Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$36$ 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?

share|cite|improve this question
What does the question has to do with set theory? – Asaf Karagila Jun 21 '11 at 14:28
This is a job for... Inclusion-Exclusion Man! – Arturo Magidin Jun 21 '11 at 17:18

If $E$ is the set of the students that passed English and $M$ are those that passe 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 student 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 this:

Perhaps I should have also added this link:

share|cite|improve this answer

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...

share|cite|improve this answer

protected by user26857 Nov 4 '15 at 16:43

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.