#23 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=298&t=12774)

The correct answer choice is (D)

If you produced a hypothetical for question #22, then you can use that hypothetical to attempt to choose an answer in this question because the hypothetical from #22 meets the condition in this question stem. However, the hypothetical above does not ultimately prove any answer choice is correct. Still, checking that hypothetical was worth the time because if it had worked, then the time savings would have been tremendous.

If M is assembled on line 1, the following setup results:

G4-Q23-d1.png

F, J, and K remain to be assigned, with the F > J rule still in effect. Thus, the following Not Laws are also in play:

G4-Q23-d2.png

Accordingly, answer choice (D) is correct.


Most students found this to be the easiest game of the December 1999 LSAT.

[studyhelp20]

Dear Power Score Support Staff,

Could you please provide an explanation to the correct answer? Thanks in advance for the help

Sincerely,
Brennan

[Robert Carroll]

Brennan,

The question sets up that M is 1, which makes G 2, and H must be 7. S is always 4, so spaces 3, 5, and 6 are available for the other variables. We can instantly get rid of answer choices (A) and (E): (A) because F cannot be 6, leaving no room for J to be after F, and (E) because M will always be one before G.

Of the answers left, answer choice (C) and (D) both refer to G, which we've already placed 2. So we think about what could follow G. J cannot, as answer choice (C) says, because there would be no room for F to precede J. Answer choice (D) looks promising - if K is right after G, we'd be able to put F 5 and J 6. In fact, this works.

For completeness's sake, answer choice (B) doesn't work because the only place F could be right before K is F 5 and K 6, but then there is no room after F for J.

Robert Carroll

[hbaezh]

Is this question best approached by testing out all of the choices once rewriting a new diagram with the extra info given in the q stem or are there obvious big scenarios to which one can eliminate choices quickly without writing each one out?


Thanks in advance

[Adam Tyson]

If by testing, hbaezh, you mean writing each answer out, then no, that would be too slow and inefficient. Instead, once you have built the local diagram that Robert described, you should just compare each answer to that diagram and eliminate anything that cannot occur within that framework. If you find yourself with multiple contender answers, THEN choose one of them to test out within the local diagram.

Testing answers is typically a last resort. Instead, prepare for the answers by following the local restrictions and try to prephrase (predict) what the answer could be.