Question

The sum of 2 numbers is 96 and their hcf is 12. How many such pairs are possible?

Answer 1

To find the H.C.F. of the given number we will follow the following steps: We divide the bigger number by smaller one. Divide smaller number in step 1 with remainder obtained in step 1. Divide divisor of second step with remainder obtained in

Answer 2

There are two such pairs (1, 7) and (3, 5) are possible.

The sum of two numbers is 96 and their HCF is 12.

We have to find the possible number of pairs.

• Here, HCF of two numbers is 12, so the numbers could be written as 12x and 12y ; where x and y are co - prime numbers.

Now the sum of numbers = 96

∴ 12x + 12y = 96

⇒12(x + y) = 96

⇒x + y = 96/12 = 8

Hence, x + y = 8 , where x and y are co - prime numbers.

If x = 1 , y = 7 and (1, 7) is co - prime numbers.

If x = 3, y = 5 and (3, 5) is co - prime numbers.

Therefore are are only two such pairs is possible that make their sum of 96 and HCF of them is 12.