TITLE: Solving a question about inclusion exclusion principle
QUESTION [5 upvotes]: I am trying to solve the following question:
There are $50$ students in a class who are given a test with $3$ questions on it: $Q_1$, $Q_2$, and $Q_3$. All the students answer at least $1$ question. If $12$ students did not answer $Q_1$, $14$ did not answer $Q_2$, $10$ did not answer $Q_3$ and $25$ answered all $3$ questions, then how many students answered exactly $1$ question?
So I define $3$ sets as such:

Students who answered $Q_1 = A$
Students who answered $Q_2 = B$
Students who answered $Q_3 = C$

I am given the following:

$|A \cup B \cup C| = 50$
$|B \cup C \cap A^c| = 12$ (Students who did not answer $Q_1$)
$|A \cup C \cap B^c| = 14$ (Students who did not answer $Q_2$)
$|A \cup B \cap C^c)| = 10$ (Students who did not answer $Q_3$)
$|A \cap B \cap C| = 25$ (Students who answered all $3$)

The principle of Inclusion-Exclusion states that:
$$|A \cup B \cup C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|$$
Substituting for the knows values:
$$50 = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + 25$$
Or $$25 = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C|$$
But now I am stuck because the statement above does not give me any information about students who answered exactly 1 question. 
How do I move ahead or where did I go wrong?
I dont want to know the answer, I just want to know how to proceed. 
Since I got stuck there I went ahead and defined $3$ more sets:

Students who answered exactly $1$ question = $D$ 
Students who answered exactly $2$ questions = $E$
Students who answered exactly $3$ questions = $F$ 

I know that $|F| = 25$
and $|D \cup E \cup F| = 50$
But now I am stuck again...

REPLY [5 votes]: I'd like to flesh out @turkeyhundt's method, which is basically what you've started out writing. Here's a Venn diagram to clarify what variable I'll assign to which question:


We're given the following:
$$a+b+c + x+y+z=25$$
$$b+c+z=12$$
$$a+c+x=14$$
$$a+b+y=10$$
We're trying to find $a+b+c$. If we add the second, third, and fourth equations above we get
$$2a+2b+2c+x+y+z=36$$
Subtracting the first equation above gives us the desired result
$$a+b+c = \boxed{11}$$

Notice that it's not possible to find the individual values of $a$, $b$, and $c$, but it's also not necessary to do that in order to solve the problem.