Seven years ago Varun's age was five times the square of Swati's age. Three years hence, Swati's age will be two-fifth of Varun's age. Find their present ages.

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Solution

Seven years ago, let Swati's age be $x$ years. Then, seven years ago Varun's age was $5 x^{2}$ years.
$∴$ Swati's present age $= (x + 7)$ years
Varun's present age $= (5 x^{2} + 7)$ years
Three years hence
Swati's age $= (x + 7 + 3) y e a r s = (x + 10) y e a r s$
Varun's age $= (5 x^{2} + 7 + 3) y e a r s = (5 x^{2} + 10) y e a r s$
According to the questions,
$x + 10 = \frac{2}{5} (5 x^{2} + 10)$
$\Rightarrow$ $x + 10 = \frac{2}{5} \times 5 (x^{2} + 2)$
$\Rightarrow$ $x +! 0 = 2 x^{2} + 4$
$\Rightarrow$ $2 x^{2} - x - 6 = 0$
$\Rightarrow$ $2 x^{2} - 4 x + 3 x - 6 = 0$
$\Rightarrow$ $2 x (x - 2) + 3 (x - 2) = 0$
$\Rightarrow$ $(2 x + 3) (x - 2) = 0$
$\Rightarrow$ $x - 2 = 0$ or $2 x + 3 = 0$
$\Rightarrow$ $x = 2$ [ $∵ 2 x + 3 \neq 0$ as $x > 0$ ]
Hence, Swati's present age $= (2 + 7) = 9$ years
and Varun's present age $= (5 \times 2^{2} + 7) = 27$ years.

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