
In planning maintenance for a city’s infrastructure, a civil engineer estimates that, starting from the present, the population of the city will decrease by $10$ percent every $20$ years. If the present population of the city is $50000$, which of the following expressions represents the engineers estimate of the population of the city t years from now ?
A)$50000{\left( {0.1} \right)^{2t}}$
B)$50000{\left( {0.1} \right)^{\dfrac{t}{2}}}$
C)$50000{\left( {0.9} \right)^{2t}}$
D)$50000{\left( {0.9} \right)^{\dfrac{t}{{20}}}}$
Answer
595.2k+ views
Hint: Break the question into small problems. Calculate the population decrement that will happen in the first $20$years, that is, it gets reduced by $10\% $. Calculate the number of $20$- years periods that will have elapsed in t years.
Complete step-by-step answer:
Present population of the city is $50000$
After $20$years, population gets decreased by $10\% $, i.e., decrease in population is $\dfrac{{10}}{{100}} \times 50000$
Population after twenty years =
$50000 - \dfrac{{10}}{{100}} \times 50000$
(Initial population – decrease in population)
$ = 50000\left( {1 - 0.1} \right)$
$ = 50000\left( {0.9} \right)$ ----(1)
Similarly,
After next $20$years, it again gets reduced by $10\% $,
This time the initial population which gets reduced to $10\% $is
$ = 50000\left( {0.9} \right)$ (Using (1))
Decrease in population is
$ = \dfrac{{10}}{{100}} \times 50000\left( {0.9} \right)$
Population after forty years is
$
50000\left( {0.9} \right) - \dfrac{{10}}{{100}} \times 50000\left( {0.9} \right) \\
= 50000\left( {0.9} \right)\left( {1 - 0.1} \right) \\
\\
$
$ = 50000{\left( {0.9} \right)^2}$ ----(2)
There are $2$ twenty- year periods in $40$years.
Let the number of twenty- year periods in t years be n.
$\therefore n = \dfrac{t}{{20}}$
We can see that in equation (2),
$n = 2$
Hence, the population after twenty years can be given by
$ = 50000{\left( {0.9} \right)^n}$
Also, we know that for t years , $n = \dfrac{t}{{20}}$.
So, the population is $50000{\left( {0.9} \right)^{\dfrac{t}{{20}}}}$.
D) is the correct answer.
Note: In these types of questions, it is always helpful to find n which is the number of $20$year periods that will have elapsed t years from present. It helps to generalise the solution.
Complete step-by-step answer:
Present population of the city is $50000$
After $20$years, population gets decreased by $10\% $, i.e., decrease in population is $\dfrac{{10}}{{100}} \times 50000$
Population after twenty years =
$50000 - \dfrac{{10}}{{100}} \times 50000$
(Initial population – decrease in population)
$ = 50000\left( {1 - 0.1} \right)$
$ = 50000\left( {0.9} \right)$ ----(1)
Similarly,
After next $20$years, it again gets reduced by $10\% $,
This time the initial population which gets reduced to $10\% $is
$ = 50000\left( {0.9} \right)$ (Using (1))
Decrease in population is
$ = \dfrac{{10}}{{100}} \times 50000\left( {0.9} \right)$
Population after forty years is
$
50000\left( {0.9} \right) - \dfrac{{10}}{{100}} \times 50000\left( {0.9} \right) \\
= 50000\left( {0.9} \right)\left( {1 - 0.1} \right) \\
\\
$
$ = 50000{\left( {0.9} \right)^2}$ ----(2)
There are $2$ twenty- year periods in $40$years.
Let the number of twenty- year periods in t years be n.
$\therefore n = \dfrac{t}{{20}}$
We can see that in equation (2),
$n = 2$
Hence, the population after twenty years can be given by
$ = 50000{\left( {0.9} \right)^n}$
Also, we know that for t years , $n = \dfrac{t}{{20}}$.
So, the population is $50000{\left( {0.9} \right)^{\dfrac{t}{{20}}}}$.
D) is the correct answer.
Note: In these types of questions, it is always helpful to find n which is the number of $20$year periods that will have elapsed t years from present. It helps to generalise the solution.
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