Question

The population of a town 3 years ago was 50,000. If the population in these three years has increased at the rate 10%, 15% and 8% respectively, find the present population.
\[\begin{array}{*{35}{l}}
   \left( \text{a} \right)\text{ 68,210} \\
   \left( \text{b} \right)\text{ 68,310} \\
   \left( \text{c} \right)\text{ 68,410} \\
   \left( \text{d} \right)\text{ 68,510} \\
\end{array}\]

Answer 1

Hint: In this we will find the population after 3 years by using the formula for the population after n years which given as follows
\[\text{Population after n years in which rate changes every year}={{P}_{n-1}}\left( 1+\dfrac{{{R}_{n}}}{100} \right)\]
\[\text{Where }n=\text{year, }P=\text{ Initial Population and }R=\text{ rate of population increased in }\!\!~\!\!\text{ }{{n}^{th}}\text{year}\text{.}\]

Complete step-by-step solution:
\[\text{Let }{{P}_{1}},\text{ }{{P}_{2}}~\text{and }{{P}_{3}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ be the population after }{{\text{1}}^{\text{st}}}\text{, }{{\text{2}}^{\text{nd}}}\text{and }{{\text{3}}^{\text{rd}}}\text{respectively}\].
\[\text{Let }{{R}_{1}},\text{ }R{{~}_{2}}\text{ and }{{R}_{3}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ be the rate of increase in population after }{{\text{1}}^{\text{st}}}\text{, }{{\text{2}}^{\text{nd}}}\text{and }{{\text{3}}^{\text{rd}}}\text{respectively}\text{.}\]
Given $P$= population 3 years ago of a town = 50,000.
By using the formula to calculate the population after $n$ year to calculate the population after 1st year which increased at rate ${{R}_{1}}$=10%.
We will find the increase in $1^{st}$ -year population at rate ${{R}_{1}}$.
Where $P=50000\text{ and }{{R}_{1}}=10%$
$\Rightarrow $${{P}_{1}}=P\left( 1+\dfrac{{{R}_{1}}}{100} \right)$
$\Rightarrow {{P}_{1}}=50000\left( 1+\dfrac{10}{100} \right)=50000\left( 1+\dfrac{1}{10} \right)=50000\left( \dfrac{11}{10} \right)$
$\Rightarrow $${{P}_{1}}$$=5000\times 11=55000$
$\Rightarrow $${{P}_{1}}=55000$.
Again, by using the formula to calculate the population after $n$ year to calculate the population after $2^{nd}$ year which increased at rate ${{R}_{2}}$=15% and taking consideration of population after 1st year.
We will find the increment in the 2nd-year population at a rate of ${{R}_{2}}$.
Where${{P}_{1}}=55000\text{ and }{{R}_{2}}=15%$.
$\Rightarrow $${{P}_{2}}={{P}_{1}}\left( 1+\dfrac{{{R}_{2}}}{100} \right)$
$\Rightarrow {{P}_{2}}=55000\left( 1+\dfrac{15}{100} \right)=55000\left( 1+\dfrac{3}{20} \right)=55000\left( \dfrac{23}{20} \right)$
$\Rightarrow $${{P}_{2}}=2750\times 23=63250$
$\Rightarrow $${{P}_{2}}=63250$.
Again, by using the formula to calculate the population after $n$ year to calculate the population after 3rd year which increased at rate ${{R}_{3}}$=8% and taking consideration of population after 2nd year.
We will find the increase in the 3rd-year population at rate ${{R}_{3}}$.
Where${{P}_{2}}=63250\text{ and }{{R}_{3}}=8%$.
$\Rightarrow $\[{{P}_{3}}={{P}_{2}}\left( 1+\dfrac{{{R}_{3}}}{100} \right)\]
\[\Rightarrow {{P}_{3}}=63250\left( 1+\dfrac{8}{100} \right)=63250\left( 1+\dfrac{2}{25} \right)=63250\left( \dfrac{27}{25} \right)\]
$\Rightarrow $${{P}_{3}}$$=2530\times 27=68310$
$\Rightarrow $${{P}_{3}}$ = 68310.
Hence, the population after 3 years is 68,310.
The option (b) is right.

Note: In this problem, we should pay attention to calculate the population in each year by taking into consideration the population of previous since the rate of population changes every year. Also, for successive increment, we can use the formula by multiplying all the formula of increment for n yeas like ${{P}_{3}}=P\cdot \left( 1+\dfrac{{{R}_{1}}}{100} \right)\cdot \left( 1+\dfrac{{{R}_{2}}}{100} \right)\cdot \left( 1+\dfrac{{{R}_{3}}}{100} \right) $