The number of students enrolled in a school is 2000. If the enrolment increases by 5% every year, how many students will be there after two years?

Hint: We know that, ${\text{amount}} = p{\left( {1 + \dfrac{{rate}}{{100}}} \right)^{time}}$. We are given the values of number of students, and rate of increase and the number of years. Therefore we can find the number of students after 2 years. By using the above mentioned formula we will find the required solution.

Complete step by step answer:

The number of students enrolled in a school is 2000. The enrolment increases by 5% every year.
Let the current number of the student be ′p′
Therefore \[p = 2000\]
Rate of increment (r)$ = 5\% $
Time (n) \[ = 2\]years
We need to find the number of students after two years. Let’s denote that by A.
Now we know that, ${\text{amount}} = p{\left( {1 + \dfrac{{rate}}{{100}}} \right)^{time}}$
i.e. $A = p{\left( {1 + \dfrac{r}{{100}}} \right)^n}$
On substituting the values we get,
$ \Rightarrow A = 2000 \times {\left( {1 + \dfrac{5}{{100}}} \right)^2}$
On simplification we get,
$ \Rightarrow A = 2000 \times {\left( {\dfrac{{21}}{{20}}} \right)^2}$
On expanding the square we get,
$ \Rightarrow A = 2000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}}$
On simplification we get,
$ \Rightarrow A = 5 \times 21 \times 21$
On multiplying we get,
$ \Rightarrow A = 2205$
Therefore, the number of students after two years will be 2205.

Note: Here increment is given so we use ${\text{amount}} = p{\left( {1 + \dfrac{{rate}}{{100}}} \right)^{time}}$, but if decrement is given so our rate will be negative, so the formula will change to, ${\text{amount}} = p{\left( {1 - \dfrac{{rate}}{{100}}} \right)^{time}}$. So, one should see what is asked to calculate and then use the formula accordingly, and solve accordingly.