RBSE Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6

Rajasthan Board RBSE Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 Textbook Exercise Questions and Answers.

RBSE Class 6 Maths Solutions Chapter 3 Playing With Numbers Ex 3.6

Question 1.
Find the H.C.F. of the following number:
(a) 18, 48
Answer:
Factors of 18 = 2 × 3 × 3
Factors of 48 = 2 × 2 × 2 × 2 × 3
∴ H.C.F. (18, 48) = 2 × 3 = 6

(b) 30,42 
Answer:
Factors of 30 = 2 × 3 × 5
Factors of 42 = 2 × 3 × 7
∴ H.C.F. (30, 42) = 2 × 3 = 6

(c) 18,60
Answer:
Factors of 18 = 2 × 3 × 3
Factors of60 = 2 × 2 × 3 × 5
∴ H.C.F. (18, 60) = 2 × 3 = 6

(d) 27,63
Answer:
Factors of 27 = 3 × 3 × 3
Factors of 63 = 3 × 3 × 7
∴ H.C.F. (27, 63) = 3 × 3 = 9

(e) 36,84 
Answer:
Factors of 36 = 2 × 2 × 3 × 3
Factors of84 = 2 × 2 × 3 × 7
∴ H.C.F. (36,84) = 2 × 2 × 3 = 12

(f) 34, 102
Answer:
Factors of 34 = 2 × 17
Factors of 102 = 2 × 3 × 17
∴ H.C.F. (34,102) = 2 × 17 = 34

(g) 70, 105, 175 
Answer:
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) 91, 112, 49
Answer:
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) 18, 54, 81
Answer:
Factors of 18 = 2 × 3 × 3
Factors of54 = 2 × 3 × 3 × 3
Factors of 81 = 3 × 3 × 3 × 3
∴ H.C.F. = 3 × 3 = 9

(j) 12, 45, 75
Answer:
Factors of 12 = 2 × 2 × 3
Factors of 45 = 3 × 3 × 5
Factors of 75 = 3 × 5 × 5
∴ H.C.F. = 1 × 3 = 3

Question 2.
What is the H.C.F. of two consecutive:
(a) numbers?
Answer:
H.C.F. of two consecutive numbers is 1.

(b) even numbers?
Answer:
H.C.F. of two consecutive even numbers is 2.

(c) odd numbers?
Answer:
H.C.F. of two consecutive odd numbers is 1.

Question 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 H.C.F. of 4 and 15 is 0. Is the solution correct? If not, what is the correct H.C.F.?
Answer:
No. The correct H.C.F. is 1.