#18 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=10395)

The correct answer choice is (C)

This is the most difficult question of the game. The wording of the conditions in the question stem is quite tricky, and you must carefully consider the meaning of each statement. The phrase, “P serves on every subcommittee on which F serves,” means that when F serves on a subcommittee, then P also serves on that subcommittee. This relationship is diagrammed as:

• F P

The second condition is worded in a similar fashion, and can be diagrammed as:

• G P

Consequently, since F and G must serve on at least one subcommittee each, we know the following:

J05_Game_#4_#18_diagram 1.png

Answer choice (A): This answer is incorrect because if M serves on the same subcommittees as F and G, then H and I would be forced to serve on the same subcommittee, a violation of the third rule.

Answer choice (B): This answer is incorrect because there is not enough room for M to serve with both H and I (M can serve on one subcommittee with either H or I, but the other two subcommittees already have P and F, or P and G, and hence there is not enough space for two more members on those subcommittees.).

Answer choice (C): This is the correct answer. H could be the member who serves on two committees.

Answer choice (D): This answer is incorrect because if P serves on all three subcommittees then F would have to serve on three subcommittees, a violation of the numerical distribution; if P serves on two subcommittees, then F would have to serve on two subcommittees, and M would have to serve on all three subcommittees, creating a 3-2-2-1-1 distribution, which also violates the previously established numerical distribution.

Answer choice (E): This answer is incorrect for the same reasoning used to disprove answer choice (D).

[ange.li6778]

Hi, could you explain why C is correct? If P is already is already the committee member who's on 2 subcommittees (with F and G) then how can H also be on 2 subcommittees? Thanks in advance!

[Robert Carroll]

ange.li6778,

I think you're misinterpreting the situation as one where P is on exactly two committees. That's not what the diagram at the beginning of this thread says - it's merely showing that P is on at least two committees because F is on at least one committee, G is on at least one committee, those can't be the same committee (2nd rule), and P is on both of these two separate committees (local condition of the question). P is not constrained not to be on a third committee, and, in fact, P must be on all three. Initially, either M or P is the one on all three, right? But imagine M is on all three in this question. Then two committees are just FPM and GPM. So the final committee is M with...H and I together, which violates the third rule. So, in fact, P must be everywhere.

With that clarification, I think the question should make sense, but let me know if you have any other questions!

Robert Carroll