**NCERT Solutions for Class 6 Maths Chapter 3**: NCERT Solutions for Class 6 Maths Chapter 3 **Playing With Numbers** Exercise 3.6

- Class 6 Maths Chapter 3 Exercise 3.1
- Class 6 Maths Chapter 3 Exercise 3.2
- Class 6 Maths Chapter 3 Exercise 3.3
- Class 6 Maths Chapter 3 Exercise 3.4
- Class 6 Maths Chapter 3 Exercise 3.5
- Class 6 Maths Chapter 3 Exercise 3.6
- Class 6 Maths Chapter 3 Exercise 3.7

## NCERT Solutions for Class 6 Maths Chapter 3 Whole 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

**Solution:**

(A) Factors of 18=2×3×3

Factors of 48=2×2×2×2×3

H.C.F. (18,48)=2×3=6

(B) Factors of 30=2×3×5

Factors of 42=2×3×7

H.C.F. (30,42)=2×3=6

(C) Factors of 18=2×3×3

of 60=2×2×3×5

H.C.F. (18,60)=2×3=6

(D) Factors of 27=3×3×3

Factors of 63=3×3×7

H.C.F. (27,63)=3×3=9

(E) Factors of 36=2×2×3×3

Factors of 84=2×2×3×7

H.C.F. (36,84)=2×2×3=12

(F) Factors of 34=2×17

Factors of 102=2×3×17

H.C.F. (34,102)=2×17=34

(G) Factors of 70=2×5×7

Factors of 105=3×5×7

Factors of 175=5×5×7

H.C.F. =5×7=35

(H) Factors of 91=7×13

Factors of 112=2×2×2×2×7

Factors of 49=7×7

H.C.F. =1×7=7

(I) Factors of 18=2×3×3

Factors of 54=2×3×3×3

Factors of 81=3×3×3×3

H.C.F. =3×3=9

(J) Factors of 12=2×2×3

Factors of 45=3×3×5

Factors of 75=3×5×5

H.C.F =1×3=3

2. What is the HCF of two consecutive?

(A) numbers?

(B) even numbers?

(C) odd numbers?

**Solution:**

(A) H.C.F. of two consecutive numbers is 1.

(B) H.C.F. of two consecutive even numbers is 2.

(C) H.C.F. of two consecutive odd numbers is 1.

3. H.C.F. 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 the HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F.?

**Solution:**

No.

Factors of 4=2×2×1

Factors of 15=3×5×1

H.C.F =1

So, the correct H.C.F. is 1.