Exercise 3.6
Page No 63:
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
Answer:
(a) 18, 48- 21839331
- 24822421226331
48 = 2 × 2 × 2 × 2 × 3
HCF = 2 × 3 = 6
(b) 30, 42
- 230315551
- 242321771
42 = 2 × 3 × 7
HCF = 2 × 3 = 6
(c) 18, 60
- 21839331
- 260230315551
60 = 2 × 2 × 3 × 5
HCF = 2 × 3 = 6
(d) 27, 63
- 32739331
- 363321771
63 = 3 × 3 × 7
HCF = 3 × 3 = 9
(e) 36, 84
- 23621839331
- 284242321771
84 = 2 × 2 × 3 × 7
HCF = 2 × 2 × 3 = 12
(f) 34, 102
- 23417171
- 210235117171
102 = 2 × 3 × 17
HCF = 2 ×17 = 34
(g) 70, 105, 175
- 270535771
- 3105535771
- 5175535771
105 = 3 × 5 × 7
175 = 5 × 5 × 7
HCF = 5 × 7 = 35
(h) 91, 112, 49
- 79113131
- 2112256228214771
- 749771
112 = 2 × 2 × 2 × 2 × 7
49 = 7 × 7
HCF = 7
(i) 18, 54, 81
- 21839331
- 25432739331
- 38132739331
54 = 2 × 3 × 3 × 3
81 = 3 × 3 × 3 × 3
HCF = 3 × 3 = 9
(j) 12, 45, 75
- 21226331
- 345315551
- 375525551
45 = 3 × 3 × 5
75 = 3 × 5 × 5
HCF = 3
Q 2 - Ex 3.6 - Playing With Numbers - NCERT Maths Class 6th - Chapter 3
Question 2:
What is the HCF of two consecutive(a) Numbers? (b) Even numbers? (c) Odd numbers?
Page No 64:
Question 3:
HCF of co-prime numbers 4 and 15 was found as follows by factorization:4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factors, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?
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