Explanation:

- Let's assign variables to the number of students in the three classes. Let the number of students in the first class be $4x$, in the second class be $9x$, and in the third class be $11y$.

- According to the problem, the total number of students is 885. Therefore, the equation becomes: $4x+9x+11y=885$.

- There is also a ratio between the second and third class, with the ratio given as $6:11$. So, we have $\frac{9x}{11y}=\frac{6}{11}$.

- Solving $\frac{9x}{11y}=\frac{6}{11}$ gives $y=\frac{9x}{6}$.

- Substituting $y=\frac{9x}{6}$ into $4x+9x+11y=885$ and solving gives us the total students in each class.

- Solving the simultaneous equations results in the number of students in the classes being $x=18$, so the first class has $72$, the second class has $162$, and the third class has $495$.

495 students are in the class with the maximum number of students.