Question

The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man’s age at the time. Find their present ages.

Answer 1

Let the present age of the man be x years

Then present age of his son is (45 - x) years

Five years ago, man’s age = (x - 5) years

And his son’s age (45 - x - 5) = (40 - x) years

Then according to question,

(x - 5)(40 - x) = 4(x - 5)

40x - x² + 5x - 200 = 4x - 20

-x² + 45x - 200 = 4x - 20

-x² + 45x - 200 - 4x + 20 = 0

-x² + 41x - 180 = 0

x² - 41x + 180 = 0

x² - 36x - 5x + 180 = 0

x(x - 36) -5(x - 36) = 0

(x - 36)(x - 5) = 0

So, either 

x - 36 = 0

x = 36

Or

x - 5 = 0

x = 5

But, the father’s age never be 5 years

Therefore, when x = 36 then

45 - x = 45 - 36 = 9

Hence, man’s present age is 36 years and his son’s age is 9 years.