In the first year the population of a town decreased by 5% due to the Corona virus first wave. In the next year it decreased again by 5% due to the second wave and in the third year it increased by 5%. At the end of the third year the population was 9,47,625. What was the population at the beginning of the first year?

Correct option is B

​Given:
Population at the end of the third year = 9,47,625

Let the population at the beginning of the first year be P.
Concept Used:
The population decreases by 5% in each of the first two years and increases by 5% in the third year.
The population after a percentage decrease/increase can be calculated using the formula:

$$\text{Population after change} = \text{Initial Population} \times \left(1 \pm \frac{\text{Rate of Change}}{100}\right)$$​​
Solution:
Population at the end of the first year after a 5% decrease:

$$P \times \left(1 - \frac{5}{100}\right) = P \times 0.9$$​​
Population at the end of the second year after another 5% decrease:

$$P \times 0.95 \times 0.95 = P \times (0.95)^2$$​​
Population at the end of the third year after a 5% increase:

$$P \times (0.95)^2 \times \left(1 + \frac{5}{100}\right) = P \times (0.95)^2 \times 1.05$$​​
We are given that the population at the end of the third year is 9,47,625. Therefore:

$$P \times (0.95)^2 \times 1.05 = 9,47,625$$​​

$$P = \frac{9,47,625}{(0.95)^2 \times 1.05}$$​​
First, $$calculate (0.95)^2 = 0.9025$$​

$$P = \frac{9,47,625}{0.9025 \times 1.05}\\\ \\P = \frac{9,47,625}{0.947625}\\\ \\P = 10,00,000$$​​
The population at the beginning of the first year was 10,00,000.

Alternative Method: