#7 - Local, List

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15241)

The correct answer choice is (C)

If O is the only ranger assigned to area 2, Template 1 provides a suitable basis to create our local diagram. Since area 2 is closed, and P cannot be assigned to area 1, it follows that P must be assigned to area 3:

PT79_Game_#2_#7_diagram 1.png

The assignment of L is still undetermined: L can be assigned to either area 1 or 3 in compliance with the third rule. Regardless, the list provided in answer choice (C) could be the complete assignment of rangers to area 3, so (C) is correct.

Answer choices (A), (B), and (D) are incomplete, because P is missing from them. Answer choice (E) is inaccurate, because J must be assigned to area 1.

[MillsV]

Hi there - why is area 2 closed? In question 6 it showed it was possible for both P and O to be in area 2.

[Jeremy Press]

Hi Mills,

It's closed because of the language of the local condition in the question stem, which is that "Olsen is the sole ranger assigned to area 2," meaning Olsen is in area 2 and no one else is.

Keep an eye on those little words (like "sole," or "only") that they sneak into question stems on local questions--I hope this helps!

[MillsV]

Ah of course, thanks Jeremy

[lsatdaynnight]

Hi,

Shouldn't L be assigned to Area 1 because N has to be assigned to Area 3? N can only be assigned to Area 2 or 3. Since it cannot be assigned to Area 2, it must be assigned to Area 3 thereby forcing L to go in Area 1. Thank you!

[Adam Tyson]

I think you may have added an extra variable to your setup, lsatdaynnight - there is no N in this game! It's O and P that cannot be in Area 1, so if O is alone in Area 2, P gets forced into Area 3. Then the numbers come into play - Area 3 already has two things and cannot have more than 3, but O in 2 means J and K must go together. They can't be with O (because it's supposed to be alone) and they can't go in Area 3 because that would be too many things there (and also because K can never be with M, an inference based on the rule about L being with K or M but not both). So J and K have to go to Area 1.

The only rule left to apply is that L has to be with either K or else M, and that means L could be in Area 1 or else in Area 3. The correct answer has it in Area 1 (M and P being the complete and accurate list of things in Area 3), but another perfectly acceptable answer would have been L, M, and P,