In This Post we are  providing Chapter 1 Real Numbers NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2025 syllabus. These Real Numbers Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Real Number NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

 

 

Class 10 Maths NCERT Solutions: Chapter 1 – Real Numbers (Latest Textbook 2025-26)

 

This chapter explores the fundamental concepts of real numbers, including the Fundamental Theorem of Arithmetic, the nature of irrational numbers, and the decimal expansion of rational numbers.

 

Exercise 1.1 (Fundamental Theorem of Arithmetic)

 

1. Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solutions:

(i) 140=22×5×7

(ii) 156=22×3×13

(iii) 3825=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

(ii) 510 and 92

(iii) 336 and 54

Solutions:

(i) 26 and 91

– Prime factors of 26: 2×13

– Prime factors of 91: 7×13

– HCF(26, 91) = 13

– LCM(26, 91) = 2×7×13=182

– Verification:

– Product of two numbers = 26×91=2366

– LCM × HCF = 182×13=2366

– Hence, verified.

(ii) 510 and 92

– Prime factors of 510: 2×3×5×17

– Prime factors of 92: 22×23

– HCF(510, 92) = 2

– LCM(510, 92) = 22×3×5×17×23=23460

– Verification:

– Product of two numbers = 510×92=46920

– LCM × HCF = 23460×2=46920

– Hence, verified.

(iii) 336 and 54

– Prime factors of 336: 24×3×7

– Prime factors of 54: 2×33

– HCF(336, 54) = 2×3=6

– LCM(336, 54) = 24×33×7=3024

– Verification:

– Product of two numbers = 336×54=18144

– LCM × HCF = 3024×6=18144

– Hence, verified.

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

Solutions:

(i) 12, 15 and 21

– Prime factors of 12: 22×3

– Prime factors of 15: 3×5

– Prime factors of 21: 3×7

– HCF(12, 15, 21) = 3

– LCM(12, 15, 21) = 22×3×5×7=420

(ii) 17, 23 and 29

– Prime factors of 17: 17

– Prime factors of 23: 23

– Prime factors of 29: 29

– HCF(17, 23, 29) = 1

– LCM(17, 23, 29) = 17×23×29=11339

(iii) 8, 9 and 25

– Prime factors of 8: 23

– Prime factors of 9: 32

– Prime factors of 25: 52

– HCF(8, 9, 25) = 1

– LCM(8, 9, 25) = 23×32×52=1800

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

Solution:

We know that for any two positive integers a and b:

HCF$(a, b)$ × LCM$(a, b)$ = a×b

Given HCF(306, 657) = 9.

Substituting the values:

9× LCM(306, 657) = 306×657

LCM(306, 657) = 9306×657​

LCM(306, 657) = 34×657

LCM(306, 657) = 22338

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

Solution:

For a number to end with the digit 0, it must have 5 as a prime factor.

The prime factorization of 6n is (2×3)n=2n×3n.

The only prime factors of 6n are 2 and 3. There is no 5 in its prime factorization.

By the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique.

Therefore, 6n cannot end with the digit 0 for any natural number n.

6. Explain why and are composite numbers.

Solution:

A composite number is a natural number that has more than two factors (i.e., it can be divided by numbers other than 1 and itself).

  • For the first expression:

    7×11×13+13

    Taking 13 as a common factor:

    13×(7×11+1)

    13×(77+1)

    13×78

    Since 78=2×3×13, the expression is 13×2×3×13=2×3×132.

    Since it has factors other than 1 and itself (namely 2, 3, and 13), it is a composite number.

  • For the second expression:

    7×6×5×4×3×2×1+5

    Taking 5 as a common factor:

    5×(7×6×4×3×2×1+1)

    5×(1008+1)

    5×1009

    Since 1009 is a prime number, the expression has factors 5 and 1009 (besides 1 and itself).

    Therefore, 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:

To find when they will meet again at the starting point, we need to find the Least Common Multiple (LCM) of the time taken by Sonia and Ravi.

  • Time taken by Sonia = 18 minutes

  • Time taken by Ravi = 12 minutes

Prime factorization of 18: 2×32

Prime factorization of 12: 22×3

LCM(18, 12) =

Therefore, Sonia and Ravi will meet again at the starting point after 36 minutes.

 

 

Here are the NCERT Class 10 Maths Chapter 1 – Real Numbers, Exercise 1.2 Solutions (as per latest 2025 textbook):

📘
Exercise 1.2 – Solutions

Q1. Prove that √5 is irrational.

Proof:

Let us assume, to the contrary, that √5 is rational.

⇒ √5 = a/b where a and b are integers with no common factor (i.e., in lowest terms), and b ≠ 0.

Squaring both sides:

→ 5 = a² / b²
→ a² = 5b²

This implies that a² is divisible by 5, so a is also divisible by 5 (because if a² is divisible by a prime, then a is divisible by that prime).

Let a = 5k for some integer k.
Then:

a² = (5k)² = 25k²

Substitute into earlier equation:

25k² = 5b²
→ b² = 5k² ⇒ b is divisible by 5

So, both a and b are divisible by 5, which contradicts the assumption that a and b are co-prime.

✅ Hence, our assumption is wrong. So, √5 is irrational.

Q2. Prove that 3 + √25 is irrational.

Solution:

First, simplify √25:

→ √25 = 5

So, 3 + √25 = 3 + 5 = 8, which is clearly a rational number.

⚠️ So, the question might be misprinted or intended to be:
Q2. Prove that 3 + √2 or 3 + √5 is irrational.
Please confirm.

For now, assuming it’s 3 + √5:

Let us assume that 3 + √5 is rational.
Then:

→ √5 = (3 + √5) – 3 = rational – 3 = rational
So, √5 is rational, which is false (as proved in Q1).

✅ Therefore, 3 + √5 is irrational.

Q3. Prove that the following are irrational:

(i) 1/√2

Assume 1/√2 is rational.

Let 1/√2 = a/b (a, b integers, b ≠ 0, fraction in lowest form)

Then:

→ √2 = b/a

⇒ √2 is rational, which is false (√2 is known to be irrational).

✅ Therefore, 1/√2 is irrational.

(ii) √7 + √5

Assume √7 + √5 is rational.

Then:

→ (√7 + √5) = r ⇒ √7 = r – √5
Since r and √5 are rational and irrational respectively, r – √5 is irrational. But √7 is irrational.

Contradiction!

✅ Hence, √7 + √5 is irrational.

(iii) √6 + √2

Assume √6 + √2 is rational.

Then:

→ (√6 + √2) = r ⇒ √6 = r – √2
Again, r is rational, and √2 is irrational, so r – √2 is irrational, but √6 is irrational — contradiction.

✅ Therefore, √6 + √2 is irrational.


Important Links

Real Number – Quick Revision Notes

Real Number – Most Important Questions

Real Number – Important MCQs

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