#13 - Local, Could Be True, List

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=2909)

The correct answer choice is (E)

If none of the windows contains O and R, then from the second rule two of the windows contain R, and O must be used in the remaining window:

D10_game #2_#13_diagram 1.png

Of course, from the fourth rule each window must contain O or P, and so the two windows with R (and thus not O) must both contain P:

D10_game #2_#13_diagram 2.png

One RP window must contain G in order to comply with the first rule, and the other RP window must contain Y because Y must be used and every time Y is used, P is used:

D10_game #2_#13_diagram 3.png

With one of the windows established as R, P, and Y (and containing no other colors since due to the third rule G and O cannot be used in that window), answer choice (E) must be correct.

As an aside, at least G or P must used in the “third” window.

[mahsan]

Hello,

The last step in the explanation above states that one RP window must contain G in order to comply with the first rule. While that could be true, must it be true?

Couldn't we instead have:

1: O G P
2: R Y P
3: R Y P

This would still satisfy the first rule and not violate any others unless I'm missing something here.

Appreciate any help with this. Thanks!

[Paul Marsh]

Hey mahsan! Nice spot, you're right. We could have that GP block go with the O instead. But either way, we still end up with a RYP window. So our answer stays the same.

Nice one!