#5 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=11660)

The correct answer choice is (D)

The local condition specifies that R is inspected the day before F, which creates this block sequence:

J96_Game_#1_#5_diagram 1.png

Of course, you should draw out that block sequence right next to the problem. At this point, there are several different ways to proceed with this question. One approach is to check any previous hypothetical to see if it conforms to the requirements in this question. Fortuitously, the only hypothetical we have—from question #1 answer choice (B)—actually conforms to the local condition imposed in this question. Thus, we can use that hypothetical, G-H-J-Q-R-F, to help answer this question. Accordingly, we can eliminate answer choices (C) and (E) since the hypothetical proves that neither must be true. Note how the hypothetical partially agrees with answer choices (A), (B), and (D), but it does not prove that they must be true.

The other method of attack is to use the linkage involving F. Since F must be in day 1 or day 6, and in this question F is scheduled behind R, we can infer that F must be inspected on day 6. Accordingly, we can make the following hypothetical right next to the problem:

J96_Game_#1_#5_diagram 2.png

Although you may not realize it, this hypothetical solves the problem. If Q, R, and F are already placed, and G cannot be inspected third (that would violate the G3 then Q5 rule), that leaves only H or J to fill day 3. That inference is reflected in the wording of answer choice (D) and thus (D) is correct.