#3 - Global, Cannot Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=23743)

The correct answer choice is (C)

This is the most difficult question of the game, and one that is not easy to answer from a quick glance at the rules.

When you encounter a Global question with no obvious answer, remember that one approach is to refer to the hypotheticals created in other questions. For example, question #1 produced a solution that placed U at 2:00. On the strength of that answer, we can eliminate answer choice (B). The discussion in question #2 indicated that U could meet at 1:00, eliminating answer choice (A). However, none of the three remaining answer choices is obviously incorrect so you have a choice: either skip the remaining answer choices and hope that future questions provide more hypotheticals so you can come back and eliminate some answers, or make a few hypotheticals right now to solve the problem.

If you choose to wait until later to answer this question, you will find that the hypothetical from question #5 eliminates answer choice (E). At that point you could simply create a hypothetical to eliminate or confirm answer choice (C) or (D).

If you choose to make hypotheticals to work your way through the final three answer choices, you would be best served by first attacking answer choice (C) or (D), and not by starting with answer choice (E). This is because (E) would seemingly be easily eliminated by the somewhat obvious hypothetical where U and R meet at 5:00 and 6:00, and the four variables in the other sequence fill in the first four hours (as in S-W-Y-T-R-U, for example). In a moment we will create hypotheticals for both answer choices (C) and (D), but before doing so, let’s discuss the logic of why U is limited at all in this game.

At first glance, U appears to be a fairly unrestricted variable, with U’s only limitation coming from the rule involving R. Obviously, though, 6:00 is not one of the answers to this question, so there must be some further limitation on U that has thus far gone unnoticed. In examining the two chains, the one point of concern is the WY block. The block not only requires two consecutive spaces, but it also affects S and T. Although that may not appear to be of an much issue for U, if U is placed at 3:00, there is not enough room for all the variables:

• Step 1. U is placed at 3:00; R must meet at 4:00, 5:00, or 6:00:
  
  pt45_d04_g1_q3a.png
  Step 2. Consider the other sequence:
  
  Because there are three meetings after Patterson’s meeting with S, normally the latest that S could meet is 3:00. But, since U already occupies the 3:00 meeting, S must be somewhere before U. Because this placement leaves only one open time slot before U, it must be that the WY block comes somewhere after U, and, of course, T is somewhere after the WY block. This chain of inferences results in a scenario where R, W, Y, and T must all come somewhere after U, but there are only three spaces for the four variables:
  
  
  pt45_d04_g1_q3b.png
  
  Thus, we cannot create a workable solution when U is placed third, and answer choice (C) is correct.
  
  By the way, a similar type of logic holds for R: if R is placed in the fourth position, a workable solution to the game cannot be created.

Answer choice (A): This answer choice is incorrect. Patterson can meet with U at 1:00, as shown by the following hypothetical: U-R-S-W-Y-T.

Answer choice (B): This answer choice is incorrect. Patterson can meet with U at 2:00, as shown by the following hypothetical: S-U-W-Y-T-R.

Answer choice (C): This is the correct answer choice. Although it is proven by the previous discussion, this answer choice is difficult to arrive at without testing a few solutions.

Answer choice (D): This answer choice is incorrect. Patterson can meet with U at 4:00, as shown by the following hypothetical: S-W-Y-U-T-R.

Answer choice (E): This answer choice is incorrect. Patterson can meet with U at 5:00, as shown by the following hypothetical: S-W-Y-T-U-R.