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 Q and Z are two new sets, both supersets of H, and N(Q) and N(Z) are the number of elements of the sets Q and Z respectively, then:

• A. N(H) is the smallest of all
• B. N(E) is the smallest of all
• C. N(C) is the greatest of all
• D. Either N(H) or N(E) is the smallest

Hint:

H and E both sit below every other set but are never compared with each other, so the overall smallest must be one of the two, even though we cannot say which.

Answer 1

The Correct Option is D

Step 1: Rebuild the size chain, ignoring Q and Z for a moment.
From the passage: \(N(H) \le N(B)\), \(N(E) \le N(B)\), and \(N(B) \le N(D) \le N(G) \le N(F) \le N(A) \le N(C)\). So among the original eight sets, both H and E sit at or below every other set in the chain, since B, D, G, F, A and C are all reached by following the chain upward from B.

Step 2: See what Q and Z actually add.
Q and Z are only said to be supersets of H, meaning \(N(H) \le N(Q)\) and \(N(H) \le N(Z)\), with no upper limit given. Since they only contain H, and H is already at or below everything else, Q and Z can never be smaller than H, but they could grow arbitrarily large, even bigger than C.

Step 3: Check each option.
Option (A), "N(H) is the smallest of all": this can fail if N(E) happens to be smaller than N(H), since the passage never compares E and H directly. Not guaranteed.
Option (B), "N(E) is the smallest of all": by the same logic, this fails if N(H) turns out smaller than N(E) instead. Not guaranteed.
Option (C), "N(C) is the greatest of all": this held true among the original eight sets, but once Q and Z exist with no upper bound beyond containing H, either one of them could be built larger than C, so C is no longer guaranteed to be the greatest.
Option (D), "Either N(H) or N(E) is the smallest": every other set, B, D, G, F, A, C, sits at or above B, and both H and E sit at or below B, so neither B nor anything above it can undercut both H and E at once. Q and Z, being supersets of H, can never be smaller than H either. So no matter which of H or E is actually the smaller one, the overall smallest count among every set in play belongs to one of these two.

Final Answer:
Since either H or E must hold the smallest count, without our being able to say which, option (D) is the statement that always holds. \[ \boxed{\text{Option (D)}} \]