The population of a city increases each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find :

(i) its population 2 years hence

(ii) its population 2 years ago.

(i) By formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = 6760000\Big(1 + \dfrac{4}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\dfrac{104}{100}\Big)^2 \\[1em] = 6760000 \times \Big(\dfrac{26}{25}\Big)^2 \\[1em] = 6760000 \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] = \dfrac{6760000 \times 676}{625} \\[1em] = 10816 \times 676 \\[1em] = 7311616.$

Hence, the population after 2 years = 7311616.

(ii) Let population 2 years ago be P and present population will be the final population.

By formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$\Rightarrow 6760000 = P \times \Big(1 + \dfrac{4}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\dfrac{104}{100}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \Big(\dfrac{26}{25}\Big)^2 \\[1em] \Rightarrow 6760000 = P \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] \Rightarrow 6760000 = P \times \dfrac{676}{625} \\[1em] \Rightarrow P = \dfrac{6760000 \times 625}{676} \\[1em] \Rightarrow P = 625 \times 10000 \\[1em] \Rightarrow P = 6250000.$

Hence, the population 2 years ago was 6250000.