#10 - Local, Must 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)

This is one of the key questions of the game. The initial approach taken by most students is to consider the implications of R not being reduced. When R is not reduced, L and M must be reduced, and when L is reduced, P is not reduced. This provides sufficient information to eliminate answer choice (C). At this point the diagram next to the question looks like this:

Oct 96_M12_game#2_L5_explanations_game#4_#10_diagram_1.png

But this leaves four answer choices in contention with no obvious path towards the correct solution. However, there are several approaches to finding the correct answer:

1. Based on our discussion of the reduction of three of the five expenditures G, N, S, P, and W, when P is not reduced then G, W, and N or S must be reduced:

Oct 96_M12_game#2_L5_explanations_game#4_#10_diagram_2.png

Consequently, only answer choice (A) must be true.

2. Another approach is to make a few hypotheticals based on L and M being selected. The various hypotheticals can then be used to eliminate answer choices.

3. Since making new hypotheticals is useful, checking the hypotheticals created in questions #6 and #7 to see if they apply to question #10 might be even better. Although the W-M-P-R-S hypothetical in question #7 answer choice (E) is inapplicable since R is reduced, the hypothetical in question #6 answer choice (A) meets the criteria in question #10. By applying the G-L-M-N-W solution, we can eliminate answer choices (B), (C), (D), and (E), leaving only answer choice (A).

Remember, always check back to earlier problems to see if you already have enough information to solve the current problem. Of course, only applicable work can be used. Do not forget that you should only use work that you are fully confident is correct. That is why answering List questions correctly is so important!

[uhinberg]

I think that there is a simpler inference that can be made to make solving 10 easier. N has two either or rules, one with R and one with S, and P/L has another either/or rule. As long as R and S are uncertain, you cannot diagram both the N rules, but once you know that R is not reduced, you can infer that another of the not reduced spaces must be N or S, and the third one must be P or L. Everything else must be in the reduced group. And voila, you can see that A is the only answer that must be true.

[Francis O'Rourke]

Hi Uhinberg,
Everything you wrote here is correct. The only thing that I want to add is that while either P or L must be not reduced, it's often a better choice to immediately diagram the more certain inferences, if only to minimize the amount of uncertainty that you deal with:
if R is not reduced M & L are reduced P is not reduced.

[KhaliaCWilliams]

Hi!

Im kind of confused on how we can come to the conclusion that G must be reduced according to the game.

I understand that if R isn't reduced, LM have to be. Since L is reduced P can't be reduced leaving GNSW however I do not understand how to go on from there.

How can we infer that G is absolutely reduced?

Thank You!

[Dave Killoran]

Hi Khalia,

It's a numbers game to make this inference! It works like this:

• From the last rule, when R is not reduced, them L and M must be reduced
  
  From the third rule, when L is reduced, then P cannot be reduced.
  
  From our setup discussion, exactly three of G, N/S, P, and W must be reduced. So, when P is not reduced you must select three from G, N/S, and W. Thus, means G and W are reduced, and then you choose between N or S.

That chain forces answer choice (A) to be correct, hence our answer

If that isn't clear, trying not reducing G and see what happens. It won't work!

Thanks!