**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 C**lass 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. **

**(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 6^{n}

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 6^{n }must be divisible by 2 and 5

But both 6^{n }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.**

__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

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

= **2 ^{2 }**

**× 3**

^{2 }**= 4 ****× 9 = 36**

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

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