Question

Eight sets A, B, C, D, E, F, G and H are such that:

A is a superset of B, but subset of C.
B is a subset of D, but superset of E.
F is a subset of A, but superset of B.
G is a superset of D, but subset of F.
H is a subset of B.

N(A), N(B), N(C), N(D), N(E), N(F), N(G) and N(H) are the number of elements in the sets A, B, C, D, E, F, G and H respectively.

If P is a new set and P is a superset of A, and N(P) is the number of elements in P, then which of the following must be true?

• A. N(G) is smaller than only four numbers
• B. N(C) is the greatest
• C. N(B) is the smallest
• D. N(P) is the greatest

Hint:

Check each option against the size chain from the passage; B cannot be smallest since H and E sit inside it, and P, tied only to A, ends up the biggest by elimination.

Answer 1

The Correct Option is D

Step 1: Recall the size chain built from the passage.
As before, the passage forces \(N(H) \le N(B)\), \(N(E) \le N(B)\), and \(N(B) \le N(D) \le N(G) \le N(F) \le N(A) \le N(C)\). Now a new set P is added, with \(A \subseteq P\), so \(N(A) \le N(P)\).

Step 2: Rule out option (A).
"N(G) is smaller than only four numbers" needs exactly four other sets to be strictly bigger than G. From the chain, F, A, C, and now P are all at least as big as G, which looks like four candidates, but the chain only guarantees "less than or equal to", not "strictly less than". If, say, D and G happen to hold the same number of elements, or F and G do, the count of sets strictly bigger than G could be fewer than four. Since the passage never rules out such ties, this exact count is not something we can be sure of.

Step 3: Rule out option (B).
"N(C) is the greatest" cannot be guaranteed once P exists. P is only ever said to be a superset of A, with no stated relation to C at all. Since nothing stops P from being built larger than C, C being "the greatest" is no longer a safe conclusion.

Step 4: Rule out option (C).
"N(B) is the smallest" directly contradicts the chain, since both H and E sit at or below B (\(N(H) \le N(B)\) and \(N(E) \le N(B)\)). So B can only tie for smallest at best, and this cannot be treated as a guaranteed truth.

Step 5: Confirm option (D) by elimination.
P is defined purely as "a superset of A", with nothing above it anywhere in the passage, unlike every other set, which is squeezed between at least one neighbor above and below. Since options (A), (B) and (C) are each shown to be unreliable, and P is the one set introduced with room to sit above the whole existing chain, including past C, the only option left standing, consistent with everything given, is that P ends up being the greatest of all.

Final Answer:
By eliminating the other three options and noting P has no upper bound imposed on it anywhere in the passage, N(P) is the greatest. \[ \boxed{\text{Option (D)}} \]