#9 - 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 (B)

After two questions on the easier side, the test makers dramatically increase the difficulty with this question. The question stem first specifies that two of the windows are made with exactly two colors, and then asks for the possible color combination of one of those windows.

From our knowledge of the powerful fourth rule, answer choices (A) and (D) can immediately be eliminated because neither contains O or P. This leaves three answer choices, and the savvy test taker may realize that with two windows limited to just two colors and the second rule establishing that R must appear in two windows, that answer choice (B) is likely to be correct because it is the only remaining answer choice that contains R. However, let’s turn to the templates to determine which answer choice is correct.

Template #2 is not possible because adding the two R’s would create at least two windows with at least three colors, a violation of the condition in the question stem. Thus, Template #1 must be the applicable template for this question. When the two R’s are added to the template, one must go with the O/P option, and the other must go with either the PY or PG blocks:

D10_game #2_#9_diagram 1.png

As O and R is one possible complete color combination, answer choice (B) is proven correct.

[Paola]

Hello!

I am really stuck on question 9. I have read through the previous posts, and the answers given make a lot of sense to me, but I still can't figure out numbers 9.

My set up has split options with P/O for each window and not blocks for YO and YG.

With number 9, I am not being able to understand why the answer has to be B. I narrowed it down to B and C. If two windows have to be two colors each (but not the same two colors, I assume), I inferred that those windows would be either P/O with R, P with G, or P with O. Am I missing an important inference?

[Claire Horan]

Hi Paola,

R must be used twice according to the global rules. (C) would give you the first window having only orange and purple, meaning the other two windows would each have to include rose glass. We know that one of the windows has at least green and purple, and that Y can't go with green. But this second window has to have at least three colors of glass because R needs to be used in both the second and third window.

So, the third window would have to include R and only one other color to meet the local condition of having two windows with exactly two colors. Y hasn't been used yet, so we have to use it, but Y O P. (I wonder if this inference is what you were missing. You get this inference by combining the third and fourth rules and their contrapositives.) The third window, than would have R, Y, and P, and the local condition of having two windows with exactly two colors cannot be fulfilled.