Draw a Venn diagram with three subsets:  Math, Science, and English.  Start in the center with students that like all three subjects.  Next, look at students that liked two subjects.  Be sure to subtract out the ones already counted in the middle.  Then, look at the students that only like one subject.  Be sure to subtract out the students already accounted for.  Once all of the subsets are filled, look at those students who don’t like any of these subjects.  To find the students who don’t like any of these subjects add all of the students who like at least one subject from the total number of students surveyed, which is 50. 

M = math

S = science

E = English

M∩S∩E = 3        

M∩S = 5 (but 3 are already accounted for) so 2 for M and S ONLY

M∩E = 7 (but 3 are already accounted for) so 4 for M and E ONLY

S∩E = 8 (but 3 are already accounted for) so 5 for S and E ONLY

M = 14 (but 3 + 2 + 4 are already accounted for) so 5 for M ONLY

S = 20 (but 3 + 2 + 5 are already accounted for) so 10 for S ONLY

E = 28 (but 3 + 4 + 5 are already accounted for) so 16 for E ONLY

Therefore, the students already accounted for is 3 + 2 +4 + 5 + 5 + 10 + 16 = 45 students

So, those students who don’t like any of these subjects are 50 – 45 = 5 students

A = {2, 4, 6, 8, 10, 12}

B = {3, 6, 9, 12, 15, 18}

The intersection of a set means that the elements are in both sets: A∩B = {6, 12}

This problem can be solved two ways, with a formula or with reason.

Using the formula, the intersection of the Venn diagram for which classes students take is:

By using reason, it is clear that 60 + 70 is greater than 100 by 30. It is assumed that this extra 30 students come from students who were counted twice because they took both classes.

To find the intersection of a set of numbers, we need to find all numbers that are in both sets (definition of intersection). We can see that all the primes less than  that are odd are also odd numbers less than , therefore they are in both sets and in the intersection. Thus the answer:

 means the complement of  union  - this is everything in the universe that is not in  or . First we need to find :

 - we find this by knowing anything in either set is in the union. Next we need to find things in the universe that are NOT in :

Since the universe is integers less than or equal to 10, we see that the only ones that fit that description that are not in  are 

First, subtract the students that are in neither class; 75 – 30 = 45 students.

Thus, 45 students are enrolled in Chemistry, Physics, or both.  Of these 45 students, we know 40 are in Chemistry, so that leaves 5 students who are enrolled in Physics only; with 15 total students in Physics, that means 10 must be in Chemistry as well. So 10 students are in both Physics and Chemistry.