#9 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8627)

The correct answer choice is (A)

When L is reduced, P cannot be reduced. Consequently, any answer choice that contains P can be eliminated, and answer choices (B) and (E) can be discarded. When S is reduced, N cannot be reduced and it follows that answer choices (C) and (D) are incorrect. Thus, answer choice (A) is correct.

[sroos1987]

So... Either I'm really struggling, or this question is broken.

So I understand the explanation above insomuch that it quickly eliminates B/E and then C/D, leaving A as the only option, and how on test day you would stop there, answer A and move on. But on practice days, I keep trying to do the problem out and it seems to me that option A should also be eliminated?

GIVEN:
S and L are reduced
INFERRED:
Since S is reduced, N is not reduced. (rule 2)
Since N is not reduced, R is reduced. (rule 2)
Since both L and R are reduced, M is not reduced. (rule 4)

RESULT:
Reduced : L (given) S (given) R (inferred) __ __
Not reduced: M (inferred) N (inferred) __
INFERRED:
Since L is reduced, P is not reduced (rule 3)
RESULT:
R: L S R __ __
N: M N P

Therefore, when L and S are reduced, the outlay HAS to be:
R: L S R G W
N: M N P

Which would result in all five answers, including the allegedly correct one (M/G could both be reduced), being impossible.

What am I missing here? I can't imagine there's an error this significant?

[Dave Killoran]

So... Either I'm really struggling, or this question is broken.

So I understand the explanation above insomuch that it quickly eliminates B/E and then C/D, leaving A as the only option, and how on test day you would stop there, answer A and move on. But on practice days, I keep trying to do the problem out and it seems to me that option A should also be eliminated?

GIVEN:
S and L are reduced
INFERRED:
Since S is reduced, N is not reduced. (rule 2)
Since N is not reduced, R is reduced. (rule 2)
Since both L and R are reduced, M is not reduced. (rule 4)

Hi S,

Your issue is right here: "Since N is not reduced, R is reduced." The rule only says this: "If N is reduced, neither R nor S is reduced." Isolating that to N and S, both can't be selected for reduction ( N R ). But where does it say at least one of the two has to be reduced? It doesn't, hence the problem

In this question, only two solutions exist: L - M - G - S - W or L - R - G - S - W, with the difference between the two bolded.

I hope that helps. Thanks!