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 and 48.
Factors of 18 = 2 x 3 x 3
Factors of 48 = 2 x 2 x 2 x 2 x 2 x 3
Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6.
(b) 30 and 42.
Factors of 30 = 2 x 3 x 5
Factors of 42 = 2 x 3 x 7
Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6.
(c) 18 and 60.
Factors of 18 = 2 x 3 x 3
Factors of 60 = 2 x 2 x 3 x 5
Here, the common factors are 2 and 3.
Hence, the HCF of 18 and 60 = 2 x 3 = 6.
(d) 27 and 63.
Factors of 27 = 3 x 3 x 3
Factors of 63 = 3 x 3 x 7
Here, the common factor is 3 (occurring twice).
Hence, the HCF = 3 x 3 = 9.
(e) 36 and 84.
Factors of 36 = 2 x 2 x 3 x 3
Factors of 84 = 2 x 2 x 3 x 7
Here, the common factors are 2, 2, and 3.
Hence, the HCF = 2 x 2 x 3 = 12.
(f) 34 and 102.
Factors of 34 = 2 x 17
Factors of 102 = 2 x 3 x 17
Here, the common factors are 2 and 17.
Thus, HCF is 2 x 17 = 34.
(g)70, 105, and 175.
Factors of 70 = 2 x 5 x 7
Factors of 105 = 3 x 5 x 7
Factors of 175 = 5 x 5 x 7
Here, common factors are 5 and 7.
Hence, the HCF of 70, 105 and 175 is 5 x 7 = 35.
(h)91, 112, and 49.
Factors of 91 = 7 x 13
Factors of 112 = 2 x 2 x 2 x 2 x 7
Factors of 49 = 7 x 7
Here, the common factor is 7.
Hence, the HCF = 7.
(i) 18, 54, and 81.
Factors of 18 = 2 x 3 x 3
Factors of 54 = 2 x 3 x 3 x 3
Factors of 81 = 3 x 3 x 3 x 3
Here, the common factor is 3 (occurring twice).
Thus, the HCF = 3 x 3 = 9.
(j) 12, 45, and 75.
Factors of 12 = 2 x 2 x 3
Factors of 45 = 3 x 3 x 5
Factors of 75 = 3 x 5 x 5
Here, the common factor is 3.
Hence, the HCF = 3.
Ex 3.6 Question 2.
What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution:
(a) Common factor of two consecutive numbers is always 1.
Hence, the HCF = 1.
(b) Common factors of two consecutive even numbers are 1 and 2.
Hence, the HCF = 1 x 2 = 2.
(c) Common factor of two consecutive odd numbers is 1.
Hence, the HCF = 1.
Ex 3.6 Question 3.
HCF of co-prime numbers 4 and 15 was found as follows by factorization:
4 = 2 x 2 and 15 = 3 x 15. Since there are no common prime factors, 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.
The correct HCF of 4 and 15 is 1.
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