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.

Which of the following could be TRUE, but not necessarily TRUE?

• A. N(A) is the greatest of all
• B. N(G) is greater than N(D)
• C. N(H) is the least of all
• D. N(F) is less than or equal to N(H)

Hint:

Check whether each statement is forced by the chain, contradicted by it, or genuinely open; H versus E is never compared, so H being the smallest is a real toss-up.

Answer 1

The Correct Option is C

Step 1: Recall the chain of set sizes.
From the passage: \(N(H) \le N(B)\), \(N(E) \le N(B)\), and \(N(B) \le N(D) \le N(G) \le N(F) \le N(A) \le N(C)\).

Step 2: Test option (A), "N(A) is the greatest of all".
The chain shows \(N(A) \le N(C)\) always, directly from the passage's statement that A is a subset of C. So A can never be strictly bigger than C. This statement is impossible, not just uncertain, so it is ruled out.

Step 3: Test option (B), "N(G) is greater than N(D)".
The passage directly states that G is a superset of D, so \(N(D) \le N(G)\) is exactly the fact already handed to us, not something left open. Since this just restates a given fact, it does not fit what the question is asking for.

Step 4: Test option (C), "N(H) is the least of all".
Both H and E sit at or below every other set in the chain, but the passage never compares H and E with each other. If \(N(H) \le N(E)\), then H really is the smallest of all, so the statement can be true. But if instead \(N(E) < N(H)\), then E is the smallest and H is not, so the statement can also be false. Since both outcomes are genuinely possible given the passage, this statement could be true without being necessarily true.

Step 5: Test option (D), "N(F) is less than or equal to N(H)".
The chain gives \(N(H) \le N(B) \le N(D) \le N(G) \le N(F)\), so \(N(H) \le N(F)\) always. For the reverse, \(N(F) \le N(H)\), to also hold, every single link in that long chain would have to collapse into an exact equality all at once, an extremely restrictive coincidence the passage gives no reason to expect, unlike the balanced, always-open uncertainty seen in option (C).

Final Answer:
Option (C) is the one statement that is genuinely undecided by the passage since H and E are never directly compared, so it could be true without being necessarily true. \[ \boxed{\text{Option (C)}} \]