90% of students face problems in A, 80%in B, 75%in C and 70% in D. What is the minimum percentage of people facing problems in all the subjects. The complete question is
In a school , 90% of students face problems in A, 80%in B, 75%in C and 70% in D. What is the minimum percentage of people facing problems in all the subjects.
It looks fairly trivial but I am confused on how to proceed. Can anyone help?
 A: Let's assume the school has 20 students, so that we can work with numbers instead of percentages.  Then $18$ are in set $A$, $16$ in $B$, $15$ in $C$, and $14$ in $D$.  One way to solve this is to go through the subjects one by one.
Let's start with $A$ and $B$.  We know that 18 students have problems in $A$, and $16$ students have problems in $B$.  We want to find the minimum number who have problems in both subjects.  The minimum occurs when we make the intersection as small as possible, which occurs when there are $14$ students in $A \cap B$, $4$ students in $A \setminus B$, and $2$ students in $B \setminus A$.  Now notice that out of these $20$ students, we only need to worry about the $14$, since they are the only ones who have a chance at being in all four sets.
So we continue with $14$ students in $A \cap B$ and $15$ students in $C$.  By the same type of argument, the minimum number of students in $A \cap B \cap C$ is $9$.  Finally, there are $14$ students in $D$ and $9$ in $A \cap B \cap C$, so the minimum number of students in $A \cap B \cap C \cap D$ is $3$, which is $15$% of the $20$ students.

EDIT: Maybe it is easier to look at the percentages not in each set.  Then we have $10\%$ not in $A$, $20\%$ not in $B$, $25\%$ not in $C$, and $30\%$ not in $D$.  The intersection is minimized when there is no overlap among the percentages listed above.  In otherwords, the full $10 + 20 + 25 + 30 = 85\%$ are not in all four, so that $15\%$ are in all four.

EDIT 2: To check that the minimum in $A \cap B$ is indeed 14, it may help to look at the following Venn diagram.  Call the intersection "brown".

There are 18 students in orange + brown, so there are 2 students in blue + white (white is outside both A and B).  There are 16 students in blue + brown, so there are 4 students in orange + white.  Therefore, there are at least 0 students in white, and at most 2 students in white.  If there are 0 students in white, there are 2 in blue, 4 in orange, and 14 in brown.  If there is 1 student in white, there is 1 in blue, 3 in orange, and 15 in brown.  If there are 2 students in white, there are 0 in blue, 2 in orange, and 16 in brown.  So you can see that we minimize brown when we choose zero in white.  (By the way, this is the basis of the alternate proof in the first edit.)
