The population of a town increased by 10% and 20% in two successive years, but decreased by 25% in the third year. Find the ratio of the population in the third year and the population 3 years back.

Population increases by 10% in the first year.

Population increases by 20% in the second year.

Population decreases by 25% in the third year.

Population after n years with successive percentage changes is calculated using:

​$$\text{New population} = P \times \left(1 + \frac{\text{rate}_1}{100} \right) \times \left(1 + \frac{\text{rate}_2}{100}\right) \times \left(1 - \frac{\text{rate}_3}{100}\right)$$ 

Solution:   

Let the Population be 100.

Using the formula: 

Population after 3 year;

​$$\text{New population} = 100 \times \left(1 + \frac{10}{100} \right) \times \left(1 + \frac{20}{100}\right) \times \left(1 - \frac{25}{100}\right) \\ \ \\ = 100 \times \left(1 + \frac{10}{100} \right) \times \left(1 + \frac{20}{100}\right) \times \left(1 - \frac{25}{100}\right) \\ \ \\ = 100 \times \left( \frac{110}{100} \right) \times \left( \frac{120}{100}\right) \times \left( \frac{75}{100}\right) \\ \ \\ = 99$$   

Thus, the ratio between Population in 3rd year  and Population 3 year back is = 99 : 100.