#### class ten maths NCERT Sol chap 1 Real Number

## Chapter 1: Real Number NCERT Solutions

**TEXTBOOK’S EXERCISE – 1.1 **

**TEXTBOOK’S EXERCISE – 1.2 **

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

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

**Sol. **(i) 140

Hence, 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7

(ii) 156

Hence, 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13

(iii) 3825

Hence, 3825 = 3 × 3 × 5 × 5 × 17

= 32 × 52 × 17 (iv) 5005

Hence, 5005 = 5 × 7 × 11 × 13

(v) 7429

Hence, 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 **

**Sol. **(i) 26 and 91

26 = 2 × 13

91 = 7 × 13

HCF of 26 and 91= 13

LCM of 26 and 91 = 2 × 7 × 13 = 182

**Verification
: **

LCM × HCF = 182 × 13 = 2366

Product of two numbers = 26 × 91 = 2366 **Verified. **

(ii) 510 and 92

So, 510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

∴ HCF of 510 and 92 = 2

∴ LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

**Verification
: **

LCM × HCF = 2 × 23460 = 46920

Product of two
numbers = 510 × 92 = 46920 **Verified. **

(iii) 336 and 54

336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF of 336 and 54 = 2 × 3 = 6

LCM of 336 and 54

= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024

**Verification
: **

LCM × HCF = 6 × 3024 = 18144

Product of two numbers = 336 × 54 = 18144 **Verified**

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

**Sol.
**We know that

HCF × LCM = Product of the two numbers

HCF (306, 657) = 9 [Given]

⇒ 9 × LCM = 306 × 657

∴ LCM = 306 x657 /9 = 22338