Question

Population of a town three years ago was $60,000$. If the annual increase during three successive years be at the rate of $4\%,5\%$ and $6\%$ respectively, find its present population.

Answer 1

Population of the town $=60,000$.

Rate of increase for the first year $\left(p\right)=5\%$

Rate of increase for the second year $\left(q\right)=4\%$

Rate of increase for the third year $\left(r\right)=3\%$

Time $=3$ years

Now,

Present population $=P\times (1+\frac{p}{100})\times (1+\frac{q}{100})\times (1+\frac{r}{100})=60000\times (1+\frac{4}{100})\times (1+\frac{5}{100})\times (1+\frac{6}{100})=60000\times \frac{104}{100}\times \frac{105}{100}\times \frac{106}{100}=6\times 21\times 26\times 21.2=69451.2$