# NCERT Solutions For Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.6

The NCERT Solutions For Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.6 explains the method of solving problems related to finding the HCF of a given set of numbers. These NCERT Solutions are prepared by the subject experts at BYJUâ€™S to help students ace the exam. Set of examples are available in the NCERT textbook, before the exercise, to help students get a hold of the concepts. The faculty has followed the same method as of the examples to solve the questions present in the exercise. The solutions are created to help students avoid any sort of confusion that could arise while solving the problems. The exercise-wise problems can be solved and revised by the students with the help of theÂ NCERT Solutions Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.6.

## NCERT Solutions for Class 6 Chapter 3: Playing with Numbers Exercise 3.6 Download PDF

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### Access NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Exercise 3.6

1. Find the HCF of the following numbers :

(a) 18, 48

(b) 30, 42

(c) 18, 60

(d) 27, 63

(e) 36, 84

(f) 34, 102

(g) 70, 105, 175

(h) 91, 112, 49

(i) 18, 54, 81

(j) 12, 45, 75

Solutions:

(a) 18, 48

18 = 2 Ã— 3 Ã— 3

48 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

HCF = 2 Ã— 3 = 6

Therefore the HCF of 18, 48 is 6

(b) 30, 42

30 = 2 Ã— 3 Ã— 5

42 = 2 Ã— 3 Ã— 7

HCF = 2 Ã— 3 = 6

Therefore the HCF of 30, 42 is 6

(c) 18, 60

18 = 2 Ã— 3 Ã— 3

60 = 2 Ã— 2 Ã— 3 Ã— 5

HCF = 2 Ã— 3 = 6

Therefore the HCF of 18, 60 is 6

(d) 27, 63

27 = 3 Ã— 3 Ã— 3

63 = 3 Ã— 3 Ã— 7

HCF = 3 Ã— 3 = 9

Therefore the HCF of 27, 63 is 9

(e) 36, 84

36 = 2 Ã— 2 Ã— 3 Ã— 3

84 = 2 Ã— 2 Ã— 3 Ã— 7

HCF = 2 Ã— 2 Ã— 3 = 12

Therefore the HCF of 36, 84 is 12

(f) 34, 102

34 = 2 Ã— 17

102 = 2 Ã— 3 Ã— 17

HCF = 2 Ã— 17 = 34

Therefore the HCF of 34, 102 is 34

(g) 70, 105, 175

70 = 2 Ã— 5 Ã— 7

105 = 3 Ã— 5 Ã— 7

175 = 5 Ã— 5 Ã— 7

HCF = 5 Ã— 7 = 35

Therefore the HCF of 70, 105, 175 is 35

(h) 91, 112, 49

91 = 7 Ã— 13

112 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 7

49 = 7 Ã— 7

HCF = 7

Therefore the HCF of 91, 112, 49 is 7

(i) 18, 54, 81

18 = 2 Ã— 3 Ã— 3

54 = 2 Ã— 3 Ã— 3 Ã— 3

81 = 3 Ã— 3 Ã— 3 Ã— 3

HCF = 3 Ã— 3 = 9

Therefore the HCF of 18, 54, 81 is 9

(j) 12, 45, 75

12 = 2 Ã— 2 Ã— 3

45 = 3 Ã— 3 Ã— 5

75 = 3 Ã— 5 Ã— 5

HCF = 3

Therefore the HCF of 12, 45, 75 is 3

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2. What is the HCF of two consecutive

(a) numbers?

(b) even numbers?

(c) odd numbers?

Solutions:

(a) The HCF of two consecutive numbers is 1

Example: The HCF of 2 and 3 is 1

(b) The HCF of two consecutive even numbers is 2

Example: The HCF of 2 and 4 is 2

(c) The HCF of two consecutive odd numbers is 1

Example: The HCF of 3 and 5 is 1

3. HCF of co-prime numbers 4 and 15 was found as follows by factorisation:

4 = 2 Ã— 2 and 15 = 3 Ã— 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Solutions:

### Access Other Exercise Solutions of Class 6 Maths Chapter 3: Playing with Numbers

Exercise 3.1 Solutions

Exercise 3.2 Solutions

Exercise 3.3 Solutions

Exercise 3.4 Solutions

Exercise 3.5 Solutions

Exercise 3.7 Solutions