LSAT 20, Logic Game 2, Question 10, LSATHacks

If R isn’t reduced, we need L and M. If we have L then we don’t have P. So we can say that we definitely don’t have P and R and that we definitely do have L and M.

That means we need three more out of G, W, N and S.  We know that we can’t N and S together; we can only have one of them.

We therefore need both G and W.

A is CORRECT as G has to be reduced.

B could be true but it could be S that is reduced instead.

C must be false. L causes P to be out.

D could be true but it could be N that’s reduced instead.

E must be false. W is always reduced.

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