Express 3825 as $3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$.

Answer (iii): $3^2 \times 5^2 \times 17$

**Part (iv)**

1. Divide 5005 by the prime factor 5 to get 1001.
2. Divide 1001 by 7 to get 143, divide 143 by 11 to get 13, and 13 by 13 to get 1.
3. Express 5005 as $5 \times 7 \times 11 \times 13$.

Answer (iv): $5 \times 7 \times 11 \times 13$

**Part (v)**

1. Divide 7429 by 17 to get 437.
2. Divide 437 by 19 to get 23, and 23 by 23 to get 1.
3. Express 7429 as $17 \times 19 \times 23$.

Answer (v): $17 \times 19 \times 23$

**Answer:** (i) $2^2 \times 5 \times 7$, (ii) $2^2 \times 3 \times 13$, (iii) $3^2 \times 5^2 \times 17$, (iv) $5 \times 7 \times 11 \times 13$, (v) $17 \times 19 \times 23$

> Common mistake: Stopping the prime factorisation before all factors are prime numbers, or writing composite numbers instead of primes.

### Question 2

*3 marks · Short answer*

Find the LCM and HCF of the following pairs of integers and verify that $\text{LCM} \times \text{HCF} = \text{product of the two numbers.}$
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

**Part (i)**

1. $26 = 2 \times 13$ and $91 = 7 \times 13$
2. $\text{HCF}(26, 91) = 13$ and $\text{LCM}(26, 91) = 2 \times 7 \times 13 = 182$
3. $\text{HCF} \times \text{LCM} = 13 \times 182 = 2366$ and product $= 26 \times 91 = 2366$, hence verified.

Answer (i): HCF = 13, LCM = 182, Product = 2366

**Part (ii)**

1. $510 = 2 \times 3 \times 5 \times 17$ and $92 = 2^2 \times 23$
2. $\text{HCF}(510, 92) = 2$ and $\text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$
3. $\text{HCF} \times \text{LCM} = 2 \times 23460 = 46920$ and product $= 510 \times 92 = 46920$, hence verified.

Answer (ii): HCF = 2, LCM = 23460, Product = 46920

**Part (iii)**

1. $336 = 2^4 \times 3 \times 7$ and $54 = 2 \times 3^3$
2. $\text{HCF}(336, 54) = 2 \times 3 = 6$ and $\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 3024$
3. $\text{HCF} \times \text{LCM} = 6 \times 3024 = 18144$ and product $= 336 \times 54 = 18144$, hence verified.

Answer (iii): HCF = 6, LCM = 3024, Product = 18144

**Answer:** LCM and HCF found and verified for all pairs.

> Common mistake: Confusing the powers of common and non-common prime factors when calculating LCM and HCF.

### Question 3

*3 marks · Short answer*

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

**Part (i)**

1. $12 = 2^2 \times 3$, $15 = 3 \times 5$, $21 = 3 \times 7$
2. $\text{HCF}(12, 15, 21) = 3$
3. $\text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 420$

Answer (i): HCF = 3, LCM = 420

**Part (ii)**

1. $17 = 17$, $23 = 23$, $29 = 29$
2. $\text{HCF}(17, 23, 29) = 1$
3. $\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 11339$

Answer (ii): HCF = 1, LCM = 11339

**Part (iii)**

1. $8 = 2^3$, $9 = 3^2$, $25 = 5^2$
2. $\text{HCF}(8, 9, 25) = 1$
3. $\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 1800$

Answer (iii): HCF = 1, LCM = 1800

**Answer:** Found HCF and LCM for the given sets of integers.

> Common mistake: Taking common factors when none exist, resulting in an HCF other than 1.

### Question 4

*3 marks · Short answer*

Given that $\text{HCF} (306, 657) = 9$, find $\text{LCM} (306, 657)$.

**Solution**

1. We know that for two positive integers $a$ and $b$, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.
2. Substituting the given values, $9 \times \text{LCM}(306, 657) = 306 \times 657$.
3. $\text{LCM}(306, 657) = \frac{306 \times 657}{9}$.
4. $\text{LCM}(306, 657) = 34 \times 657 = 22338$.

