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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 hardest question of the game, with fewer than 50% of students answering this question correctly. The key to the question is realizing that the question stem contains the word “only.” The use of this word in the question stem indicates that the test makers are looking for a list where each piece of mail could be the only piece of mail addressed to J. Thus, the answers do not list all of the pieces of mail that could be addressed to J, but rather the pieces of mail that could be addressed to her in scenarios where she receives only one piece of mail. This is one reason F does not appear in any choice—the appearance of F would automatically eliminate the answer in easy fashion (probably too easy of a fashion for Law Services’ taste).

A question like this is most easily solved when you have created as many game scenarios as possible, so the best strategy with this game is to skip it, and then return at the end after you have solved the other questions (and hopefully created applicable hypotheticals that can then be used to solve this question more quickly; question #12, for example, helps to show that L could be the sole piece of mail addressed to J).

Let’s consider two other approaches to this question. The first involves looking at the relationships and abstractly solving the problem. With this approach, we would focus on the second rule, which ends with P being addressed to Jana. The actions of this rule prohibit the two randoms—M and S—from being the sole piece of mail addressed to Jana. If either M or S were the sole piece of mail addressed to Jana, then L would need to be assigned to Rini. However, this would enact the second rule, forcing P to be assigned to Jana. Clearly, then, neither M nor S can be the sole piece of mail addressed to Jana, which knocks out answer choices (C), (D), and (E).

The difference between answer choices (A) and (B) is the presence of L in answer choice (B). A mental calculation shows that L could be the sole piece of mail addressed to Jana, because F, P, and S could conveniently be addressed to Georgette, and M could be addressed to Rini.

The process described above is admittedly difficult, but sometimes considering the different solution avenues is helpful because many questions are solved not with just one path of attack but with a combination of approaches.

Another method of solving this question is to use hypotheticals to confirm and eliminate variables. For example, our inference involving the second rule sets the stage for a hypothetical where P is the only piece of mail addressed to Jana:
  • When L is addressed to R, then P is addressed to J, allowing F and S to be addressed to G. M, a random, can then be addressed to R or G:
June 06_M12_game#2_L5_explanations_game#2_#9_diagram_1.png
June 06_M12_game#2_L5_explanations_game#2_#9_diagram_1.png (2.69 KiB) Viewed 2133 times
Establishing that P can be the sole piece of mail addressed to Jana eliminates answer choices (C) and (D), neither of which contain P.

As the three remaining answers only address L or M (or neither), we do not have to worry about S. This is odd because S is a random, and one would normally expect that a random would be a likely variable that could be the sole piece of mail addressed to Jana. This suggests that instead of examining M—the other random—first, we should instead examine L first. Indeed, L can be the only piece of mail addressed to J, as shown by the following hypothetical:
June 06_M12_game#2_L5_explanations_game#2_#9_diagram_2.png
June 06_M12_game#2_L5_explanations_game#2_#9_diagram_2.png (2.37 KiB) Viewed 2133 times
This information eliminates answer choice (A), leaving only M under consideration.

As mentioned before, neither of the two randoms—M or S—can be the sole piece of mail addressed to Jana. If M or S were the sole piece of mail addressed to Jana, then L would have to be assigned to Rini, enacting the second rule and forcing P to be assigned to Jana.

Perhaps the easiest way to handle this question is by having a key inference handy: Jana must receive either L or P at all times. This is simply a function of the contrapositive of the second rule: if P is not addressed to Jana, then L cannot be addressed to Rini. But, since L can only be addressed to Jana or Rini (first rule), it logically follows that if P is not addressed to Jana, then the L would be addressed to Jana (PJ :arrow: LJ). And inversely, if L is not addressed to Jana, then P must be (LJ :arrow: PJ). Either way, Jana must receive at least one of L or P, automatically eliminating any answer choice that contains F, M or S.

Regardless of which method or combination of methods you use to solve the problem, the correct answer choice is (B).
  • Posts: 33
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I wanted to get someone's thoughts on an approach I took with question #9. It seems this type of questions set up usually trips me up. When you see a question like this one, I should (1) look through previous work to see if there is a particular piece of mail that contradicts what the questions is asking for (cannot be "the only piece of mail deliver to J"). Hopefully that would eliminate some options. (2.) Work out the remaining ones to come to the correct answer. Does that sound correct?

Another note, which I think further complicated my approach above is that the question asked for "any one of which could",meaning that even though one piece of mail being deliver only to J did not work for one set up, if it worked in any others it was a valid for the question. It just seems like a lot of work. Am I missing something that could have saved me time on the question?

Thank you,
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 Dave Killoran
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Hi Ann,

Just fyi, we moved this question over to the Question #9 discussion, and then we'll post an answer. That way we will be able to keep all of the discussion related to that question together in a consolidated form :-D

 Jonathan Evans
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Great question! This is a somewhat frustrating situation because of the difficulty with assigning distributions (how many items go where) to specific groups without violating conditional clues.

For instance, given that we know that we have either a 3-1-1 distribution or a 2-2-1 distribution, we know that in either case F will go into a group with at least one other item. However, now we have to pay attention both to the L, M not law restriction for the G group and the LR :arrow: PJ conditional.

Your approach is valid, and if nothing else you could observe previous work to establish what could be alone in the J group. Then you could brute force your way through the remaining items, testing out each remaining possibility alone to see whether it works. I understand how frustrating and time-consuming this process can be.

For a faster approach, consider applying the distribution templates to this particular scenario. Then add the conditional clue and its contrapositive to seal the deal:
  1. LR :arrow: PJ
  2. PG :arrow: LJ
  3. PR :arrow: LJ
Note that it seems as though P or L necessarily end up in J a lot of the time. This is the trick to this question. You should note that it seems probably if not necessary that given only one in the J group it will very likely be P or L. This is because from (1) above we see that given that L is not in J, P must be in J. This is because if L is not in J, it can only be in R (L cannot be in G). Therefore, P is in J and we're done with that possibility.

Now, consider the other possibility that P is not in J. Then P must be in either G or R. In either case from the contrapositives (2) and (3) above, L is in J.

So, yes, there is a faster way! Note that since this is a Global question, there will always be a method to combine the clues to generate an inference that solves the problem. However, there are multiple valid ways to solve these problems. If you miss a connection, don't be afraid to use some process of elimination skills and try possibilities out. The process can go a lot faster than you think!

I hope this helps!

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