Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

Given :

Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son.

To find :

We have to find the present ages of father and son.

Solution : 

Let the present ages of the son and the father be $x$ and $y$ respectively.

This implies,

Age of the son after 2 years $= x+2$

Age of the father after 2 years $= y+2$.

Age of the son 2 years ago $= x-2$.

Age of the father 2 years ago $= y-2$.

Therefore,

$y+2 = 3(x+2)+8$

$y+2 = 3x+6+8$

$y = 3x+14-2$

$y = 3x+12$.....(i)

$y-2=5(x-2)$

$y-2=5x-10$

$(3x+12)-2=5x-10$    (From (i))

$3x+12-2=5x-10$

$5x-3x=10+10$

$2x=20$

$x=\frac{20}{2}$

$x=10$

$\Rightarrow y=3(10)+12=30+12=42$

The present age of the son is $10$ years and the present age of the father is $42$ years. 

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Updated on: 10-Oct-2022

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