KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

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Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

Question 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) 18, 48
18 = 1, 2, 3, 6, 9, 18
48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Common factors = 1, 2, 3, 6
H.C.F of 18 & 48 = \(\boxed { 6 } \)

b) 30, 42
30 = 1, 2, 3, 5, 6, 10, 15, 30
42 = 1, 2, 3, 4, 6, 7, 14, 21, 42
Common factors = 1, 2, 3, 6
∴ H.C.F. = \(\boxed { 6 } \)
∴ H.C.F of 30 & 42 is \(\boxed { 6 } \)

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

c) 18 & 60
18 = 1, 2, 3, 6, 9, 18
60 = 1, 2, 3, 4, 6, 10, 12, 15, 30, 60
Common factors = 1, 2, 3, 6
∴ H.C.F. of 18 & 60 is \(\boxed { 6 } \)

d) 27 & 63
27 = 1, 3, 9, 27
63 = 1, 3, 7, 9, 21, 63
∴ Common factors = 1, 3, 9
∴ H.C.F. of 27 & 63 is = 9.

e) 36 & 84
36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
∴ Common factors = 1, 2, 3, 4, 6, 12
∴ H.C.F. of 36 & 84 is 12.

f) 34 & 102
34 = 1, 2, 17, 34
102 = 1, 2, 3, 6, 17, 34, 51, 102
∴ Common factors = 1, 2, 17, 34
∴ H.C.F. of 34 & 102 is 34.

g) 70, 105, 175
70 = 1, 2, 5, 7, 10, 14, 35, 70
105 = 1, 3, 5, 7, 10, 14, 35, 105
175 = 1, 5, 7, 25, 35, 175
∴ Common factors 1, 5, 7, 35
∴ H.C.F. of 70, 105, 175 is 35

h) 91, 112, 49
91 = 1, 7, 13, 91
112 = 1, 2, 4, 7, 8, 14, 16, 28, 56, 112
49 = 1, 7, 49
∴ Common factors = 1, 7
∴ H.C.F. of 91, 112, 49 is 7

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

i) 12, 45, 75
12 = 1, 2, 3, 4, 6, 12
45 = 1, 3, 5, 9, 15, 45
75 = 1, 3, 5, 15, 26, 75
∴ Common factors 1, 3
∴ H.C.F of 12, 45, 75 is 3.

Question 2.
What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?
Find the HCF of the following:
(i) 24 and 36
(ii) 15, 25 and 30
(iii) 8 and 12
(iv) 12, 16 and 28
Solution:
(i) 1 e.g., HCF of 2 and 3 is 1.
(ii) 2 e.g., HCF of 2 and 4 is 2.
(iii) 1 e.g., HCF of 3 and 5 is 1.

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

To do the prime factorization of 84, successively divide 84 by prime numbers.

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, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?
Solution:
No. The answer is not correct. 1 is the correct HCF.

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