Question

The Population of a village declines by 5% every year. If the present population of the village is 1444, find the population of the village a year back.

Answer 1

Hint: As we can see, the rate of decrement is 5%. Similar in a way of compound interest getting an principle amount of the currently. The difference here is that in the case of people, the rate is negative.

Complete step-by-step answer:
 To get compound interest in the form of the population. Let’s assume that 2 years back. There was a P population of a village and in a time period of 2 years, it declined to 1444.
Which seems like $\left( {1 - \dfrac{5}{{100}}} \right) \times \;total\;population$
Next year the remaining will be 5% decrement on the remaining which is
=$\left( {1 - \dfrac{5}{{100}}} \right) \times \left( {1 - \dfrac{5}{{100}}} \right) \times \;total\;population$
Similarly, next year $\left( {1 - \dfrac{5}{{100}}} \right) \times \left( {1 - \dfrac{5}{{100}}} \right) \times \left( {1 - \dfrac{5}{{100}}} \right)\;total\;population$
\[
{\rm{So}}\,{\text{Population}}\,{\rm{Remained}}\,{\rm{ = }}\,{\text{P}}{\left( {1 - \dfrac{{{\rm{Rate}}}}{{{\rm{100}}}}} \right)^{{\rm{Time}}}}\\
{\text{So}}\,{\text{we}}\,{\text{have,}}\,\\
\Rightarrow 1444 = {\rm{P}}\,{\left( {1 - \dfrac{5}{{100}}} \right)^2}\\
\Rightarrow {\rm{P}}\,{\rm{ = }}\,\dfrac{{{\rm{1444}}}}{{{{\left( {\dfrac{{{\rm{19}}}}{{{\rm{20}}}}} \right)}^2}}}\\
 = \dfrac{{20 \times 20 \times 1444}}{{19 \times 19}} \\
= 1600.\]
Hence, the total population 2 years back was 1600.

Note: As population was constantly decreasing by 5% next year, the remaining 95%. Again, next year the remaining will be 5% decrease on the remaining. It works in a similar way to the Rate of interest in Compound form where we also get interest on the amount of interest we obtain. This is so because the principal value increases. We needed to acquire the knowledge of compound Interest to deal with such a type of question, where the next data is dependent on the obtained data on previously.