LSAT 79, Logic Game 3, Question 16, LSATHacks

This question places V and Z together. When a question gives you a new rule, you see which other rules are affected. 

Rules 3 and 4 are relevant. Rule 3 says you can’t place V and Y together. Since VZ are together, that means you can’t place Z and Y together on this question.

And that affects rule 4. We either must place both Z and Y in P, or neither of them. So on this question, we can’t put Z or Y in P. So VZ must go in group L, or group M. We can make two scenarios for that:

In both scenarios, Y goes in the other open group (out of L or M: it can’t go in P). Next, you should ask what this affects. The people left to place are TS, and R. It’s hard to quickly place them for sure. But, one thing we can say is that TS can not go in M – there’s not enough space. Rule 1 says M can only have two assistants. So, C is CORRECT.

As for the other answers: you can easily eliminate those with R. R is random, so it can go in any group that isn’t full.

And Smith can go in both labor and pricing, as those groups can hold up to three people. So it’s no trouble to do this, for example:

You can also move up one of either TS or R to L. This shows A and B are possible.

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