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

Class 6, Maths, Chapter 3, Exercise 3.6 Solutions

Q.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

Ans:

The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 a

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 b

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 c

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 d

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 e

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 f

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 g

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 h

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 i

NCERT Solutions Class 6 Maths Ch. 3 Playing with Numbers Exe. 3.6 Q.1 j

Q.2. What is the HCF of two consecutives

(a) numbers?

(b) even numbers?

(c) odd numbers?

Ans:

(a) The HCF of two consecutive numbers is 1

Example(i): The HCF of 2 (2=1×2) and 3 (3 = 1×3) is 1.

Example(ii): The HCF of 15 (15 = 1 x 3 x 5) and 16 (16 = 1x2x2x2x2) is 1.

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

Examples:

(a) HCF of 2 (2= 1×2)  and 4 (4= 1x 2 x 2) is 2.

(b) HCF of 6 (6 = 1×2 x 3) and 8 (8 = 1 x 2 x 2 x 2) is 2

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

Example: The HCF of 3 (3 = 1×3) and 5 (1×5) is 1

Q.3. HCF of co-prime numbers 4 and 15 was found as follows by factorisation: 4 = 2 × 2 and 15 = 3 × 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?

Ans:

No, the answer is not correct. The correct HCF (Highest Common Factor) of co-prime numbers 4 and 15 is 1.

Co-prime numbers are two numbers that do not have any common factor except 1.

In this case,

Prime factors of 4 = 2 x 2,

Prime factors of 15 = 3 x5.

As there is no common prime factor between them, the HCF of 4 and 15 is 1.

So, the correct answer is HCF (4, 15) = 1.

NCERT Solutions For Class 6 Maths, Chapter 3 Playing With Numbers (All Exercises)