NCERT Solutions class 10 Maths Exercise 1.2 Ch 1 Real Numbers

NCERT Solutions for Class 10 Maths Exercise 1.2 Chapter 1 Real Numbers – FREE PDF Download

NCERT Class 10 Maths Ch 1 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter 1 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Real Numbers solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

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

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

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

2. Find the LCM and HCF of the following pairs of integers and verify that product of the two numbers.

(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Ans. (i) 26 and 91


HCF (26, 91) = 13
LCM (26, 91) = = 182
Product of two numbers 26 and 91 = 2366
= 2366
Hence, product of two numbers =  

(ii) 510 and 92

HCF (510, 92) = 2

LCM (510, 92) = = 23460

Product of two numbers 510 and 92 = = 46920

= 46920

Hence, product of two numbers = 

(iii) 336 and 54

336= 

54 = 

HCF (336, 54) = 

LCM (336, 54) = = 3024

Product of two numbers 336 and 54 = = 18144

= 18144

Hence, product of two numbers = 

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

Ans. (i) 12, 15 and 21 

HCF (12, 15, 21) = 3

LCM (12, 15, 21) = = 420

(ii) 17, 23 and 29

HCF (17, 23, 29) = 1

LCM (17, 23, 29) = = 11339

(iii) 8, 9 and 25

HCF (8, 9, 25) = 1

LCM (8, 9, 25) = = 1800

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

Ans. HCF (306, 657) = 9 

We know that, LCM × HCF = Product of two numbers

LCM (306, 657) = 22338

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

Ans. If any number ends with the digit 0, it should be divisible by 10. 

In other words, it will also be divisible by 2 and 5 as 

Prime factorisation of 

It can be observed that 5 is not in the prime factorisation of.

Hence, for any value of n, will not be divisible by 5.

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

6. Explain why and are composite numbers.

Ans. 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

= 

= 

= 

The given expression has 6 and 13 as its factors other than 1 and number itself.

Therefore, it is a composite number.

= 

= = 

1009 cannot be factorized further

Therefore, the given expression has 5 and 1009 as its factors other than 1 and number itself.

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?

Ans. 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. And 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. 

And, 

LCM of 12 and = 36

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