Question

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 $2005$?

Answer 1

Given:

The population of a place increased to $54,000$ in $2003$

i)

To find population in $2001$

Here,

New population $A=54,000$

Time from $2001$ to $2003$ is $\left(n\right)=2$ yrs

$R=5\%$

Original population $P=?$

$A=P(1+\frac{R}{100}{)}^2$

$\Rightarrow 54000=P(1+\frac{5}{100}{)}^2$

$\Rightarrow 54000=P(\frac{105}{100}{)}^2$

$\Rightarrow 54000\times (\frac{100}{105}{)}^2=P$

$\Rightarrow \frac{54000\times 100\times 100}{105\times 105}=P$

$\Rightarrow 48979.59=P$

Since the population is always a whole number.

So, $P=48980$ [Since, rounded to tens.]

Therefore, Population in $2001$ was $48,980$.

ii)

To find population in $2005$.

Here, $P=54000$

$R=5\%$

Population in $2005$ is $A=?$

From $2003$ to $2005$, time $\left(n\right)=2$

$A=P(1+\frac{R}{100}{)}^2$

$\Rightarrow A=54000(1+\frac{5}{100}{)}^2$

$\Rightarrow A=54000(\frac{105}{100}{)}^2$

$\Rightarrow A=\frac{54000\times 105\times 105}{100\times 100}$

$\Rightarrow A=5.4\times 105\times 105$

$\Rightarrow A=59535$

Therefore, Population in $2005$ will be $59,535$.