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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.