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Question

Five years later, the father's age will be three times the age of his son. Five years ago, father was seven times as old as his son. Find their present ages.


Solution

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

After five years, the father’s age will be $$\left( {x + 5} \right)$$ and the age of son will be $$\left( {y + 5} \right)$$, then,

$$x + 5 = 3\left( {y + 5} \right)$$

$$x - 3y - 10 = 0$$              ....  (1)

Five years ago, the age of the father was $$\left( {x - 5} \right)$$ and son’s age was $$\left( {y - 5} \right)$$, so,

$$\left( {x - 5} \right) = 7\left( {y - 5} \right)$$

$$x - 7y + 30 = 0$$             ....  (2)

From subtracting equation (1) from (2),

$$-3y+7y-10-30=0 \Rightarrow y=10 $$

Putting $$y=10$$ in (1),  

$$x-3(10)-10=0 \Rightarrow x=40 $$

Therefore, the present age of father is 40 years and that of son is 10 years.


Mathematics

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