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Question
Two years ago, a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.
Solution
Let the age of son 2 years ago be x years.
Then, father’s age 2 years ago = 3x2 years
Present age of son = (x + 2) years
Present age of father = (3x2 + 2) years
3 years hence:
Son’s age = (x + 2 + 3) years = (x + 5) years
Father’s age = (3x2 + 2 + 3) years = (3x2 + 5) years
From the given information,
3x2 + 5 = 4(x + 5)
3x2 – 4x – 15 = 0
3x2 – 9x + 5x – 15 = 0
3x(x – 3) + 5(x – 3) = 0
(x – 3)(3x + 5) = 0
x = 3,
Since, age cannot be negative.
So, x = 3.
Present age of son = (x + 2) years = 5 years
Present age of father = (3x2 + 2) years = 29 years
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