Class 10 Math (NCERT) Chapter 1 Exercise 1.2 Solutions in (Hindi & English)

Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 Solutions are given below in both Hindi and English medium. Exercise 1.2 has been solved by the method given by the book so that all of you can understand this solution easily and you get full marks in the board exam and all of you will be easy to prepare. Below in this page, first in English medium and then in Hindi medium, the exercises / questions are solved. Hope this solution will help you in your preparation and you will like it.

Class 10 Math Solutions Chapter 1 Exercise 1.2

Ex 1.2 Question 1 (Class 10 Math)

Express each number as a product of its prime factor.

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(V) 7429

Solutions:

Ex 1.2 Question 2 ( Class 10 Math)

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) By prime factorisation, we get:

26 = 2×13

91 = 7×13

HCF of 26 and 91 =13

And LCM of 26 and 91 = 2×7×13

= 182

Now, HCF×LCM=182×13=2366 ...(i)

Product of numbers= 26×91= 2366 ...(ii)

From (i) and (ii), we get:

HCF × LCM = product of number

Hence, verified.

(ii) By prime factorisation, we get:

510= 2×3×5×17

92=2×2×23

HCF of 510 and 92 = 2

And LCM of 510 and 92

=2²×3×5×17×23=23460

Now, HCF×LCM=2×23460=46920 ...(i)

Product of numbers

= 510×92=46920 ...(ii)

From (i) and (ii), we get:

LCM×HCF= product of numbers

Hence, verified.

(iii) By prime factorisation, we get:

336=2×2×2×2×3×7

54=2×3×3×3

HCF of 336 and 54 = 2×3 = 6

and LCM of 336 and 54 = 2⁴×3³×7

=3024

Now, LCM×HCF = 3024×26=18144 ...(i)

Product of numbers

=336×54=18144 ...(ii)

From (i) and (ii), we get:

LCM×HCM= product of number

Hence, verified.

Ex 1.2 Question 3 (Math Class 10)

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) By prime factorization, we get:

12= 2×2×3

15=3×5

21=3×7

HCF of 12, 15 and 21 = 3

And LCM=2×2×3×5×7=420.

(ii) By prime factorisation, we get:

17=17×1

23=23×1

29=29×1

HCF of 17, 23 and 29= 1

and LCM = 17×23×29= 11339.

(iii) By prime factorisation, we get:

8=2×2×2×1

9=3×3×1

25=5×5×1

HCF of 8, 9 and 25 = 1

and LCM of 8, 9 and 25= 2³×3²×5²

=1800.

Ex 1.2 Question 4 (Math class 10)

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

Solution:

Ex 1.2 Question 5 (Class 10 Math)

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

Solution:

Ex 1.2 Question 6 (Class 10 Math)

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

Solution:

Ex 1.2 Question 7 (class 10 Math)

There is a circular path around a sport field. Sonia takes 18 minute to drive one round of the field, while Ravi takes 12 minute 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: