Question

The population of a town increases by \[5\% \] annually. If the population in 2009 is 1,38,915, what was it in 2006?
A.100,000
B.110,000
C.120,000
D.130,000

Answer 1

Hint: Here, we will assume the population in 2006 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 present population is 1,38,915 and the rate of increment \[r\] is \[5\].
Let us assume that the population in 2006 is \[x\].
We know that the time from 2006 to 2009 is 3 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
\[{\text{Present Population}} = x \times {\left( {1 + \dfrac{5}{{100}}} \right)^3}\]
Substituting the value of present population in the above equation, we get
\[
   \Rightarrow 1,38,915 = x \times {\left( {\dfrac{{100 + 5}}{{100}}} \right)^3} \\
   \Rightarrow 1,38,915 = x \times {\left( {\dfrac{{105}}{{100}}} \right)^3} \\
   \Rightarrow 1,38,915 = {\left( {\dfrac{{21}}{{20}}} \right)^3}x \\
   \Rightarrow 1,38,915 = \dfrac{{9261}}{{8000}}x \\
 \]
Multiplying the above equation by 8000 on each of the sides, we get
\[
   \Rightarrow 1,38,915 \times 8000x = \left( {\dfrac{{9261}}{{8000}}} \right) \times 8000 \\
   \Rightarrow 1111320000x = 9261 \\
 \]
Dividing the above equation by 9261 on both sides, we get
\[
   \Rightarrow \dfrac{{1111320000}}{{9261}} = \dfrac{{9261x}}{{9261}} \\
   \Rightarrow 120000 = x \\
   \Rightarrow x = 120000 \\
 \]
Thus, the population in 2006 was 1,20,000

Hence, the option C 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. Make sure to not mistake the values between the old and new population while assigning it in the formula. Students must know that formula for getting the increase in population is \[{\text{Present population = Population }}n{\text{ years ago}} \times {\left( {1 + \dfrac{r}{{100}}} \right)^n}\], but if need to find the decrease in the population, then we use the formula, \[{\text{Present population = Population }}n{\text{ years ago}} \times {\left( {1 - \dfrac{r}{{100}}} \right)^n}\].