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Question

Ten years ago, father was the twelve times as old as his son and ten years later will be twice as old as his son will be. Find their present ages.


Solution

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

Ten years ago, Father's age $$=(x-10)$$ years
Son's age $$=(y-10)$$ years
$$\therefore x-10=12(y-10)\Rightarrow x-12y+110=0$$ ...(i)

Ten years later, Father's age $$=(x+10)$$ years
Son's age $$=(y+10)$$
$$\therefore x+10=2(y+10)\Rightarrow x-2y-10=0$$ ...(ii)

Subtracting (ii) from (i), we get
$$-10y+120=0\Rightarrow 10y=120\Rightarrow y=12$$

Putting $$y=12$$ in (i), we get
$$x-144+110=0\Rightarrow x=34$$

Thus, the present age of the father is $$34$$ years and the present age of the son is $$12$$ years.

Mathematics

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