Question

The population of a town 3 years ago was 25000. If the population in these three years has increased at the rate of 10%, 15% and 8% respectively, find the present population.

Answer 1

Hint: In this problem, the initial population, three years ago, is given and we have to find the present population. The percentage increase in each of the three years is also given. Since, the percentage increase in each year is different, we can find the present population using the formula,
$P={{P}_{0}}\left( 1+\dfrac{{{r}_{1}}}{100} \right)\left( 1+\dfrac{{{r}_{2}}}{100} \right)\left( 1+\dfrac{{{r}_{3}}}{100} \right)......\left( 1+\dfrac{{{r}_{n}}}{100} \right)$.

Complete step-by-step answer:
In such questions, if the percentage increase is the same for all the years, then the present population is found using the formula:
$\Rightarrow P={{P}_{0}}{{\left( 1+\dfrac{r}{100} \right)}^{n}}.........(i)$
Here,
$P=$ Present population
${{P}_{0}}=$ Initial population
$r=$ Percentage increase in each year
$n=$ Total number of years from ${{P}_{0}}$ to $P$
If the percentage increase is not same then the present population is found using the formula:
$\Rightarrow P={{P}_{0}}\left( 1+\dfrac{{{r}_{1}}}{100} \right)\left( 1+\dfrac{{{r}_{2}}}{100} \right)\left( 1+\dfrac{{{r}_{3}}}{100} \right)......\left( 1+\dfrac{{{r}_{n}}}{100} \right).............(ii)$
Here, ${{r}_{1}},{{r}_{2,}}{{r}_{3}}.....{{r}_{n}}$ are the percentage increase in each year from ${{P}_{0}}$ to $P$.

In this question, the initial population before three years is given and the percentage increase in these three years is given. We need to find the present population.
The percentage increase in each year is given as 10%,15%, and 8%.
 Since the percentage increases are different for each year, we will be using equation (ii) to solve this problem. In this question \[n=3\]. Therefore, we will have ${{r}_{1}},{{r}_{2}}$ and ${{r}_{3}}$.
$\Rightarrow P={{P}_{0}}\left( 1+\dfrac{{{r}_{1}}}{100} \right)\left( 1+\dfrac{{{r}_{2}}}{100} \right)\left( 1+\dfrac{{{r}_{3}}}{100} \right)......(iii)$
The given values are:
$\begin{align}
& P=25000 \\
& {{r}_{1}}=10\% \\
& {{r}_{2}}=15\% \\
& {{r}_{3}}=8\% \\
\end{align}$
Substituting the above values in equation (iii) we get,
$\Rightarrow P=25000\left( 1+\dfrac{10}{100} \right)\left( 1+\dfrac{15}{100} \right)\left( 1+\dfrac{8}{100} \right)$
Simplifying the above equation,
\[\begin{align}
& \Rightarrow P=25000\left( \dfrac{110}{100} \right)\left( \dfrac{115}{100} \right)\left( \dfrac{108}{100} \right) \\
& \Rightarrow P=25{0}{00}\left( \dfrac{11{0}}{1{00}} \right)\left( \dfrac{115}{1{00}} \right)\left( \dfrac{108}{100} \right) \\
& \Rightarrow P=25\left( 11 \right)\left( 115 \right)\left( \dfrac{108}{100} \right).......(iv) \\
\end{align}\]
Further simplifying equation (iv) we get,
\[\begin{align}
& \Rightarrow P=\dfrac{25\times 11\times 115\times 108}{100} \\
& \Rightarrow P=\dfrac{\overset{5}{\mathop{{25}}}\,\times 11\times 115\times \overset{54}{\mathop{{108}}}\,}{\underset{\underset{10}{\mathop{{20}}}\,}{\mathop{{100}}}\,} \\
& \Rightarrow P=\dfrac{{5}\times 11\times 115\times \overset{27}{\mathop{{54}}}\,}{\underset{{2}}{\mathop{{10}}}\,} \\
& \Rightarrow P=11\times 115\times 27 \\
& \Rightarrow P=34155 \\
\end{align}\]

$\therefore $ the present population $P$ is obtained as 34155.
Hence, the present population is 34155.

Note:In this problem, we have used a direct formula to find the solution. You can also solve this problem by finding the increased population in each year and then by adding it to the initial population. But this process can be tedious sometimes when you have a large number of years to calculate. Also, check if the percentage increases are the same or different.
Also note, the values for${{r}_{1}},{{r}_{2}}$ and ${{r}_{3}}$are substituted as percentages itself.