Alright, my guess is the question is saying, for some fixed value of $p$ and $q$, what is the maximum and minimum number of students we can have. If so, I agree the maximum number is $pq$, as we just don't let any of the students overlap in any committees. It's also been awhile since I've done anything like this, so hopefully I'm not making a silly mistake!
For the minimum, you want to maximize the overlap. Lets start with the simplest(?) case, where $p=5$ and $q=6$. Call the committees $p_1,\ldots,p_5$. The maximum overlap that can occur, is 4 students from one committee each being members of the other 4 committees, hence with the maximum overlap possible, the number of students we require is \begin{align*} 30 - 4 - 3 - 2 - 1 = 20. \end{align*}
What if we change $q = 7$? Nothing changes, as the committees have overlapped as much as possible already, so for any $q \geq 6$, the number of students is \begin{align*} pq - \sum_{i=1}^{p-1} i = pq - \frac{p(p-1)}{2}. \end{align*}
Of course this only "half" of what we want. What if we leave $q$ at 6, but set $p=6$? Now there is an entire other committee where we can overlap students, so we overlap as many more as possible, we get that the minimum number of students required is \begin{align*} 36 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*}
As the number of committees increases, we may be able to overlaps more students. Suppose $p=7$ and $q=6$. We find the minimum number of students required is \begin{align*} 42 - 6 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*} The same as above! But if we add one more committee, we cannot overlap any of the present students, so we must have another $q$ added in. In general, set $m = \lfloor p/(q+1) \rfloor$. We require, minimally \begin{align*} m\left(q(q+1) - \sum_{i=1}^q i\right) + \left(rq - \sum_{i=1}^{r-1} i\right) \end{align*} students, where $r$ is the remainder when $p$ is divided by $q+1$. Of course, the sums can be evaluated to give you an expression without the $i$'s if you'd like.