#17 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=1452)

The correct answer choice is (B)

If the display includes two green toys, then the five colors are fully determined, and from the fourth rule P cannot be displayed:

Jun 09__M12_game#3_L11_explanations_game#3_#17_diagram_1.png

Thus, P is not displayed, and from the fifth rule either U or V is not displayed. As two dinosaurs are thus effectively removed, the remaining five must be displayed: I, L, S, T, and V/U. This lineup allows for several combinations of mauve dinosaurs, and thus there are still multiple solutions available. However, there is now sufficient information present to answer the question.

Answer choice (A): This answer cannot occur as the five colors on display are set, and no yellow toy is included.

Answer choice (B): This is the correct answer choice. If T is green, then the two mauve dinosaurs are L and V.

Answer choice (C): The analysis of the toys concluded that L must be displayed, and thus this answer choice can never occur and is incorrect.

Answer choice (D): The analysis of the toys concluded that T must be displayed, and thus this answer choice can never occur and is incorrect.

Answer choice (E): The analysis of the toys concluded that exactly one of U and V must be displayed, and thus this answer choice can never occur and is incorrect.

[Jon Denning]

First, it will be helpful to refer to this previous post regarding the setup: lsat/viewtopic.php?f=156&t=1452.

If there are exactly two green toys, then the five colors are fully determined (MMRGG), and from the fourth rule P cannot be displayed since no dinos are yellow. Further, either U or V is not displayed (fifth rule). So the five dinos must be I, L, S, T and V/U. While there are still several combinations possible, you can quickly eliminate the wrong answers, and see why B is possible: T can be green, with the two mauve dinosaurs as L and V.

So tricky game, but hopefully this helps you better understand it.

Thanks!

Jon

[Basia W]

Hello!

17e) Because of rule 5 I know that neither v is included then u is not and vice versa (resulting in a double not arrow). How is it possible then, that they are both out? I inferred that one of them had to be in the "in" pile and one had to be in the "out" pile at all times.


Thank you for your time!

Best,

Basia

[Emily Haney-Caron]

Hi Basia,

Thanks for your question. Rule 5 should not be diagrammed with a double not arrow because it does not say "if and only if." Instead, you would diagram it this way:
V notU
The contrapositive is:
U notV
Therefore, they can both be OUT, they just can't both be IN. Does that help?

[Basia W]

Hello,

Yes it does- So when one is selected the other is not and it is possible for them both not to be selected or "in" from what I understand.

Best,

Basia