To fall in the set \(\displaystyle A \cup B \cup C'\), a restaurant would have to be in the union of three sets - that is, any one or more. The three sets are:

\(\displaystyle A\), the set of all restaurants in Bancroft that accept the America's Best credit card;

\(\displaystyle B\), the set of all restaurants in Bancroft that accept the Bank Zero credit card; and

\(\displaystyle C '\), the complement of the set of all restaurants in Bancroft that accept the Chargitall credit card, or, equivalently, set of all restaurants in Bancroft that do not accept the Chargitall credit card.

The complement of this set, \(\displaystyle (A \cup B \cup C' ) '\), is the set of all restaurants that belong to none of \(\displaystyle A , B, C'\), so the correct choice must be a restaurant that belongs to  \(\displaystyle A', B',\) and \(\displaystyle C\). The restaurant must accept Chargitall, and must not accept the other two cards.

The only one of the five restaurants fitting this description is Benny's.

The opposite of the statement \(\displaystyle v = A ' \cap B'\) is the statement 

\(\displaystyle v = (A ' \cap B' )'\),

which, by DeMorgan's Law, is equivalent to

\(\displaystyle v = (A ' )' \cup ( B' )'\), or

\(\displaystyle v =A \cup B\)

Victor belongs to either or both of \(\displaystyle A\), the set of all people who like cola, and \(\displaystyle B\), the set of all people who like milk. Therefore, Victor likes cola, milk, or both.

We can best answer this question using a Venn diagram. We want to examine the union of two sets:

\(\displaystyle A \cap B\), the intersection of \(\displaystyle A\) and \(\displaystyle B\), which is the region the two circles have in common, and

\(\displaystyle C'\), the complement \(\displaystyle C\), which is the region outside the circle representing this set.

Since we are examing the union, we want to shade in everything in one or both of these two sets. This is shown below:

If Tommy likes Frank Sinatra, The Beatles, and Elvis Presley, \(\displaystyle t\) falls in all three circles, placing him in Region (i).

If Tommy does not like Frank Sinatra, The Beatles, or Elvis Presley, \(\displaystyle t\) falls in none of the three circles, placing him in Region (ii).

If Tommy likes Frank Sinatra, but neither The Beatles nor Elvis Presley, \(\displaystyle t\) falls in circle A, but neither B nor C, placing him in Region (iii).

If Tommy likes Frank Sinatra and The Beatles, but not Elvis Presley, \(\displaystyle t\) falls in circles A and B, but not C, placing him in Region (iv).

If Tommy likes Frank Sinatra and Elvis Presley, but not The Beatles, \(\displaystyle t\) falls in circles A and C, but not B, placing him in Region (v).

In all of the first four cases, \(\displaystyle t \in (A \cap B) \cup C'\) is a true statement, since the region is part of the grayed area. In the fifth case, \(\displaystyle t \in (A \cap B) \cup C'\) is false. Therefore, the only impossible statement among the choices is "Tommy likes Frank Sinatra and Elvis Presley, but not The Beatles."

Assume Statement 1 alone. \(\displaystyle A \cap B \cap C\) is the intersection of the sets \(\displaystyle A\), \(\displaystyle B\), and \(\displaystyle C\). Since Brittany falls in this intersection, she falls in all three sets, and it follows that she likes all three of James Blunt, John Legend, and Pharrell Williams.

Assume Statement 2 alone. \(\displaystyle A \cup B \cup C\) is the union of the three sets. Since Brittany falls in this union, she falls in one, two, or all three sets, which means that she likes at least one of James Blunt, John Legend, and Pharrell Williams. However, more information is needed before it can be established whether or not she likes all three.

In set notation, "Jeremy is not a dancer, but he is a singer" can be stated as

\(\displaystyle j \in A' \cap B\)

That is, Jeremy falls in the intersection of the complement of the set of dancers \(\displaystyle A\) and the set of singers \(\displaystyle B\).

The opposite of this is that \(\displaystyle j\) is in the complement of this set:

\(\displaystyle j \in (A' \cap B) '\)

By DeMorgan's law, this is equivalent to saying

\(\displaystyle j \in (A' ) ' \cup B'\), or

\(\displaystyle j \in A \cup B'\).

