**Go To: Exercise 1.1 Exercise 1.3 Exercise 1.4 **

**NCERT Solutions Maths Class 10 Exercise 1.2**

**Q1.
Express each number as product of its prime factors:**

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

**Solution:
**(i) 140 = 2 × 2 × 5 × 7 = 2^{2}
× 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2^{2} × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3^{2} × 5^{2} × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

**Q2.
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: ** (i) 26
= 2 × 13

91 =7 × 13

HCF = 13

LCM =2 × 7 × 13 =182

Product of HCF and LCM = 13 × 182 = 2366

Product of two numbers = 26 × 91 = 2366

Hence, HCF × LCM = product of two numbers

(ii) 510 = 2 × 3 × 5 × 17

92 =2 × 2 × 23

HCF = 2

LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of HCF and LCM = 2 × 23460 = 46920

Product of two numbers = 510 × 92 = 46920

Hence, HCF × LCM = product of two numbers

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 = 6

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

Product of HCF and LCM = 6 × 3024 = 18144

Product of two numbers = 336 × 54 =18144

Hence, HCF × LCM = product of two numbers

**Q3.
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
= 2 × 2 × 3

15 =3 × 5

21 =3 × 7

HCF = 3

LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

HCF = 1

LCM = 1 × 17 × 19 × 23 = 11339

(iii) 8 = 2 × 2 × 2

9 = 3 × 3

25 = 5 × 5

HCF = 1

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

**Q4. Given that HCF (306, 657) = 9, find LCM
(306, 657).**

**Solution: **We know that, HCF
× LCM = product of
number

9 × LCM = 306 × 657

Thus, LCM = (306 × 657) / 9 = 22338

**Q5. Check whether 6^{n} can end with the digit 0 for any natural number n.**

**Solution:
**If any digit has
last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.

So, the value of 6* ^{n}* should
be divisible by 2 and 5 both.

6* ^{n}*
is divisible by 2 but not divisible by 5.

So, it can not end with 0.

**Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5
× 4 × 3 × 2 × 1 + 5 are composite numbers.**

**Solution:
**7 × 11 × 13 + 13

Taking 13 common, we get

13(7 × 11 +1 )

13(77 + 1 )

13(78)

It is product of two numbers and both numbers are more than 1, so it is a
composite number.

Taking 5 common, we get

5(7 × 6 × 4 × 3 × 2 × 1 + 1)

5(1008 + 1)

5(1009)

It is product of two numbers and both numbers are more than 1, so it is a
composite number.

**Q7. 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:
**Sonia and Ravi
will be meeting again after LCM of both the values at the starting point. Let
us find the LCM of 18 and 12.

18 = 2 × 3 × 3

12 = 2 × 2 × 3

LCM = 2 × 2 × 3 × 3 = 36

Therefore, they will be meeting together at the starting point after 36
minutes.