**Answer:** 22338

> Common mistake: Multiplying the numbers first and making calculation errors while dividing by HCF.

### Question 5

*3 marks · Short answer*

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

**Solution**

1. If the number $6^n$, for any $n$, were to end with the digit zero, then it would be divisible by 5.
2. That is, the prime factorisation of $6^n$ would contain the prime 5.
3. However, $6^n = (2 \times 3)^n = 2^n \times 3^n$; so the only primes in the factorisation of $6^n$ are 2 and 3.
4. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of $6^n$.
5. Therefore, there is no natural number $n$ for which $6^n$ ends with the digit zero.

**Answer:** No, $6^n$ cannot end with the digit 0 for any natural number $n$.

> Common mistake: Stating that $6^n$ can end with zero because 6 is a multiple of 2 and 3 without mentioning the necessity of factor 5.

### Question 6

*3 marks · Short answer*

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

**Solution**

1. Given numbers can be simplified by taking out common factors as per the distributive property.
2. For the first number: $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 \times 1 + 1) = 13 \times (77 + 1) = 13 \times 78$.
3. Since $13 \times 78$ has factors other than 1 and itself, it is a composite number.
4. For the second number: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$.
5. Since $5 \times 1009$ has factors other than 1 and itself, it is a composite number.

**Answer:** Both numbers are composite because they can be expressed as a product of factors other than 1 and themselves.

> Common mistake: Multiplying out the entire expression fully instead of factoring out the common term.

### Question 7

*3 marks · Short answer*

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**

1. Time taken by Sonia to drive one round = 18 minutes
2. Time taken by Ravi to drive one round = 12 minutes
3. The time when they meet again at the starting point is given by the LCM of 18 and 12
4. Find the prime factorisation of 18 and 12: $18 = 2 \times 3^2$ and $12 = 2^2 \times 3$
5. Calculate the LCM of 18 and 12: $\text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$
6. Sonia and Ravi will meet again at the starting point after 36 minutes

**Answer:** 36 minutes

> Common mistake: Finding the HCF instead of the LCM of the two numbers.

## EXERCISE 1.2

### Question 1

*3 marks · Proof*

Prove that $\sqrt{5}$ is irrational.

**Solution**

1. Let us assume, to the contrary, that $\sqrt{5}$ is rational.
2. That is, we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \frac{a}{b}$.
3. Suppose $a$ and $b$ have a common factor other than 1, then we divide by the common factor, and assume that $a$ and $b$ are coprime.
4. So, $b\sqrt{5} = a$.
5. Squaring on both sides, and rearranging, we get $5b^2 = a^2$.
6. Therefore, $a^2$ is divisible by 5, and by Theorem 1.2, it follows that $a$ is also divisible by 5.
7. So, we can write $a = 5c$ for some integer $c$.
8. Substituting for $a$, we get $5b^2 = 25c^2$, that is, $b^2 = 5c^2$.
9. This means that $b^2$ is divisible by 5, and so $b$ is also divisible by 5 (using Theorem 1.2).
10. Therefore, $a$ and $b$ have at least 5 as a common factor.
11. But this contradicts the fact that $a$ and $b$ are coprime.
12. This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.
13. So, we conclude that $\sqrt{5}$ is irrational. Hence proved.

**Answer:** $\sqrt{5}$ is irrational.

> Common mistake: Forgetting to state that $a$ and $b$ are coprime after cancelling common factors.

### Question 2

*3 marks · Proof*

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

**Solution**

1. Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $3 + 2\sqrt{5} = \frac{a}{b}$.
3. Rearranging this equation, we get $2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$.
4. Therefore, $\sqrt{5} = \frac{a - 3b}{2b}$.
5. Since $3$, $a$, and $b$ are integers, $\frac{a - 3b}{2b}$ is rational, and so $\sqrt{5}$ is rational.
6. But this contradicts the fact that $\sqrt{5}$ is irrational.
7. This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.
8. So, we conclude that $3 + 2\sqrt{5}$ is irrational. Hence proved.

