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Question
Two years ago, man’s age was three times the square of his son’s age. In three years’ time, his age will be four time his son’s age. Find their present ages.
Solution
Let son’s age 2 years ago be x years. Then,
Man’s age 2 years ago=`3x^2` years
Sons present age=`(x+2)` years
Man’s present age=`(3x^2+2)` years
In three years time,
Son’s age=`(x+2++3)`years =`(x+5)` years
Man’s age =`(3x^2+2+3)` years =`(3x^2+5)` years
According to the given condition,
Man’s age =`4xx` Son’s age
∴`3x^2+5=(x+5)`
⇒`3x^2+5=4x+20`
⇒`3x^2-4x-15=0`
⇒`3x^2-9x-5x-15=0`
⇒`3x(x-3)+5(x-3)=0`
⇒`(x-3)(3x+5)=0`
⇒`x-3=0 or 3x+5=0`
⇒`x=3 or x=-5/3`
∴ `x=3` (Age cannot be negative)
Son’s present age =`(x+2) years=(3+2) years=5years`
Man’s present age=`(3x^2+2)years=(3xx9+2)years=29 years`
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