Question

The population of a place increased to $54000$ in $2003$ at a rate of $5$ % per annum

1. find the population in $2001$
2. what would be its population in $2005$?

Answer 1

(i) It is given that, population in the year $2003$ = $54,000$
Therefore,
$54000$ = $(Population\; in \;2001)$ $\bigg(1 + \frac{5}{100}\bigg)^2$

$Population \;in \;2001$ = $54000 × \frac{20}{21} × \frac{20}{21}$ = $48979.59$

Thus, the population in the year $2001$ was approximately $48,980$.

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(ii) $Population\; in\; 2005$ = $54000\bigg(1+\frac{5}{100}\bigg)^2$

$54000\bigg(1+\frac{1}{20}\bigg)^2 = 54000 × \frac{21}{20} × \frac{21}{20} = 59,535$

Thus, the population in the year $2005$ would be $59,535$.