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.

The statement says all seniors take statistics so if you are a senior you are in statistics automatically.  It also said some baseball team members are seniors which means at least some teammates must be in statistics.

 represents the intersection of the two sets. In other words, we want all the elements that appear in both sets. The elements that appear in both sets are 2, 4, 6, and 10.

Therefore, 

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.

The intersection of the Venn Diagram is only the numbers in both circles.

The section in the middle contains the answer set. 

Thus the intersection is, .

 represents the intersection of the two sets A and B. The intersection of A and B is the set of elements that are contained in both A and B. 

The elements that appear in A and B are 4, 5, and 154. 

Therefore, the intersection of A and B is . 

To find how many plots grow both carrots and corn, we subtract . 

That means that fifty plots grow corn only, while thirty plots grow carrots only. 

The intersection of 2 sets A and B is the set of the items that are included in both sets. The items that appear in both A and B are 4 and 35. 

Therefore, the intersection of A and B is .

The shaded portion of the Venn diagram is  - the set of all elements in  but not .

The following elements of  yield a remainder of 1 when divided by 4, and therefore comprise set :

If 3 is subtracted from each, what results are the elements whose division by 4 yields a remainder of 1, and thus, elements of .

The following elements of  yield a remainder of 1 when divided by 4, and therefore comprise set :

The elements in  that are also elements in  are 1, 13, 25, and 37 - four elements out of ten. Therefore, the set  comprises six elements.

The average of two numbers can be calculated as the sum of those numbers divided by 2. B would thus be calculated as (A + C)/2, and D would be calculated as (C + E)/2. To find the average of those values, you would add them up and divide by 2: