# Six years hence a man’s age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Given :

Six years hence a man’s age will be three times the age of his son and three years ago he was nine times as old as his son.

To find :

We have to find their present ages.

Solution :

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

This implies,

Age of the son after 6 years $= x+6$

Age of the father after 6 years $= y+6$.

Age of the son 3 years ago $= x-3$.

Age of the father 3 years ago $= y-3$.

Therefore,

$y+6 = 3(x+6)$

$y+6 = 3x+18$

$y = 3x+18-6$

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

$y-3=9(x-3)$

$y-3=9x-27$

$(3x+12)-3=9x-27$ (From (i))

$3x+12-3=9x-27$

$9x-3x=9+27$

$6x=36$

$x=\frac{36}{6}$

$x=6$

$\Rightarrow y=3(6)+12=18+12=30$

The present age of the son is $6$ years and the present age of the father is $30$ years.

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