# Class 6 Maths Chapter 3 Exercise 3.6 Solution – Playing With Numbers NCERT

NCERT Class 6 Math Exercise 3.6 Solution of Chapter 3 Playing with Numbers with explanation. Here we provide Class 6 Maths all Chapters in Hindi for cbse, HBSE, Mp Board, UP Board and some other boards.

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NCERT Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.6 Solution in english Medium.

## Class 6 Math Chapter 3 Exercise 3.6 Solution

Question 1. Find the HCF of te 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, 75
I. 18, 54, 81
J. 12, 45, 75

A.
Factors of 18 = 2x3x3
Factors of 48 = 2x2x2x2x3
HCF ( 18, 48) = 2×3 = 6

B.
Factors of 30 = 2x3x5
Factors of 42 = 2x3x7
HCF (30, 42) = 2×3 = 6
C.
Factors of 18 = 2x3x3
Factors of 60 = 2x2x3x5
HCF (18, 60) = 2×3 = 6
D.
Factors of 27 = 3x3x3
Factors of 63 = 3x3x7
HCF (27, 63) = 3×3 = 9
E.
Factors of 36 = 2x2x3x3
Factors of 84 = 2x2x3x7
HCF (36, 84) = 2x2x3 = 12
F.
Factors of 34 = 2×17
Factors of 102 = 2x3x17
HCF (34, 102) = 2X17 = 34
G.
Factors of 70 = 2x5x7
Factors of 105 = 3x5x7
Factors of 175 = 5x5x7
HCF (70, 105, 175) = 5X7 = 35
H.
Factors of 91 = 7×13
Factors of 112 = 2x2x2x2x7
Factors of 49 = 7×7
HCF (91, 112, 49) = 7
I.
Factors of 18 = 2x3x3
Factors of 54 = 2x3x3x3
Factors of 81 = 3x3x3x3
HCF (18, 54, 81) = 3×3 = 9
J.
Factors of 12 = 2x2x3
Factors of 45 = 3x3x5
Factors of 75 = 3x5x5
HCF (12, 45, 75) = 3

Question 2. What is the HCF of two consecutive :
A. numbers?
B. even numbers?
C. odd numbers?
A. HCF of two consecutive numbers is 1.
B. HCF of two consecutive even numbers is 2.
C. HCF of two consecutive numbers is 1.

Question 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 m so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?