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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.
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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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