The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005?

Hint: Here, we will use the formula of compound ratio to calculate the population in the years 2001 and 2005 i.e.., $New.P = Old.P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$.First we calculate population in 2001 with respect to given population in 2003 and similarly calculate population in 2005 with respect to 2003 using above formula we get desired answer.

Complete step-by-step answer:
1. Find the population in 2001.
It is given that the population in 2003 is 54,000 and we are supposed to find the population in 2001, the difference between 2001 and 2003 is 2 years. Therefore, the time is 2 years, and the rate at which the population is increasing per year is 5%.
Therefore, Rate of interest, $R = 5\% $.
We are going to use the formula $New.P = Old.P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$ to find the population in 2001.
Since, we do not have the value of the original population let us assume that to be P.
Therefore from the given data,
\[
   \Rightarrow 54000 = P{\left( {1 + \dfrac{5}{{100}}} \right)^2} \\
   \Rightarrow 54000 = P{\left( {1 + \dfrac{1}{{20}}} \right)^2} \\
   \Rightarrow 54000 = P{\left( {\dfrac{{21}}{{20}}} \right)^2} \\
   \Rightarrow P = \dfrac{{54000 \times 440}}{{441}} \\
   \Rightarrow P = 48979.59 \\
 \]
The population in 2001 would have been around 48,980.

2. What would be its population in 2005?
In this part, we are supposed to find the population in 2005.
Same formula used in the previous part will be used here i.e. $New.P = Old.P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$
Here, we do not have the value of the new population, let us assume that to be P.
The value of n is 2 since the time difference is 2 and the rate of interest(R) is 5%
Therefore, the above formula becomes
\[
   \Rightarrow P = 54000{\left( {1 + \dfrac{5}{{100}}} \right)^2} \\
   \Rightarrow P = 54000{\left( {1 + \dfrac{1}{{20}}} \right)^2} \\
   \Rightarrow P = 54000{\left( {\dfrac{{21}}{{20}}} \right)^2} \\
   \Rightarrow P = \dfrac{{54000 \times 441}}{{400}} \\
   \Rightarrow P = 59,395 \\
 \]
Therefore, the population in 2005 will be 59,395.

Note: Make sure to take note of the original and the new population values and make sure you do not mismatch. Here when we are finding the population in the year 2001, we have to find the old population and in the case of finding the population in the year 2005, we have to find the new population with respect to the year 2003.Student should remember the formula $New.P = Old.P{\left( {1 + \dfrac{R}{{100}}} \right)^n}$ for solving these types of problems.