#4 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15932)

The correct answer choice is (E)

The condition in the question stem creates an SR block. Adding this to the third rule, we arrive at the following super-block:

pt44_o04_g1_q4a.png

The size of this block presents immediate problems for the three Fs, and you must consider the placement options of the block prior to attacking the answer choices. First, the block cannot be placed at the beginning or the end of the series of meetings because that would not leave sufficient room to separate the three meetings with F (if meetings 1-3 are occupied by the block, then only meetings 4-7 are available for the three Fs; if meetings 5-7 are occupied by the block, then only meetings 1-4 are available for the three Fs). Consequently, the TSR block must be placed in either meetings 2-4, 3-5, or 4-6. We can also deduce that F must be Garibaldi’s first meeting because only F, R, or T could be Garibaldi’s first meeting, and with T and R involved in the block neither can be first. With this fact in hand, you should quickly create hypotheticals reflecting each placement option:

• Option #1: TSR as the second, third, and fourth meetings
  
  When the TSR block is placed as the second, third, and fourth meetings, the remaining two Fs must be placed as the fifth and seventh meetings, leaving M as the sixth meeting.
  
  
  pt44_o04_g1_q4b.png
  Option #2: TSR as the third, fourth, and fifth meetings
  
  Hopefully you realized that this placement option would be unworkable prior to drawing out the diagram. When the TSR block is placed as the third, fourth, and fifth meetings, there is no way to place the other two Fs so that they do not violate the second rule. Consequently, TSR cannot be placed in this position and a viable hypothetical cannot be created.
  
  
  pt44_o04_g1_q4c.png
  Option #3: TSR as the fourth, fifth, and sixth meetings
  
  When the TSR block is placed as the fourth, fifth, and sixth meetings, the remaining two Fs must be placed as the third and seventh meetings, leaving M as the second meeting.
  
  
  pt44_o04_g1_q4d.png

Thus, the condition in the question stem only allows for two solutions to the question:

pt44_o04_g1_q4e.png

Answer choices (A), (B), (C), and (D): Each of these answer choices could be true. However, this is a Must be true question, and thus each of these answers is incorrect.

Answer choice (E): This is the correct answer choice. As proven by the two solutions above, F is always the last meeting.