**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?

**Solution:**

Let the population of the town be 50000 .

Rate of increase for the first year, $p=5 \%$

Rate of increase for the second year, $q=4 \%$

Rate of increase for the third year, $r=3 \%$

Time $=3$ years

Now, present population $=\left\{P \times\left(1+\frac{p}{100}\right) \times\left(1+\frac{q}{100}\right) \times\left(1+\frac{r}{100}\right)\right\}$

$=\left\{50000 \times\left(1+\frac{5}{100}\right) \times\left(1+\frac{4}{100}\right) \times\left(1+\frac{3}{100}\right)\right\}$

$=\left\{50000 \times\left(\frac{100+5}{100}\right) \times\left(\frac{100+4}{100}\right) \times\left(\frac{100+3}{100}\right)\right\}$

$=\left\{50000 \times\left(\frac{105}{100}\right) \times\left(\frac{104}{100}\right) \times\left(\frac{103}{100}\right)\right\}$

$=\left\{50000 \times\left(\frac{21}{20}\right) \times\left(\frac{26}{25}\right) \times\left(\frac{103}{100}\right)\right\}$

$=(21 \times 26 \times 103)$

$=56238$

Therefore, the present population of the town is 56238 .