Can an algorithm that yields $O(n^2)$ answers run in $O(n)$ time? My question may actually be more broadly described as: can I use the fact that an algorithm is expected to return $(O(f(n))$ answers to show that it may never run better than $O(f(n))$? I would intuitively assume yes, but I am not sure. 

Examples of what I'm talking about is algorithms to calculate all the paths that pass through a set of points. I can easily calculate a higher bound $O(f(n))$ on how many those paths are. Will that tell me the algorithm must be no better than $\Omega(f(n))$?

Thanks
Can an algorithm that yields $O(n^2)$ answers run in $O(n)$ time?


Yes.  The following algorithm yields $O(n^2)$ answers (remember what $O(\cdot)$ means!) but it runs in time $O(n)$.

halt;


You probably meant to ask if an algorithm that yields $\Theta(n^2)$ answers can run in time $O(n)$. The answer to that is no. Whatever exact definition you choose for "yielding an answer", yielding one answer must take at least one computation step in any reasonable model of computation, so writing $\Theta(n^2)$ answers must take $\Omega(n^2)$ steps.