Discussion
Statements: S = A; R < T; R ≥ S ; P ≥ Q; Q > A
Conclusions:
I. P > S
II. Q > S
III. T < A
IV. A ≥ R
Answer: C
Correct answer: Only I and II are true
The statements give \( S = A \), \( Q > A \), \( P \ge Q \), \( R \ge S \) and \( R < T \). Merging S with A gives two branches rising from that point: one through Q to P, the other through R to T.
II. \( Q > S \): \( Q > A = S \). True.
I. \( P > S \): extend it - \( P \ge Q > A = S \), strict in the middle. True.
III. \( T < A \): follow the other branch. \( R \ge S = A \) and \( T > R \), so \( T > R \ge A \), giving \( T > A \). The conclusion is definitely false.
IV. \( A \ge R \): we have \( R \ge A \), which permits equality but does not require it, so the reverse relation cannot be asserted. Not determined - and note this is not a complementary pair with anything else on offer, so no “either…or” answer arises.
Only I and II follow.
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