the sum of two numbers is 288 and their HCF is 16. How many pairs of such members can be formed?

Correct option is A

Given:

The sum of two numbers is 288 and their HCF is 16

Solution:

Let the ratio of number be x : y

So the numbers will be 16x & 16y 

Now ATQ,

16x + 16y = 288

=> 16(x + y) = 288

=> x + y = 18  (x and y will be prime numbers)

Pairs of x, y can be (1, 17) (5, 13) (7, 11)

Hence, only 3 pairs are possible.

Option (a) is the correct answer.