#15 - Local, Must Be True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=10807)

The correct answer choice is (B)

This question stipulates that only the 2-2-2 distribution applies. Since the selection of variables for aisle 3 is restricted to exactly two of M, P, and R, there would only be three combinations of variables available for that aisle—MP, MR, and PR. A PR block, however, would violate the first rule of the game. Likewise, an MR block in aisle 3 cannot be satisfied in a 2-2-2 distribution, because it forces P to be alone in aisle 2:

oct12_game_3_#15_diagram_1.png

Thus, we can infer that if aisle 3 contains exactly two sections, those sections must be M and P:

oct12_game_3_#15_diagram_2.png

Next, recall that R must be located in the same aisle as either F or M (first rule). With M unavailable, R and F must form a block in aisle 1 or 2. This leaves H and S to form the last vertical block, in either aisle 1 or 2, proving answer choice (B) correct:

oct12_game_3_#15_diagram_3.png

[TheKingLives]

I apologize if this question has an obvious answer, but if M and R are in aisle 3, couldn't aisle 1 have H and F and aisle 2 have S and P? I got stuck here during the game. Thank you!

H S M
F P R
1 2 3

[Adam Tyson]

You're violating the third rule there, your Highness! S must be in a lower numbered aisle than P, so they cannot be in the same aisle as each other. Make sure to always run your answer choices through all the rules, like a filter! That's how you can avoid mistakes like that one.

[TheKingLives]

Ah, of course! Sorry, I connected my S-P rule to my F-M, F-P rule and it must have slipped past me. Will do!