NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 1 Real Numbers Class 10 NCERT Solutions Ex 1.2. 

• Real Numbers Class 10 Ex 1.1
• Real Numbers Class 10 Ex 1.2
• Real Numbers Class 10 Ex 1.3
• Real Numbers Class 10 Ex 1.4

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	Real Numbers
Exercise	Ex 1.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Solution:

Concept Insight:   Since the given number needs to be expressed as the product of prime factors so in order to solve this problem knowing prime numbers is required. Do not forget to put the exponent in case a prime number is repeating.

Question 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

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

Solution:

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

Solution:

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

Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorization must include primes 2 and  5 both Prime factorization of 6^_n = (2 x 3)^_n. By Fundamental Theorem of Arithmetic Prime factorization of a number is unique. So 5 is not a prime factor of 6^_n.
Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6^_n cannot end with the digit 0 for any natural number n.

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

Solution:
Numbers are of two types – prime and composite. Prime numbers have only two factors namely 1 and the number itself whereas composite numbers have factors other than 1 and itself.
It can be observed that
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6
The given expression has 6 and 13  as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x  4 x 3 x 2 x 1 + 1)
= 5 x (1008 + 1)
= 5 x 1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Question 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:
It can be observed that Ravi and Sonia do not take the same amount of time 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.
i.e When  Sonia completes one round then ravi completes 1.5 rounds.   So they will meet first time at the time which is a common multiple of the time taken by them to complete 1 round
i.e LCM of 18 minutes and 12 minutes.
Now 18 = 2 x 3 x 3 = 2 x 3^₂
And, 12 = 2 x 2 x 3 = 2^₂ x 3
LCM of 12 and 18 = product of factors raised to highest exponent = 2² x 3² = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

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