Sum of two numbers A and B is 208 (A > B) and HCF of A and B is 16. If A is a perfect square, then what is the ratio of the difference between both the numbers to the difference between LCM and HCF of both the numbers?

Sum of two numbers A and B is 208 (A > B) and HCF of A and B is 16. If A is a perfect square, then what is the ratio of the difference between both the numbers to the difference between LCM and HCF of both the numbers? Correct Answer 1 : 7

Given:

HCF of A and B is 16.

Formula Used:

First number × Second number = HCF × LCM

Calculation:

Let A be ‘16a’ and B be ‘16b’.

⇒ 16a + 16b = 208

⇒ (a + b) = 13

Since, ‘a’ and ‘b’ are co-prime.

So, possible values of ‘a’ and ‘b’ = (12, 1), (11, 2), (10, 3), (9, 4), (8, 5) and (7, 6)

⇒ Possible values of A = 16a = 192 or 176 or 160 or 144 or 128 or 112

Since, A is a perfect square.

⇒ A = 144, then B = 64

Now,

144 × 64 = 16 × LCM

⇒ LCM = 576

Difference between both the numbers = 144 – 64 = 80

Difference between LCM and HCF of both the numbers = 576 – 16 = 560

∴ Required ratio = 80 : 560