Question

The population of a city in Jan 1987 was 20,000. It increased at the rate of 2% p.a. Find the population of the city in Jan 1997.

Answer 1

Hint: If the rate of interest is annual then in such cases we use the formula for compound interest to solve such questions. Calculate the amount which becomes the population in Jan 1987 putting time (t) to get to the desired answer.

Complete step-by-step answer:
If the principal = P, rate of interest per unit time = r %, number of units of time = n, the amount = A
Then
$\text{Amount} = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$
Similarly population in Jan 1997 is the amount , rate = 2% , time(n) = (1997-1987) = 10 years,
P = Population in 1987.
Therefore population in 1997 = $20000{\left( {1 + \dfrac{2}{{100}}} \right)^{10}}$
$ = 20000{\left( {\dfrac{{102}}{{100}}} \right)^{10}} = 20000 \times {1.02^{10}} = 24380$

Note: Computation of compound interest by using growing principal becomes lengthy and complicated when the period is long. You should understand the principle of using compound interest in such types of questions. Remember to round off the figure at the end as the population has to be a whole number.