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Question
The sum of 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.
Solution
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 is36 years and his son’s age is 9 years.
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