Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.
Given :
Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then.
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 10 years $= x+10$
Age of the father after 10 years $= y+10$.
Age of the son 10 years ago $= x-10$.
Age of the father 10 years ago $= y-10$.
Therefore,
$y+10 = 2(x+10)$
$y+10 = 2x+20$
$y = 2x+20-10$
$y = 2x+10$.....(i)
$y-10=12(x-10)$
$y-10=12x-120$
$(2x+10)-10=12x-120$ (From (i))
$2x+10-10=12x-120$
$12x-2x=120$
$10x=120$
$x=\frac{120}{10}$
$x=12$
$\Rightarrow y=2(12)+10=24+10=34$
The present age of the son is $12$ years and the present age of the father is $34$ years.
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