Question

The ages of two persons differs by 16 years. If 6 years ago, the elder one will be 3 times as old as the younger one.Find the sum of their present ages?

Answer 1

hopefully this will help you

Answer 2

Let the age of the person be X

Then the older one's age be (X+16) (1)
==> 6 years ago
$3(x - 6) = (x + 16 - 6)$
$3x - 18 = x + 10$
$3x - x = 10 + 18$
$2x = 28$
$x = \frac{28}{2} = 14$

Now sub X= 14 in (1),
14 +16 =30

The sum of their ages is 30 +14 = 44
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