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

Here we are giving you the solution of NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.2 . You can also download Free NCERT Solutions for Class 10 Maths Chapter 1 PDF on our website . NCERT Maths class 10 chapter 1 exercise 1.2 solutions This post has been prepared by the experienced teachers of EducationForIndia.com . Detailed answers of all the questions in Class 10th Maths Chapter 1

### Class 10 Maths Chapter 1 Exercise 1.2

Exercise 1.2 class 10th Question 1 :

Express each number as a product of its prime factors:

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

Solution:

(i) 140

140=2×70

=2×2×35

=2×2×5×7

=2×2×5×7×1

(ii) 156

156=2×78

=2×2×39

=2×2×3×13

=2×2×3×13×1

(iii) 3825

3825=3×1275

=3×3×425

=3×3×5×85

=3×3×5×5×17

=3×3×5×5×17×1

(iv) 5005

5005=5×1001

=5×7×143

=5×7×11×13

=5×7×11×13×1

(v) 7429

7429=17×437

=17×19×23

=17×19×23×1

Exercise 1.2 class 10th 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

Solution:

LCM × HCF = Product of the two numbers, this is the formula we are going to use.

First of all, we have to find the LCM and HCF of the given pairs of the integers so let’s find the prime factors of the given pairs of numbers.

1) 26=2×13 and 91=1×91

HCF=1 and LCM=2×13×91=2366

verification: 1×2366=26×91⇒2366=2366

2) 510=2×3×5×17 and 92=22×23

HCF=2 and LCM=22×3×5×17×23=23460

verification: 2×23460=92×510⇒46920=46920

3) 54=2×33 and 336=24×3×7

HCF=6 and LCM=24×33×7=3024

verification: 6×3024=336×54⇒18144=18144

Exercise 1.2 class 10th Question 3 :

Find the LCM and HCF of the following integers by applying the prime factorization method.

यह भी पढ़े-  NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4

(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Solution:

We will solve this by using the prime factorize method

(i) 12, 15 and 21

Factor of 12=2×2×3
Factor of 15=3×5
Factor of 21=3×7
HCF (12,15,21) =3
LCM (12,15,21) =2×2×3×5×7=420

(ii) 17, 23 and 29

Factor of 17=1×17
Factor of 23=1×23
Factor of 29=1×29
HCF (17,23,29) =1
LCM (17,23,29) =1×17×23×29=11,339

(iii) 8, 9 and 25

Factor of 8=2×2×2×1
Factor of 9=3×3×1
Factor of 25=5×5×1
HCF (8,9,25) =1
LCM (8,9,25) =2×2×2×3×3×5×5=1,800

Exercise 1.2 class 10th Question 4 :

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

Solution:

Given that HCF = 9 and the numbers are 306 and 657.
LCM =?

We know that,

LCM × HCF= Product of the two numbers

LCM= \displaystyle \frac{{Product\text{ }of\text{ }two\text{ }numbers}}{{HCF}}

So, the product of the two numbers will be =306×657=201042

LCM = \displaystyle \frac{{201042}}{9}

So, the LCM is 22338

Exercise 1.2 class 10th Question 5 :

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

Solution:

Given, A number 6n

So here we have to look that whether it can end with 0 or not?

It is known that when any number is multiplied by 5 or 10 or by the multiples of 10 ends with zero that means it is divisible by 0.

So, the factor of 10 will be 2×5,

The condition is that the value of 6n must be divisible by 2 and 5

But both 6n is divisible by 2 and not divisible by 5

That’s why it cannot end with a zero.

Exercise 1.2 class 10th Question 6 :

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

यह भी पढ़े-  NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

Solution:

For solving this equation must remember that the prime numbers are the whole numbers whose only factors are 1 and the number itself. The composite numbers are positive integers that can contain factors other than 1 and itself.

Now, simplify 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5.

By the process of simplifying, we can see that both the numbers have more than two factors. Hence, if the number has more than two factors, it will be known as composite.

It is also to be noted that,

7 × 11 × 13 + 13 = 13 (7 × 11 + 1)

= 13(77 + 1)

= 13 × 78

= 13 × 13 × 6 × 1

= 13 × 13 × 2 × 3 × 1

The given number has 2, 3, 13, and 1 as its factors. So, it is a composite number.

Now, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 × 1009 × 1

1009 cannot be factorized more. Therefore, the given expression has 5,1009 and 1 as its factors. Therefore, it is a composite number.

Exercise 1.2 class 10th 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:

Given, the time taken by Sonia to drive one round is18 minutes

The time is taken by Ravi to drive one round is12 minutes

यह भी पढ़े-  NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Here, according to the question time taken by both to meet again will be the LCM of 18 and 12

Let’s find it using the prime factorize method, As we know, LCM is the product of the greatest power of each prime factor

= 22 × 32

= 4 × 9 = 36

Ans. The time is taken by both to meet again = LCM of 18 and 12 = 36 minutes.

आपको यह सभी पोस्ट Video के रूप में भी हमारे YouTube चैनल  Education 4 India पर भी मिल जाएगी।

Scroll to Top