At the beginning of the school year, the ratio of the number of Grade \(10\) students to the number of Grade \(9\) students at CEMC H.S. was \(15:16\). By the end of the year, there were \(30\) more Grade \(10\) students and there were \(20\) fewer Grade \(9\) students, and the ratio of the number of Grade \(10\) students to the number of Grade \(9\) students was now \(11:10\).
How many Grade \(9\) students and how many Grade \(10\) students were there at the beginning of the school year?
Originally, the ratio of the number of Grade \(10\) students to the number of Grade \(9\) students was \(15:16\). Therefore, we can let the number of Grade \(10\) students at the beginning of the year be \(15n\) and the number of Grade \(9\) students at the beginning of the year be \(16n\), for some integer \(n\).
Thus, at the end of the year, there were \(15n+30\) Grade \(10\) students and \(16n - 20\) Grade \(9\) students.
Since the ratio of the number of Grade \(10\) students at the end of the year to the number of Grade \(9\) students at the end of the year is \(11:10\), we have \[\begin{aligned} \frac{15n+30}{16n-20}&=\frac{11}{10}\\ 150n+300&=176n - 220\\ 520&=26n\\ n&=20 \end{aligned}\] Therefore, there were \(16n = 16(20) =320\) Grade \(9\) students and \(15n = 15(20) = 300\) Grade \(10\) students at the beginning of the school year.