## NCERT solutions for class 10 Maths

**Chapter 1**

**Real Numbers**

**Exercise 1.2**

**Ex 1.2 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 = 2 × 2 × 5 × 7 = 2² × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

**Ex 1.2 Question 2.**

**Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers:**

**(i) 26 and 91**

**(ii) 510 and 92**

**(iii) 336 and 54**

**Solution:**

(i) 26 and 91

26 = 2 ×13

91= 7 × 13

HCF = 13

LCM = 2 × 7 × 13

Product of two numbers = 26 × 91 = 2366

HCF × LCM = 13 × 182 = 2366

Hence, product of two numbers = HCF × LCM

(ii) 510 and 92

510 = 2 × 3 × 5 × 17

9 = 2 × 2 × 23

HCF = 2

LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of two numbers = 510 × 92 = 46920

HCF × LCM = 2 × 23460 = 46920

Hence, product of two numbers = HCF × LCM

(iii) 336 and 54

336 = 2 × 2 × 2 × 3 × 7 ⇒ 336 = 2^{4 }× 3 × 7

54 = 2 × 3 × 3 × 3 ⇒ 2 × 3³

HCF = 2 × 3 = 6

LCM = 2^{4 } × 3³ × 7 = 3024

Product of two numbers = 336 × 54 = 18144

HCF × LCM = 6 × 3024 = 18144

Hence, product of two numbers = HCF × LCM

**Ex 1.2 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:**

(i) 12, 15 and 91

12 = 2² ×3

15 = 3 × 15

91 = 3 × 7

HCF = 3

LCM = 2² × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

HCF = 1

LCM = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

8 = 2 × 2 × 2

9 = 3 × 391= 5 × 5

25 = 5 × 5

HCF = 1

LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

**Ex 1.2 Question 4**

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

**Solution:**

HCF (306, 657) = 9

Product of two numbers = HCF × LCM

∴ LCM × HCF = 306 × 657

LCM = 306 × 657 / HCF = 306 × 657 / 9

LCM = 22338

**Ex 1.2 Question 5.**

**Check whether 6 ^{n} can end with the digit 0 for any natural number n.**

**Solution:**

If any number ends with number 0, it should be divisible by 10 0r in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorization of 6n = (2 ×3)n

It can be observed that 5 is not in the prime factorization of 6n.

Hence, for any value of n, 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

**Ex 1.2 Question 6. **

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

**Solution:**

**Ex 1.2 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:**

LCM of 12 minutes and 18 minutes.

18 = 2 × 3 × 2

12 = 2 × 2 × 3

LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

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