# NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.6

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

## 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.