Two years ago, the population of a village was 4000. During next year it increased by 6% but due to an epidemic, it decreased by 5% in the following year. What is its population now?

Given,

P = 4000

r₁ = 6%

r₂ = 5%

Given,

The population increased by 6% in first year and decreased by 5% in second year.

By formula,

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

Substituting the values in formula,

$\text{Population after 2 years }= 4000 \times \Big(1 + \dfrac{6}{100}\Big) \times \Big(1 - \dfrac{5}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{100 + 6}{100}\Big) \times \Big(\dfrac{ 100 - 5}{100}\Big) \\[1em]\ = 4000 \times \Big(\dfrac{106}{100}\Big) \times \Big(\dfrac{95}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{53}{50}\Big) \times \Big(\dfrac{19}{20}\Big) \\[1em] = \dfrac{4000 \times 53 \times 19}{50 \times 20} \\[1em] = 4 \times 53 \times 19 \\[1em] = 4028.$

Hence, the present population of the village = 4028.