
The population of a city is \[20000\]. Find the population of the city after \[3\] year if the population increases by \[5\% \] every year.
Answer
414.9k+ views
Hint: In this question, we have to find the population of the city after \[3\] year. We will use the concept of percentage to find the population after three years. First of all, we find \[5\% \] of population and add the result to the population to get the population after \[1\] year. We will again find \[5\% \] of the population after \[1\] year and add to the same to get the population after \[2\] year. Continuing the same way, we can find the population after any given number of years.
Complete answer:
Consider the given question,
The total population of the city \[ = \]\[20000\]
We are given that every year population increases by \[5\% \]
Hence Total increase in population in \[{1^{st}}\] year \[ = \] \[5\% {\text{ of }}20000\]
i.e., \[ \Rightarrow \dfrac{5}{{100}} \times {\text{20000 = 1000}}\]
Therefore, total population after \[{1^{st}}\] year \[ = \]\[20000 + 1000 = 21000\]
Now again population increases by \[5\% \]
Therefore, total increase in \[{2^{nd}}\] year \[ = \]\[5\% {\text{ of }}21000\]
i.e., \[ \Rightarrow \dfrac{5}{{100}} \times {\text{21000 = 1050}}\]
Total population after \[2\] year \[ = \]\[21000 + 1050 = 22050\]
Again, in \[{3^{rd}}\] year, population increases by \[5\% \]
Hence total increase in \[{3^{rd}}\] year \[ = \]\[5\% {\text{ of }}22050\]
i.e., \[ \Rightarrow \dfrac{5}{{100}} \times {\text{22050 = 1102}}{\text{.5}}\]
Therefore, total population after \[3\] year \[ = \]\[22050 + 1102.5 = 23152.5\]
Hence the total population of the city after \[3\] year is \[23152.5\]
Note:
We can also find the solution to this problem using the shortcut method.
The population increases \[5\% \] every year.
Hence total population after \[1\] year \[ = \] \[\left( {100 + {\text{ }}5} \right){\text{ }}\% \]of population. i.e., \[105{\text{ }}\% {\text{ }}of{\text{ }}20000\].
Now we have to find population after \[3\] year, therefore we will find \[105\% {\text{ of }}105\% {\text{ of }}105\% {\text{ of }}20000\] to get the required population after \[3\] year.
Hence, \[ \Rightarrow \dfrac{{105}}{{100}} \times \dfrac{{105}}{{100}} \times \dfrac{{105}}{{100}} \times 20000 = 23152.5\] is the required population of the city.
Complete answer:
Consider the given question,
The total population of the city \[ = \]\[20000\]
We are given that every year population increases by \[5\% \]
Hence Total increase in population in \[{1^{st}}\] year \[ = \] \[5\% {\text{ of }}20000\]
i.e., \[ \Rightarrow \dfrac{5}{{100}} \times {\text{20000 = 1000}}\]
Therefore, total population after \[{1^{st}}\] year \[ = \]\[20000 + 1000 = 21000\]
Now again population increases by \[5\% \]
Therefore, total increase in \[{2^{nd}}\] year \[ = \]\[5\% {\text{ of }}21000\]
i.e., \[ \Rightarrow \dfrac{5}{{100}} \times {\text{21000 = 1050}}\]
Total population after \[2\] year \[ = \]\[21000 + 1050 = 22050\]
Again, in \[{3^{rd}}\] year, population increases by \[5\% \]
Hence total increase in \[{3^{rd}}\] year \[ = \]\[5\% {\text{ of }}22050\]
i.e., \[ \Rightarrow \dfrac{5}{{100}} \times {\text{22050 = 1102}}{\text{.5}}\]
Therefore, total population after \[3\] year \[ = \]\[22050 + 1102.5 = 23152.5\]
Hence the total population of the city after \[3\] year is \[23152.5\]
Note:
We can also find the solution to this problem using the shortcut method.
The population increases \[5\% \] every year.
Hence total population after \[1\] year \[ = \] \[\left( {100 + {\text{ }}5} \right){\text{ }}\% \]of population. i.e., \[105{\text{ }}\% {\text{ }}of{\text{ }}20000\].
Now we have to find population after \[3\] year, therefore we will find \[105\% {\text{ of }}105\% {\text{ of }}105\% {\text{ of }}20000\] to get the required population after \[3\] year.
Hence, \[ \Rightarrow \dfrac{{105}}{{100}} \times \dfrac{{105}}{{100}} \times \dfrac{{105}}{{100}} \times 20000 = 23152.5\] is the required population of the city.
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