$ S(x) $ is the predicate "$x$ is a student"
$F(x)$ is the predicate "$x$ is a faculty member"
$A(x,y)$ is the predicate $x$ asked $y$ a question
I need to translate this sentence into logic: Some students asked every faculty member a question.
This is the answer: $\forall y(F(y) \longrightarrow \exists x(S(x) \vee A(x,y))) $
It seems to translate into: For all $y$, if $y$ is a faculty member, then there is some $x$ where $x$ is either a student or $x$ asks $y$ a question. This means that it's possible for $x$ to not be a student, but this makes no sense. I would replace $\vee$ with $\wedge $, would I be wrong?
My second question iswhether the following would also be correct, assuming the first answer is right:
$\exists x(S(x) \longrightarrow \forall y(F(y) \vee A(x,y)))$
There exists an $x$ where if $x$ is a student, then for all $y$ where $y$ is a faculty member, or (I still think it makes more sense if you plug and and here) $x$ asks $y$ a question. It seems to say the same thing.
I think the English of "some student asked every faculty member a question" is ambiguous. One could read it that there is a single student who asked every faculty member a question, or that every faculty member got asked a question by some student, but it could be various students. The book answer (with $\wedge$ instead of $\vee$ as you suggest) uses the second reading. Your answer is getting close to the first reading because starting with $\exists x$ forces it to be the same student doing the asking. You still need to work on the connectives. The difference between $\forall x \exists y Q(x,y)$ and $\exists y \forall x Q(x,y)$ is that in the first case we can find a different $y$ that works for each $x$, while in the second it has to be the same $y$ that works with every $x$.