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First of all we have to write the numbers as product of the prime factor. Prime factor of 608 = 2^{5} * 19 i.e.,

And the prime factor of 544 = 2^{5} * 17 i.e.,

So, the HCF of 608 and 544 = common factor in both = 2^{5} = 32 Similarly, The prime factor of 638 = 2 * 11 * 29 The prime factor of 783 = 3* 3 * 3 * 29 So, the HCF of 638 and 783 = 29

Now, the prime factor of 425 = 5 * 5 * 17 The prime factor of 476 = 2*2*7*17 So, the HCF of 425 and 476 = 17

co-prime numbers with sum 8 are (1, 7) and (3, 5).

Therefore we can say that the required numbers are (27 x1, 27 x 7) and (27 x 3, 27x 5), i.e. (27, 189) and (81, 135). Out of the above two answers, the given one in the answer is the pair (27, 189).

7) Two numbers are in the ratio of 15:11. If the HCF of numbers is 13, find the numbers.

18 = 2 x3 x 3 42 = 2 x 3 x 7 24 = 2 x 2 x 2 x 3 HCF = 2 x 3 = 6

Hence the required number is 6.

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9) Three bells toll at intervals of 36 sec, 40 sec, and 48 sec respectively. They start singing together at a particular time. When will they toll next together?