The population of a town is increasing at the rate of 10% per annum. If its present population is 36300, find:

(i) its population after 2 years,

(ii) its population 2 years ago.

(i) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

Population after n years = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population after 2 years } = 36300 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{100 + 10}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] =36300 \times \dfrac{121}{100} \\[1em] =43923.$

Hence, population of the town after 2 years = 43923.

(ii) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

Population before n years = $\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Population 2 years ago }= \dfrac{36300}{\Big(1 + \dfrac{10}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{100+10}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{110}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{11}{10}\Big)^2} \\[1em] = \dfrac{36300}{\dfrac{121}{100}} \\[1em] = \dfrac{36300 \times 100} {121} \\[1em] = 30000.$

Hence, population of the town 2 years ago = 30000.