October 1996 LSAT
Section 3
Question 10

October 1996 LSAT
Section 3
Question 10

Reply

Ravi on August 1, 2019

@Lizzie-Annerino,Let's take a look at (B) and (A).

If R is not reduced, then we know that L and M must both be reduced.

If L is reduced, then we know that P is not reduced. This means that P

and R occupy 2 of the 3 "not reduced" slots.

Recall the rule that states if N is reduced, neither R nor S is

reduced. We know that R isn't reduced, so the necessary condition for

that part of the rule is satisfied. However, look at the rest of the

rule. If N is reduced, then S can't be reduced. Conversely, if S is

reduced, then N can't be reduced. This means that either N or S can't

be reduced.

From this, we know that both W and G must be reduced.

(A) says G is reduced, and we know that must be true, so it's the

correct answer choice.

(B) says that N is not reduced. It's possible that N is reduced,

however. We just know that either N or S is not reduced, but it's

possible that N is the one that's reduced and S isn't reduced, so we

can get rid of (B).

Does this make sense? Let us know if you have any other questions!

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