Question

The percentage increase in the total number of students of a school over that in the previous year.

Year	Percentage increase
1999-2000	20%
2000-2001	30%
2001-2002	10%

Find the effective percentage increase in the number of students from \[1998 - 1999\] to $2000 - 2001$.

Answer 1

Hint: The percentage increase in every academic year is given. So we can calculate the effective percentage increase in two years finding the increase in number in two years with respect to the initial number of students. Since the initial number of students is not given, we can call it with some variable (say $n$).

 Formula used:
${\text{Effective percentage increase = }}\dfrac{{{\text{increase in number }}}}{{{\text{original number }}}} \times 100$

Complete step-by-step answer:
We are given the percentage increase of students in three years.
We have to calculate the effective percentage increase.
Let the total number of students of the school in the academic year $1998 - 1999$ be $n$.
And the total number of students in the following years will be ${n_1},{n_2},{n_3}$ respectively.
It is given that there is an increase of $20\% $ in the year $1999 - 2000$.
So, number of students in the year $1999 - 2000$ will be ${\text{n}} + (20\% {\text{ of n)}}$
Number of students in the year $1999 - 2000$,${n_1} = n + \dfrac{{20}}{{100}}n = n + \dfrac{n}{5} = \dfrac{{6n}}{5}$
$ \Rightarrow {n_1} = \dfrac{{6n}}{5}$
It is given that there is an increase of $30\% $ in the year $2000 - 2001$.
So, number of students in the year $2000 - 2001$ will be $\dfrac{{6n}}{5} + (30\% {\text{ of }}\dfrac{{6n}}{5}{\text{)}}$
$ \Rightarrow {n_2} = \dfrac{{6n}}{5} + (\dfrac{{30}}{{100}} \times \dfrac{{6n}}{5})$
Simplifying we get,
$ \Rightarrow {n_2} = \dfrac{{6n}}{5} + \dfrac{{18n}}{{50}} = \dfrac{{60n}}{{50}} + \dfrac{{18n}}{{50}}$
$ \Rightarrow {n_2} = \dfrac{{78n}}{{50}}$
Therefore the total increase in these two years is given by subtracting the number of students in the year $1998 - 1999$ from the number of students in the year $2000 - 2001$.
$ \Rightarrow {\text{Increase in number of students = }}{n_2} - n$
Substituting we get,
$ \Rightarrow {\text{Increase in number of students = }}\dfrac{{78n}}{{50}} - n = \dfrac{{78n - 50n}}{{50}} = \dfrac{{28n}}{{50}}$
${\text{Effective percentage increase = }}\dfrac{{{\text{increase in number }}}}{{{\text{original number }}}} \times 100$
Substituting we get,
${\text{Effective percentage increase = }}\dfrac{{\dfrac{{28n}}{{50}}{\text{ }}}}{{{\text{n }}}} \times 100$
$ \Rightarrow {\text{Effective percentage increase = }}\dfrac{{{\text{28n }}}}{{{\text{50n }}}} \times 100 = 28 \times 2 = 56$
$\therefore $ The effective percentage increase in the number of students from $1998 - 1999$ to $2000 - 2001$ is $56\% $.

Note: In the table it is given the percentage increase in three years. But we are asked to find the effective increase in two years. So we do not need the final value ($10\% $). If we add that in calculating effective percentage, our answer will be incorrect.