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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