Solving a question about inclusion exclusion principle 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...
 A: 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. 
A: 25 students answer all three questions hence 25 students answered one or two questions, as no student didn't answer any question. But we know that there are 12 + 14 + 10 = 36 unanswered questions in total.  Hence 11 of the 25 students didn't answer two questions, as 25 + 11 = 36.
Therefore 11 answered one question; 25 - 11 = 14 answered two questions.
A: You can set up 7 linear equations.
Let's let "A only" = a
B only = b
C only = c
A and B only = x
B and C only = z
C and A only = y
then
$b+c+z=12$
$a+x+y=13$
$a+y+c=14$
$b+x+z=11$
$a+x+b=10$
$y+c+z=15$
$a+b+c+x+y+z=25$
