Question

By using the formula, find the amount and compound interest on:

Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of $5\%,4\%and3\%$ respectively, what is its present population?

Answer 1

$\text{3 years ago, the population was = 50000}\text{Rate of increase successively}(r_1,r_2,r_3)=4\%,5\%and3\%p.a.\text{Period (n) =3 years}\therefore \text{Present Population}=P(1+\frac{r_1}{100})(1+\frac{r_2}{100})(1+\frac{r_3}{100})=50000(1+\frac{5}{100})(1+\frac{4}{100})(1+\frac{3}{100})=50000\times \frac{21}{20}\times \frac{26}{25}\times \frac{103}{100}=562.38$