LSAT 61, Logic Game 3, Question 17, LSATHacks

This question places Q and S, and asks how many people can run first. Rule 1 says that T follows Q, so this question really means we have QT and S in.

That leaves only R and U. If R is in, then U is out. So, rule 2 applies (“If U is out, R goes 2nd”) and we get this scenario:

R is second, so only slots 3-4 have space for QT. S is left to go first.

So, we have proven that S can go first. Who else? 

Well, if U is in instead of R, then we could place the runners in this order:

(There is another possible order. We could so SUQT or even SQTU. But those add nothing, since we already proved S can go first)

So, only S and Q can go first. B is CORRECT.

Why can’t R, T and U go first?

• R can’t go first because rule 3 forces it second if U is out. (This question allows only one of R/U to be in, since QT and S are already in)
• T can’t go first because rule 1 places Q directly in front of T. (And this question places Q in).
• If U were first, S would have to go third, and that doesn’t leave two spaces open for QT. (S can’t go 2nd or 4th)

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