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#27048
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13972)

The correct answer choice is (B)

This is the fourth List question in this game. This question is identical to question #10, but this time the focus is on R, not J.

As shown in the discussion of the second rule, when L is addressed to R, the consequence is that F or S or both must be addressed to G. If both L and S are addressed to R, then F must be addressed to G, but there is no other piece of mail available to be addressed to G, creating a violation of the third rule (the other two variables are P and M; P must be addressed to J when L is addressed to R, and M can never be addressed to G according to the first rule). Therefore, answer choice (B) cannot occur and is correct.
 frank.dawson
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#38419
Hi,
would you please explain if L and S are pieces of R and M and P are pieces of J then why F can not be piece of G and if that is correct why B is the correct answer choice ? Answer choice E also besides B appears to have only S as a piece of G. I am still not clear why answer choice B is the correct one. Would you please explain in detail ?
Thank you
 Adam Tyson
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#38507
Hey there Frank! Take a look at the last rule of the game - F cannot be the only piece of mail addressed to anyone. If L and S go to R and M and P go to J, then F goes to G and is all alone. Whoops, can't let that happen!

The problem with B is that if L goes to R, P must go to J per the 2nd rule. What's left to go to G? M, S, and F. But M cannot go to G per the first rule, so that leaves only F and S. Since F cannot be there ALONE, S must go to G, and that means it cannot go to R. The second rule ends up telling us that when L goes to R, P goes to J, and when P goes to J, S MUST go to G.

The issue is not that F cannot go to G, but that F cannot go to G ALONE. It needs a partner, and in this case S has to be that partner.

I hope that helps!
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 tyracompagnon
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#106369
What strategy would you suggest for going through questions like this faster? I wasn't able to make many inferences with the given rules so the strategy I used was to plug in the options into my diagram.
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 Jeff Wren
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#106400
Hi tyra,

The first point I always make when discussing this game is that it is harder than it looks.

There are some games that you can tell right away are going to be difficult (and those are probably best to skip and save for last), but some games (like this one) can be deceptive.

When you read through this game, it doesn't sound bad at all.

There are five pieces of mail spread across three people. There are three rules that seem pretty straightforward/easy to understand.

However, these seemingly simple rules lead to a few tricky inferences that many students miss. Without these inferences, the questions become much more difficult and require a lot of time to "brute force" them by testing each answer.

One key inference is that Jana must get either (meaning at least one of) the letter or the postcard. (For an explanation of this inference, please see the complete game setup linked above.)

Another key inference that is indirectly shown in the game setup, but may not be entirely clear, is that Georgette must receive either (meaning at least one of) the postcard or the survey. The main diagram in the setup shows "F/P/S" as the possible pieces of mail that Georgette can have. While this is technically true, it may lead some students to think any of those three would work. However, since the flyer (F) must go with at least one other piece of mail, Georgette must have at least one of P or S (whether or not Georgette also has the flyer).

Now question 11 is asking for a list of two pieces of mail that CANNOT go to Rini. According to our main diagram/setup, there are really no restrictions on what pieces of mail can go to Rini. This should suggest that the correct answer isn't really so much about Rini as the other two people. For example, if Georgette needs to get either P or S, then Rini can't get both P and S. Similarly, if Jana needs to get either L or P, then Rini can't get both L and P.

Unfortunately, neither of these two combinations (P and S) or (L and P) appears as an answer. However, Answer B (L and S) turns out to be just as good. With both of these inferences, we can infer that Rini cannot get both L and S because that would force P to go to both Georgette and Jana, and P can't go twice.

Without these inferences, this game is much harder and requires more time/effort.

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