Question

The population of a place increased to 54000 in 2003 at a rate of \[5\% \] per annum. Find the population is in 2001.
A. 48980(approx.)
B. 46480(approx.)
C. 49910(approx.)
D. 47970(approx.)

Answer 1

Hint: Here, we will assume the population in 2006 with any variable and then use the formula, \[{\text{Present population = Population }}n{\text{ years ago}} \times {\left( {1 + \dfrac{r}{{100}}} \right)^n}\], where \[n\] is the time in years and \[r\] is the rate of increment. Then we will substitute the given values in the above formula and simplify to find the required value.

Complete step by step answer:

It is given that the population in 2003 is 54000 and the rate of increment \[r\] is \[5\].

Let us assume that the population in 2001 is \[x\].

We know that the time from 2001 to 2003 is 2 years.

Now we will use the formula to calculate the present population, \[{\text{Present population = Population }}n{\text{ years ago}} \times {\left( {1 + \dfrac{r}{{100}}} \right)^n}\], where \[n\] is the time in years and \[r\] is the rate of increment.


Substituting the above values of \[r\] and \[n\] in the above formula, we get
\[{\text{Present Population}} = x \times {\left( {1 + \dfrac{5}{{100}}} \right)^2}\]

Substituting the value of the present population in the above equation, we get

\[
   \Rightarrow 54000 = x \times {\left( {\dfrac{{100 + 5}}{{100}}} \right)^2} \\
   \Rightarrow 54000 = x \times {\left( {\dfrac{{105}}{{100}}} \right)^2} \\
   \Rightarrow 54000 = {\left( {\dfrac{{21}}{{20}}} \right)^2}x \\
   \Rightarrow 54000 = \dfrac{{441}}{{400}}x \\
 \]

Multiplying the above equation by 400 on each of the sides, we get

\[
   \Rightarrow 54000 \times 400x = \left( {\dfrac{{441}}{{400}}} \right) \times 400 \\
   \Rightarrow 21600000x = 441 \\
 \]

Dividing the above equation by 441 on both sides, we get

\[
   \Rightarrow \dfrac{{21600000}}{{441}} = \dfrac{{441x}}{{441}} \\
   \Rightarrow 48979.59 = x \\
   \Rightarrow x = 48980{\text{ (approx.)}} \\
 \]

Thus, the population in 2001 is 48980 (approx.).

Hence, option A is correct.

Note: In this question, we assume the value of the present population with any of the variables and then use it to find the original value will be the easiest way to find the required answer. Also, we will use the formula for calculating the present population, \[{\text{Present population = Population }}n{\text{ years ago}} \times {\left( {1 + \dfrac{r}{{100}}} \right)^n}\], where \[n\] is the time in years and \[r\] is the rate of increment and substitute the values carefully or else the students may go wrong.