The population of town decrease by $12$ % during $1998$ and then increased by $8$ % during $1999$ . find the population of the town ,at the beginning of $1998$ , if at the end of $1999$ its population was $285120$ .

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Solution

Let population at the beginning of $1998 = x$ .

After decreasing $12 %$ in $1998$ , population will be $= 12 %$ of $x$
$= x - \frac{12 x}{100} = \frac{88 x}{100}$

After $8 %$ increase in $1999$ , population will be $= \frac{88 x}{100} + 8 %$ of $\frac{88 x}{100}$
$= \frac{88 x}{100} [1 + \frac{8}{100}] = \frac{88 x}{100} [\frac{108}{100}]$

$\Rightarrow \frac{(88) (108)}{10000} x = 285120$
$\Rightarrow x = \frac{285120 \times 10000}{88 \times 108} = 300000$

$\Rightarrow$ Population, at the beginning of $1998$ was

$= 300000$

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The population of town decrease by 12 % during 1998 and then increased by 8 % during 1999. find the population of the town ,at the beginning of 1998, if at the end of 1999 its population was 285120.

The population of town decrease by $12$ % during $1998$ and then increased by $8$ % during $1999$ . find the population of the town ,at the beginning of $1998$ , if at the end of $1999$ its population was $285120$ .