#14 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=11429)

The correct answer choice is (E)

Irrespective of the Local conditions, answer choice (C) can be eliminated by applying the Not Laws. The conditions produce the following diagram:

Dec 94_M12_game#3_L7_explanations_game#1_#14_diagram_1.png

Once L is scheduled for day 2, O cannot be scheduled for day 2, and, consequently O cannot be scheduled for any of the days. The other lecture affected by the local conditions is P. As in question #2, P cannot be scheduled on day 1, since if P is scheduled for day 1 then both F and H have to be scheduled for day 2. Since H is already scheduled for day 3, answer choice (D) can be eliminated.

There are now three answer choices remaining: F, N, and S. Forced to guess with just the information at hand, you might suspect that S could be scheduled for the morning of day 1 because S is a variable already confined to the morning spaces. Although answer choice (E) is correct, it is important to understand why both answer choices (A) and (B) are incorrect. Consider that for the three afternoon spaces there are initially eight lectures: F, H, L, N, O, P, S, and W. From the second rule both S and W cannot lecture in the afternoon, so the pool is now down to six: F, H, L, N, O, and P. The conditions in the question place L and H in the morning and eliminate O from scheduling, and so the candidate pool is now three: F, N, and P. Thus, F, N, and P must be the three afternoon lectures in this question. If any of the three is placed in a morning space then there would not be enough lectures for the afternoon. This illustrates a critical grouping concept: it is as important to evaluate the variables left for consideration as it is to evaluate the variables already placed.