LSAT 90, Logic Game 1, Question 3, LSATHacks

This question gives us a new rule: T is fifth. You should draw that first. From our three scenarios, T can only be 5th in scenario 3. Here’s scenario 3:

Placing T 5th forces his forces R 4th:

Almost everything is determined in this question. The only thing remaining is whether R is local or nonlocal. Since this is a could be true question, it is asking about remaining possibilities, so the right answer will almost certainly be about R/the fourth call.

A is CORRECT! The fourth call can be nonlocal, because R can always be nonlocal.

None of the other answers are possible. The first two calls have to be local: S because of rule 2, S must be different from V, which is third and nonlocal. And Q must be local because of rule 3: Q is always local. So B and C are wrong.

D and E are wrong because they contradict the new constraint that T must be 5th. For D, if S is third, then QV have to be fourth and fifth (rule 1), blocking T from being fifth.

For E, if R is first, then only S and V are left to go third. If S is third, QV are fourth and fifth. If V is third, then SQ are first and second (so R can’t be first!).

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