What will be difference in population 3 years ago and 2 years ago of a town, whose current population is 100000 and which is increasing at a rate of 25% every year?

Correct option is B

Given:
Current population = 100000
Annual increase rate = 25%
Formula Used:
​$$\text{Population }\ n\ \text{years ago} = \frac{\text{Current Population}}{\left(1 + \frac{r}{100}\right)^n}$$​
Where r is the growth rate per year.
Solution:
Population 3 years ago :
​$$= \frac{100000}{\left(1 + \frac{25}{100}\right)^3} = \frac{100000}{\left(\frac{5}{4}\right)^3} \\\ \\= \frac{100000}{\frac{125}{64}} = \frac{100000 \times 64}{125} = 51200$$​​

Population 2 years ago:
​$$= \frac{100000}{\left(1 + \frac{25}{100}\right)^2} = \frac{100000}{\left(\frac{5}{4}\right)^2}\\\ \\= \frac{100000}{\frac{25}{16}} = \frac{100000 \times 16}{25} = 64000$$​​
Difference in population = 64000 - 51200 = 12800
Alternate Solution:

Population 2 years ago = 100000 $$\times$$​ $$\left( \frac{4}{5} \right) \times \left( \frac{4}{5} \right)$$ = 64000

Population 3 years ago = 100000 $$\times$$​$$\left( \frac{4}{5} \right) \times \left( \frac{4}{5} \right)\times \left( \frac{4}{5} \right)$$= 51200

Difference in population = 64000 - 51200 = 12800