Let \(\displaystyle M,N,P,R\) be the sets of Moogs, Noogs, Poogs, and Roogs, respectively, From the premise of the problem, we have that

\(\displaystyle M \subseteq N \subseteq P \subseteq R\).

It follows by transitivity that \(\displaystyle M \subseteq P\) and \(\displaystyle M \subseteq R\).

Assume Statement 1 alone. If Stumpy is a Moog, then

\(\displaystyle \textup{Stumpy} \in M \subseteq P\).

, makng Stumpy a Poog. But since Stumpy is NOT a Poog, then it follows, contrapositively, that Stumpy is not a Moog.

Assume Statement 2 alone. If Stmpy is a Moog, then

\(\displaystyle \textup{Stumpy} \in M \subseteq R\),

making Stumpy a Roog. Since Stumpy is NOT a Roog, then it follows, contrapositively, that Stumpy is not a Moog.

If \(\displaystyle n \in (A \cup B ' ) \cap C'\), then \(\displaystyle n\) falls in the intersection of two sets. One is \(\displaystyle A \cup B '\), the union of \(\displaystyle A\), the set of people who like The Rolling Stones, and the complement of \(\displaystyle B\), the set of people not in \(\displaystyle B\) - the set of all people who do not like The Who. The other set is the complement of \(\displaystyle C\), the set of all people not in \(\displaystyle C\) - the set of people who don't like Lynyrd Skynyrd. 

Therefore, if \(\displaystyle n \in (A \cup B ' ) \cap C'\), both of the following must happen: Nancy must like The Rolling Stones or not like The Who, or both, and Nancy must not like Lynyrd Skynyrd.

From Statement 1 alone, Nancy does not like The Who, but it is not clear who else she likes. She is in \(\displaystyle A \cup B '\), but it is not clear whether she is in \(\displaystyle C\) the set of people who don't like Lynyrd Skynyrd, so the question of whether she falls in \(\displaystyle (A \cup B ' ) \cap C'\) is open. 

From Statement alone, since Nancy likes Lynyrd Skynyrd, she cannot be in \(\displaystyle (A \cup B ' ) \cap C'\). This makes \(\displaystyle n \in (A \cup B ' ) \cap C'\) false.

The opposite of the statement \(\displaystyle k \in A ' \cup B\) is the statement

\(\displaystyle k \in (A ' \cup B )'\),

which, by DeMorgan's Law, is equivalent to

\(\displaystyle k \in ( A ' ) ' \cap B '\)

or

\(\displaystyle k \in A \cap B '\)

This means that Kim is in the intersection of \(\displaystyle A\) and \(\displaystyle B'\), or, equivalently, she is in both \(\displaystyle A\), the set of people who can swim, and \(\displaystyle B '\), the complement of the set of people who can ride a bicycle. Therefore, Kim can swim, but she cannot ride a bicycle.

\(\displaystyle j \in A \cap (B' \cup C)\), which is the intersection of \(\displaystyle A\) and \(\displaystyle B' \cup C\). It follows that \(\displaystyle j \in A\) and \(\displaystyle j \in B' \cup C\).

\(\displaystyle j \in A\), so it follows that Julianna likes Band A. The three choices that state that she does not can be eliminated.

\(\displaystyle j \in B' \cup C\). This is the union of \(\displaystyle B'\), the complement of \(\displaystyle B\) - that is, the set of people not in \(\displaystyle B\) - and \(\displaystyle C\). It follows that either Julianna does not like Band B, does like Band C, or both. Therefore, it is not true that she likes Band B and does not like Band C. This can be eliminated.

The only possible choice is that Julianna likes Band A and Band C, but not Band B. 

In set theory, "Wendy likes neither catfish nor flounder" can be stated as

\(\displaystyle w \in C' \cap F'\).

The opposite of this is that \(\displaystyle w\) is in the complement of this set:

\(\displaystyle w \in (C' \cap F')'\).

By DeMorgan's law, this is equivalent to saying

\(\displaystyle w \in (C' )' \cup ( F')'\), or

\(\displaystyle w \in C \cup F\).