Discussion
STATEMENT :
R > B ≥ S ≥ D ; S ≥ D ; S ≥ O ≥ U
CONCLUSION I : B > U
CONCLUSION II : D ≤ O
CONCLUSION III : U = B
CONCLUSION IV : R > O
Answer: C
Correct answer: Only IV and either I or III are true
The statements give \( R > B \ge S \ge D \) and \( S \ge O \ge U \), with \( S \ge D \) repeated.
IV. \( R > O \): the run \( R > B \ge S \ge O \) points one way and opens with a strict link, so \( R > O \). True.
II. \( D \le O \): D and O both hang below S on separate branches, and two elements below a common third have no order between them. Not determined.
Conclusions I and III are the interesting pair. The run \( B \ge S \ge O \ge U \) is weak at every link, so it gives \( B \ge U \) and no more. I claims \( B > U \) and III claims \( U = B \); neither can be established on its own, since all four values could be equal or B could stand well above U. But \( B \ge U \) means B is either greater than U or equal to it, with no third possibility, so one of the two must hold even though we cannot say which. That is precisely the “either…or” case.
So IV is definitely true, and either I or III is true: the keyed answer. The pattern to recognise is a complementary pair - \(>\) against \(=\), or \(<\) against \(\ge\) - arising from a chain of weak links.
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