How to solve this problem using set theory? $36$ students took English and Math test. $25$ passed English, and $28$ passed Math. $20$ passed both subjects.
a. How many students failed both subject?
b. How many students passed english only?
c. How many students passed math only?
 A: If $E$ is the set of the students that passed English and $M$ are those that passed Math then you have:
$|E|=25$, $|M|=28$, $|E\cap M|=20$
From the formula
$$|E\cup M|=|E|+|M|-|E\cap M|$$
you get that $|E\cup M|=33$, which means that 33 students passed at least one subject.
I guess you can go on from here...
If you prefer diagrams instead of formulas, you can try to draw something like the diagrams shown here (just with different numbers): Venn Diagrams: Exercises (Purplemath). To prevent the link rot, here is also Wayback Machine link.
Perhaps I should have also added the link to the Wikipedia article Inclusion–exclusion principle.
A: Denote by $S$ the set of all students, by $E$ the set of all students that passed English and by $M$ the set of all students that passed Math. Then what you know is: $|S|=36$, $|E|=25$, $|M|=28$ and $|E\cap M|=20$. Then what you are looking for is:
a. $|(S\setminus E)\cap (S\setminus M)|=$
b. $|E\setminus M|=$
c. $|M\setminus E|=$
So you can use De Morgan's laws and Martin Sleziak's answer from here on to solve...
A: Let $e$ be the number of students who passed English but not math,
$m$ the number of students who passed math but not English,
$a$ the number of students who passed both tests,
$z$ the number of students who passed neither test.
$$36=e+m+a+z$$
$$25=e+a$$
$$28=m+a$$
$$20=a$$
$4$ equations in $4$ unknowns!
Spoiler:
$$a=20$$
$$e=(e+a)-a=25-20=5$$
$$m=(m+a)-a=28-20=8$$
$$e+m+a+z=36$$
$$5+8+20+z=36$$
$$33+z=36$$
$$z=36-33=3$$