**Answer:** $3 + 2\sqrt{5}$ is irrational.

> Common mistake: Attempting to prove $\sqrt{5}$ is irrational all over again instead of using the known result.

### Question 3

*3 marks · Proof*

Prove that the following are irrationals :
(i) $\frac{1}{\sqrt{2}}$
(ii) $7\sqrt{5}$
(iii) $6 + \sqrt{2}$

**Part (i)**

1. Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $\frac{1}{\sqrt{2}} = \frac{a}{b}$.
3. Taking the reciprocal, we get $\sqrt{2} = \frac{b}{a}$.
4. Since $a$ and $b$ are integers, $\frac{b}{a}$ is rational, which implies that $\sqrt{2}$ is rational.
5. This contradicts the fact that $\sqrt{2}$ is irrational.
6. So, we conclude that $\frac{1}{\sqrt{2}}$ is irrational. Hence proved.

Answer (i): $\frac{1}{\sqrt{2}}$ is irrational.

**Part (ii)**

1. Let us assume, to the contrary, that $7\sqrt{5}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $7\sqrt{5} = \frac{a}{b}$.
3. Rearranging, we get $\sqrt{5} = \frac{a}{7b}$.
4. Since $7$, $a$, and $b$ are integers, $\frac{a}{7b}$ is rational, and so $\sqrt{5}$ is rational.
5. This contradicts the fact that $\sqrt{5}$ is irrational.
6. So, we conclude that $7\sqrt{5}$ is irrational. Hence proved.

Answer (ii): $7\sqrt{5}$ is irrational.

**Part (iii)**

1. Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $6 + \sqrt{2} = \frac{a}{b}$.
3. Rearranging, we get $\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}$.
4. Since $6$, $a$, and $b$ are integers, $\frac{a - 6b}{b}$ is rational, and so $\sqrt{2}$ is rational.
5. This contradicts the fact that $\sqrt{2}$ is irrational.
6. So, we conclude that $6 + \sqrt{2}$ is irrational. Hence proved.

Answer (iii): $6 + \sqrt{2}$ is irrational.

**Answer:** All given numbers are irrational.

> Common mistake: Rationalising the denominator for part (i) incorrectly or missing the contradiction step.

## Frequently asked questions

### How many exercises and questions are there in Class 10 Maths Chapter 1 Real Numbers for the 2026-27 session?

This chapter in the NCERT textbook has a total of 2 exercises with 10 questions in all. Exercise 1.1 contains 7 questions, and Exercise 1.2 contains 3 questions.

### What mathematical topics are covered in the exercises of this chapter?

Exercise 1.1 covers prime factorisation, HCF and LCM of three numbers, the relation between HCF and LCM, LCM word problems, and composite numbers. Exercise 1.2 focuses on proving the irrationality of numbers involving surds and square roots of primes.

### Which question type is considered the hardest in Class 10 Maths Real Numbers and how should I approach it?

Proving the irrationality of numbers involving surds in Exercise 1.2 is often found difficult by students. You can approach this by using the method of contradiction and assuming the number is rational before reaching a logical impossibility.

### How can I write answers in board exams to score full marks for these Real Numbers questions?

To get full marks, you must write all logical steps clearly, state prime factorisations correctly, and explicitly mention theorems like the Fundamental Theorem of Arithmetic. SwaVid's step-by-step solutions are available on this page to help you master this presentation format.

### Is the free PDF of NCERT Solutions for Class 10 Maths Chapter 1 available here?

Yes, SwaVid's free PDF and detailed step-by-step solutions for this chapter are provided right on this page. You can easily download or view them to assist with your practice.

## Related pages

- [Exercise 1.1 solutions](https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions/exercise-1-1)
- [Exercise 1.2 solutions](https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions/exercise-1-2)
- [Real Numbers: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/real-numbers)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student