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

Two new sets are added here, Q and Z, but both are only pinned down as supersets of H: N(Q) and N(Z) are at least N(H), with nothing said about how big they can get. To find what must be true, line up every set's position using the passage: H and E both sit inside B, and B sits inside D, which sits inside G, which sits inside F, which sits inside A, which sits inside C.

1. N(H) is the smallest of all: this fails as a certainty because E was never compared to H directly. If E turns out smaller than H, this statement breaks.
2. N(E) is the smallest of all: fails for the mirror reason, since H could turn out smaller than E instead.
3. N(C) is the greatest of all: this held for the original eight sets, but Q and Z are new sets with no upper limit placed on them beyond containing H. Either one could be built bigger than C, so C cannot be guaranteed to stay on top once Q and Z exist.
4. Either N(H) or N(E) is the smallest: since every other set, B, D, G, F, A, C, sits at or above B, and both H and E sit at or below B, neither B nor anything above it can undercut both H and E at once. And Q and Z, being supersets of H, can never be smaller than H either. So no matter which of H or E is actually smaller, the overall smallest count among every set in play belongs to one of these two.

The first two options each name one specific set as the smallest, but the passage leaves the H versus E comparison open, so neither can be asserted with certainty. The third option gets knocked out once the unrestricted Q and Z enter the picture. Only the fourth option survives, because it correctly hedges between the two candidates without needing to know which one wins.

Let's summarize:

• H and E are both below every other original set, but never compared with each other, so the smallest count is one of the two, we just cannot say which.
• Q and Z being unrestricted supersets of H means they can beat C in size, so C is no longer guaranteed to be the greatest once they exist.

So the statement that must be true is that either N(H) or N(E) is the smallest. The answer is option (D).