Question

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% and 3% respectively, what is its present population?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{population}\mathrm{of}\mathrm{the}\mathrm{town}\mathrm{be}50000. \\ \mathrm{Rate}\mathrm{of}\mathrm{increase}\mathrm{for}\mathrm{the}\mathrm{first}\mathrm{year},\mathit{}p=5\% \\ \mathrm{Rate}\mathrm{of}\mathrm{increase}\mathrm{for}\mathrm{the}\mathrm{second}\mathrm{year},q=4\% \\ \mathrm{Rate}\mathrm{of}\mathrm{increase}\mathrm{for}\mathrm{the}\mathrm{third}\mathrm{year},r=3\% \\ \mathrm{Time}=3\mathrm{years} \\ \mathrm{Now},\mathrm{present}\mathrm{population}=\left\{P\times \left(1+\frac{p}{100}\right)\times \left(1+\frac{q}{100}\right)\times \left(1+\frac{r}{100}\right)\right\} \\ =\left\{50000\times \left(1+\frac{5}{100}\right)\times \left(1+\frac{4}{100}\right)\times \left(1+\frac{3}{100}\right)\right\} \\ =\left\{50000\times \left(\frac{100+5}{100}\right)\times \left(\frac{100+4}{100}\right)\times \left(\frac{100+3}{100}\right)\right\} \\ =\left\{50000\times \left(\frac{105}{100}\right)\times \left(\frac{104}{100}\right)\times \left(\frac{103}{100}\right)\right\} \\ =\left\{50000\times \left(\frac{21}{20}\right)\times \left(\frac{26}{25}\right)\times \left(\frac{103}{100}\right)\right\} \\ =\left(21\times 26\times 103\right) \\ =56238 \\ \mathrm{Therefore},\mathrm{the}\mathrm{present}\mathrm{population}\mathrm{of}\mathrm{the}\mathrm{town}\mathrm{is}56238. \end{gathered}$$