Question 14 Marks
The population of a place increased to $54,000$ in $2003$ at a rate of $5\%$ per annum.
$(i)$ Find the population in $2001.$
$(ii)$ What would be its population in $2005?$
$(i)$ Find the population in $2001.$
$(ii)$ What would be its population in $2005?$
Answer
View full question & answer→$(i)$ Given: Population in $2003=54,000$
$\begin{aligned}
& \text { Rate }=5 \% \text { pa } \\
& \text { Time }=2003-2001=2 \text { years } \\
& \text { Population in } 2003=\text { Population in } 2001 \times\left(1+\frac{ R }{100}\right)^n \\
& 54,000=\text { Population in } 2001 \times\left(1+\frac{5}{100}\right)^2 \\
& \begin{array}{r}
\Rightarrow \quad 54,000=\text { Population in } 2001 \times\left(\frac{21}{20}\right)^2 \\
\Rightarrow \quad 54,000=\text { Population in } 2001 \times \frac{441}{400} \\
\therefore \text { Population in } 2001=\frac{54,000 \times 400}{441} \\
\quad=\frac{21,6,00,000}{441}=48,979.59 \\
\quad=48,980 \text { (approximately) }
\end{array}
\end{aligned}$
$\begin{aligned}
& \text { (ii) Population in } 2005=\text { Population in } 2003 \times\left(1+\frac{ R }{100}\right)^n \\
& =54,000 \times\left(1+\frac{5}{100}\right)^2 \\
& =54000 \times\left(\frac{21}{20}\right)^2 \\
& ={54000} \times \frac{441}{400} \\
& =135 \times 441=59,535
\end{aligned}$
$\begin{aligned}
& \text { Rate }=5 \% \text { pa } \\
& \text { Time }=2003-2001=2 \text { years } \\
& \text { Population in } 2003=\text { Population in } 2001 \times\left(1+\frac{ R }{100}\right)^n \\
& 54,000=\text { Population in } 2001 \times\left(1+\frac{5}{100}\right)^2 \\
& \begin{array}{r}
\Rightarrow \quad 54,000=\text { Population in } 2001 \times\left(\frac{21}{20}\right)^2 \\
\Rightarrow \quad 54,000=\text { Population in } 2001 \times \frac{441}{400} \\
\therefore \text { Population in } 2001=\frac{54,000 \times 400}{441} \\
\quad=\frac{21,6,00,000}{441}=48,979.59 \\
\quad=48,980 \text { (approximately) }
\end{array}
\end{aligned}$
$\begin{aligned}
& \text { (ii) Population in } 2005=\text { Population in } 2003 \times\left(1+\frac{ R }{100}\right)^n \\
& =54,000 \times\left(1+\frac{5}{100}\right)^2 \\
& =54000 \times\left(\frac{21}{20}\right)^2 \\
& ={54000} \times \frac{441}{400} \\
& =135 \times 441=59,535
\end{aligned}$