Question

The population of a town increases by $10\%$ annually. If the present population is $60,000$, what will be its population after $2$ years?

Answer 1

Population after three years will be given by the formula ,

$A=P{\left(1+\frac{r}{100}\right)}^n$

Where $A=$Population of the town after two years

$P$= Current population of the city

$r=$rate of increase of population per year

$n=$Time period

Given , $P=60,000,r=10\%,n=2years.$

Substituting the given values in the formula we get ,

$$\begin{gathered} A=60,000{\left(1+\frac{10}{100}\right)}^2 \\ =60,000{\left(1+\frac{1}{10}\right)}^2 \\ =60,000{\left(\frac{11}{10}\right)}^2 \\ =60,000\times \frac{11}{10}\times \frac{11}{10} \\ =72,600 \end{gathered}$$

Hence the population of a town after three years is $72,600$