Applications of Compound Interest
CI formulae can be applied to phenomenon of population, price change etc.
Population
If the population of a city is P and it increases with the rate of r% per annum, then
(i) Population after n years = P (1 + $\frac{r}{100})^n$
(ii) Population n years ago = $\frac{P}{(1 + \frac{r}{100})^n}$
(iii) If the rate of growth per year is $r_1$%, $r_2$%, $r_3$%,………$r_n$%, then Population after n years = P (1 + $\frac{r_1}{100}$) (1 + $\frac{r_2}{100}$) (1 + $\frac{r_3}{100}$)……(1 + $\frac{r_n}{100}$)
(This formula can also be used, if there is increase/decrease in the price of an article).
In questions involving population growth or appreciation/depreciation of assets, we will assume Compound Interest (unless stated otherwise).
In all other cases (e.g. bank loans), if nothing is mentioned, we will assume Simple Interest.
Q. Aedes Aegypti mosquitos are responsible for the spread of dengue. A colony of 25,000 such mosquitos increase at a rate of 10% of its population per hour. What will be the count of mosquitos after three hours?
(a) 33,275 (b) 33,175 (c) 27,000 (d) 27,335
Explanation:
Suppose that original population = P, Rate of population rise = r%, and Time = n periods
The given population will rise at a compounded rate, as the new mosquitos will keep adding on to the population growth.
Then the final population after n time periods = P $[1 + (r/100)]^n$
So, the population after three hours = 25000 $[1 + (10/100)]^3$ = 25000 $(11/10)^3$ = 25000 × $(1.1)^3$ = 25000 × 1.331 = 33,275
Answer: (a)