LSAT 82, Logic Game 3, Question 16, LSATHacks

This question asks what could be true if Shaw does the 4th interview. And it’s super easy if you made these upfront scenarios:

A is CORRECT. S can be fourth in the first scenario if Ikaba goes fourth. And in that scenario, F is first.

All of the other answers contradict the scenarios. Only read the rest of this explanation. if you were stuck on an answer: the analysis above is the main thing to know for solving this explanation.

The big challenge on the other answers is placing G and Ikaba. Rule 2 says they must be separate. So, for example, B and C place G 3rd and 4th, respectively. That means that Ikaba must be 1st in B, or 1st or 2nd in C.

And this leaves things very restrictive. To place Shaw 4th, you must have them interview Ikaba, G or H. They can’t interview F, because F is interviewed by Q.

And this just doesn’t work, because of rule 1: G and H must be interviewed by the same person. In B , that’s R, since G is interviewed by R. So H can’t be interviewed by Shaw.

In C, the issue is who goes third with Rivera. It can’t be Q, because Q has to interview F. So only H can go third. And therefore H and G are both interviewed by Rivera:

What about D? Here, H is 2nd. We also have to place G and Ikaba apart (rule 2). So one of G and Ikaba goes 1st, and the other goes in one of 3/4.

We could maybe make one of them S if they were 4th. The problem is we still have to place F. And they have to be investigated by Q, which means we can’t place Q 3rd. So Q/F are 4th, and once again S can’t be 4th:

Finally, E. We place Ikaba 1st. Then, there are only two ways to arrange things:

If G and H are 3rd and 4th, then they must both be interviewed by R (rules 3 and 4).

If G and Q are 3rd and 4th, then F interviews Q 4th (rule 3)

So, either way S can’t be 4th.

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