Question

The population of Andhra Pradesh in \[1988\] was ${ 5.4 \times {{10}^7}^{}}$ . If the population was growing at a constant rate of \[2.4\% \;p.a.\], what will be the population in 2008? Given that ${\left( {1.024} \right)^{20}} = 1.60694$

Answer 1

Hint: To solve this type of problem we need to know the formula of compound interest because the calculation of population uses parallel concepts as compound interest. Here this question can be solved by using the same concept.

Complete step-by-step answer:
The population of a particular year is given by
\[ \Rightarrow {P_n} = {P_0}{\left( {1 + \dfrac{r}{{100}}} \right)^n}\]
where \[{P_{0}} = \;\]Population in \[1988\]
\[ \Rightarrow {P_n} = \;\] Population in n years after \[1988\]
\[ \Rightarrow r = \;\]Population growth rate
\[ \Rightarrow n = \;\]Number of years after \[1988\]
Here the total number of years in from \[1988\] to \[2008\] is \[20\] year
\[
  { \Rightarrow {P_0} = 5.4 \times {{10}^7}^{}} \\
  {\Rightarrow n = 20} \\
  {\Rightarrow r = 2.4\% }
\]
Now calculating population in \[2008\]
Apply the formula
\[ \Rightarrow {P_n} = {P_0}{\left( {1 + \dfrac{r}{{100}}} \right)^n}\]
Put all the given values
\[
  { \Rightarrow {P_{2008}} = 5.4 \times {{10}^7}{{\left( {1 + \dfrac{{2.4}}{{100}}} \right)}^{20}}} \\
  { = 5.4 \times {{10}^7} \times 1.60694} \\
  { = 8.677 \times {{10}^7}}
\]
Hence, the population of Andhra Pradesh in 2008 will be \[8.677 \times {10^7}\].
So, the correct answer is “\[8.677 \times {10^7}\].”.

Note: We have seen in this solution that we have used the same formula which we are generally used to calculate the total amount in case of compound interest as the nature by which the population was increasing o demands it.