LSAT 23, Logic Game 4, Question 23, LSATHacks

A and B are wrong. If T goes three times in 1 or 2 then there is no space for everyone else to go in spots 1 and 2 once. We only have three spaces left and four people to place.

C is CORRECT. We could put R in 1 and T in 2 at one meeting. In the other two meetings, T could go anywhere after R as long as we didn’t put T in fifth (T can’t go first if it is after R.)

D is wrong.  All the variables are interchangeable. So if T could go after U and S then T could also go after R. It’s easier to put T after one variable than after two variables.

Therefore, D can’t be true because then C would be true also. We can’t have two correct answers.

That’s a useful trick. If you want a more real reason it’s because the earliest you could put T if you put it after S and U is third. T could never go in spots 1 or 2.

E is wrong. If you put T in 3 then one of R and U would have to go in fifth. But they could never go in first, because they can’t go before T. That doesn’t work.

But if you put T earlier you would either fill too many spots in 1 and 2 with T.

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