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Question

One year ago, a man was $$8$$ times as old as his son. Now his age is equal to square of his son's age. Find their present ages.


Solution

Let $$x$$ be the age of the son one year ago, then the age of the man was $$8x$$.

The present age of the son is $$\left( {x + 1} \right)$$ and that of the man is $$\left( {8x + 1} \right)$$. Then,

$$8x + 1 = {\left( {x + 1} \right)^2}$$

$$8x + 1 = {x^2} + 1 + 2x$$

$$x^{2}=6x$$

$$x = 0,6$$

As the age cannot be 0, so the value of $$x$$ is 6.

So, present age of son $$=(x+1)=7$$ years and present age of man $$=(8x+1)=49$$ years


Mathematics

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