The population of a town increases by 10% in the first year and 20% in the second year. If the present population is 66,000 , what was the population two years ago?

Correct option is D

Given:
First year increase = 10%
Second year increase = 20%
Present population = 66000
Formula Used:
Initial Population = Final Population × $$\frac{100}{100 + R_1} × \frac{100}{100 + R_2}$$​​
Solution:
Let the population two years ago be P.
​$$P × (1 + \frac{10}{100}) × (1 + \frac{20}{100}) = 66000 \\ \ \\P × \frac{110}{100} × \frac{120}{100} = 66000\\ \ \\P × 1.1 × 1.2 = 66000\\ \ \\P × 1.32 = 66000\\ \ \\P = \frac{66000}{1.32}\\ \ \\P = 50000$$​​
Final Answer
So the correct answer is (d)