The mode of a set, if it exists, is the value that occurs most frequently. The inequality

\(\displaystyle A = B > C > D = E > F\)

means that the set

\(\displaystyle \left \{ A, B, C, D, E,F\right \}\) 

can be rewritten as

\(\displaystyle \left \{ A, A, C, D,D,F\right \}\)

\(\displaystyle A\) and \(\displaystyle D\) occur as values twice each; the other values, \(\displaystyle C\) and \(\displaystyle F\), are unique. Therefore, the set has two modes, \(\displaystyle A\) and \(\displaystyle D\).

The mode of a set is the value that occurs most frequently in that set. Since 

\(\displaystyle A = B < C= D\), it follows that 

\(\displaystyle \left\{ A, B, C, D \right \}\)

can be rewritten as

\(\displaystyle \left\{ A, A, C, C \right \}\).

This makes \(\displaystyle A\) and \(\displaystyle C\) both modes, since both occur twice. Equivalently, since \(\displaystyle A = B\) and \(\displaystyle C= D\), \(\displaystyle B\) and \(\displaystyle D\) are modes.

Assume both statements to be true, and examine two cases.

Case 1: \(\displaystyle A = 1, B = 2, C = 3, D= 4, E= 5\)

\(\displaystyle \mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C\)

\(\displaystyle A < B < C < D< E\)

The arithmetic mean of \(\displaystyle A\) and \(\displaystyle E\) is

\(\displaystyle \frac{A+ E}{2} = \frac{1+5}{2} = \frac{6}{2} = 3 = C\)

The conditions of both statements are satisfied.

The mean of the five numbers is their sum divided by 5:

\(\displaystyle \mu = \frac{A+B+C+D+E}{5} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 = C\)

Case 2: \(\displaystyle A = 0, B = 2, C = 3, D= 5, E= 6\)

\(\displaystyle A < B < C < D< E\)

The arithmetic mean of \(\displaystyle A\) and \(\displaystyle E\) is

\(\displaystyle \frac{A+ E}{2} = \frac{0+6}{2} = \frac{6}{2} = 3 = C\)

The conditions of both statements are satisfied.

But the mean of the five numbers is

\(\displaystyle \mu = \frac{A+B+C+D+E}{5} = \frac{0+2+3+5+6}{5} = \frac{16}{5} = 3.2 \ne C\)

Therefore, the mean may or may not be equal to \(\displaystyle C\).

Suppose we know  \(\displaystyle a, b \in \left \{ 1, 3, 5 ,7, 9\right \}\), but we do not assume the second statement.

If \(\displaystyle c = 2\) and \(\displaystyle d = 4\), then \(\displaystyle A \cap B = \left \{}a, b, 2, 4 \right \}\), a four-element set. If If \(\displaystyle c = 2\) and \(\displaystyle d = 12\), then \(\displaystyle A \cap B = \left \{}a, b, 2 \right \}\), a three-element set. Therefore, we cannot make a conclusion about the size of \(\displaystyle A \cap B\). A similar argument can be used to show that assuming only the second statement also does not allow a conclusion.

If we know both statements, however, \(\displaystyle A \cap B = \left \{}a, b, c, d \right \}\), and we can prove that \(\displaystyle A \cap B\) has four elements.

The answer is that both statements together are sufficient to answer this question, but neither statement alone is sufficient.

\(\displaystyle A\) includes all multiples of 2; \(\displaystyle B\) includes all multiples of 3. \(\displaystyle A \cup B\) comprises all multiples of either 2 or 3.

Knowing \(\displaystyle n\) is a perfect square is neither necessary nor helpful, as, for example, \(\displaystyle 9 \in B \subseteq A \cup B\), but \(\displaystyle 25 \notin A \cup B\) (as 25 is neither a multiple of 2 nor a multiple of 3).

If you know that \(\displaystyle n\) is a multiple of 99, then it must also be a multiple of any number that divides 99 evenly, one such number is 3. This means \(\displaystyle n \in B \subseteq A \cup B\)

Assume both statements are true.

Consider these two cases:

Case1: \(\displaystyle A = \left \{ 1,2,3,4,5\right \}\) and \(\displaystyle B = \left \{ 5,6\right \}\)

Case 2: \(\displaystyle A = \left \{ 1,2,3,4,5,6\right \}\) and \(\displaystyle B = \left \{ 5,6,7\right \}\)

In both situations, \(\displaystyle A\) has three more elements than \(\displaystyle B\) and \(\displaystyle A\) includes exactly four elements not in \(\displaystyle B\) (1, 2, 3 and 4). However, the number of elements in the union differ in each case - in the first case, \(\displaystyle A \cup B= \left \{ 1,2,3,4,5,6\right \}\), and in the second case, \(\displaystyle A \cup B= \left \{ 1,2,3,4,5,6, 7\right \}\). 

The two statements together do not yield an answer to the question.

The question asks whether Jimmy is an element of \(\displaystyle M\).

Statement 1 alone - that Jimmy is an element of \(\displaystyle T'\) - provides insufficient information, since \(\displaystyle T'\) contains elements that are and are not elements of \(\displaystyle M\). By a similar argument, Statement 2 alone is insufficient.

Now assume both statements to be true. Then Jimmy is an element of \(\displaystyle T' \cap E'\), shaded in the Venn diagram below:

It can be seen that \(\displaystyle T' \cap E'\) shares no elements with \(\displaystyle M\), so Jimmy cannot be an element of \(\displaystyle M\). Jimmy is not a Mason.

The question asks whether or not Marty is an element of \(\displaystyle E\).

Assume Statement 1 alone. He is an element of \(\displaystyle M' \cap T'\), represented by the shaded region below:

\(\displaystyle M' \cap T'\) includes elements that are and are not elements of \(\displaystyle E\), so it cannot be determined whether or not Marty is in \(\displaystyle E\).

Assume Statement 2 alone. Then Marty has to be an element of the set represented by the shaded region below:

Since some of the set is in \(\displaystyle E\) and some is not, it cannot be determined whether or not Marty is in \(\displaystyle E\).

If both statements are known, then, since Marty is in exactly one of the three sets, and he is not a Mason or a Toastmaster, then he must be an Elk.

The question is whether or not Craig is an element of \(\displaystyle T\).

Assume both statements to be true. Craig is an element of the set \(\displaystyle M' \cap E'\), shaded in this Venn diagram:

There are elements of this set that both are and are not elements of \(\displaystyle T\). Therefore, the two statements together do not prove or disprove Craig to be an element of \(\displaystyle T\), a Toastmaster.

The question is equivalent to asking whether Jerry is an element of set \(\displaystyle M\).

The sets \(\displaystyle T\) and \(\displaystyle M\) are disjoint - they have no elements in common. From Statement 1 alone, Jerry is an element of \(\displaystyle T\), so he cannot be an element of \(\displaystyle M\). He is not a Mason.

From Statement 2 alone, Jerry is an element of \(\displaystyle E\). Since there are elements not in \(\displaystyle E\) that are and are not elements of \(\displaystyle M\), it cannot be determined whether Jerry is an element of \(\displaystyle M\) - a Mason.