#11 - 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 (A)

The question stem specifies that O is used in more windows than G. As discussed during the analysis of the third rule, G, and O are each used in one or two windows. Thus, in order to conform to the condition in this question, G must be used in exactly one window (with P in accordance with the first rule), and O must be used in exactly two windows.

At this point, many people are naturally inclined to assume that the G and O windows are separate, as in the following scenario:

D10_game #2_#11_diagram 1.png

However, this is impossible because from the third rule Y could not be used in a window. Thus, there must be an overlap and one of the windows must contain both G and O:

D10_game #2_#11_diagram 2.png

This inference proves sufficient to answer the question by using process of elimination. From the condition in the question stem, G is used in only one window, and that window must also contain P and O. Answer choices (B), (C), (D), and (E) each contain G, but none of those answers also contains P and O. Consequently, each of those answer choices is incorrect. Only answer choice (A) remains, and is thus correct.

Note that the second diagram above is incomplete. For example, the “third” window must contain Y and P (Y must be used in one of the windows and P must also be used when Y is used), and the second rule involving 2 R’s must also be accounted for. However, there is no need to continue to create the diagram when the initial inference allows you to eliminate each of the four incorrect answers.

[tld5061]

Hi,

I have a question on #11. The way I set it up was window 1: RYP. window 2: O P + maybe R. window 3: OGP + maybe R. It makes sense then the answer can be OP (without the R). Just want to make sure I'm thinking about this correctly. Thanks!

[Luke Haqq]

Hi tld5061!

Yes, it sounds like you're thinking about it correctly. The question stem asks for which of the answer choices "could" be true.

In some games, you'll have a lot of restrictive rules, or a lot of inferences you can make at the outset, that could help you knock out incorrect answer choices. That's less the case here. For this one, it appears you would have to try each answer choice--so it's serendipitous that the answer was (A). Since your reasoning was "It makes sense then the answer can be OP (without the R)" (emphasis added), that is enough for you to select it as correct and move on to the next question.

If it hadn't been answer (A), the method you could apply involves tests the other answer choices. If you're looking at a could-be-true question, four answer choices must be false. For example, answer (B) supposes the complete color combination of one window is G, P, R. We know that must be false because we have to use all 5 colors, because O is occupying the other 2 windows, and because this leaves no room for Y because Y and O cannot be together.

[cnoury1221]

Hello,

I am wondering if someone could offer clarification on the point that because answer choices B,C,D, and E each contain G but not O and P, they are eliminated. I am not sure why they are eliminated. I know one window has to have G and P, but I am stuck as to how these answer choices can be eliminated with ease.

Thank you!

Carolyn

[Adam Tyson]

It's all about the numbers here, Carolyn! We know that we cannot have three windows with Orange, because then we would be unable to have a window with Yellow. But we also know we need more Orange than Green, per the question stem. That means two windows with Orange and one with Green.

From the original setup, we have a window that has Green and Purple, and which cannot have Yellow. That has to be the only window with Green in it. We also know that we have another window that does have Yellow, which cannot have Green or Orange in it, and because that window does not have Orange it must have Purple (the contrapositive of the last rule). So where are we going to put our two Oranges, if there is none in the window that has Yellow? We have to put Orange in the other two windows, INCLUDING the window that has Green and Purple. We are guaranteed, then, that one window has GPO (and possibly R), another window that has YP (and possibly R), and the third window has O and neither Y nor G (so it must have R or P or both).

This is where we get the idea that any answer with G must have P and O, and maybe R, or else it's out. OP is possible because that third window, the one with neither G nor Y, could be just those two, with the two Rs going to the other two windows.

[cnoury1221]

I am late, but thank you!!!!

It's all about the numbers here, Carolyn! We know that we cannot have three windows with Orange, because then we would be unable to have a window with Yellow. But we also know we need more Orange than Green, per the question stem. That means two windows with Orange and one with Green.

From the original setup, we have a window that has Green and Purple, and which cannot have Yellow. That has to be the only window with Green in it. We also know that we have another window that does have Yellow, which cannot have Green or Orange in it, and because that window does not have Orange it must have Purple (the contrapositive of the last rule). So where are we going to put our two Oranges, if there is none in the window that has Yellow? We have to put Orange in the other two windows, INCLUDING the window that has Green and Purple. We are guaranteed, then, that one window has GPO (and possibly R), another window that has YP (and possibly R), and the third window has O and neither Y nor G (so it must have R or P or both).

This is where we get the idea that any answer with G must have P and O, and maybe R, or else it's out. OP is possible because that third window, the one with neither G nor Y, could be just those two, with the two Rs going to the other two windows.

[ncolicci11]

Note that the second diagram above is incomplete. For example, the “third” window must contain Y and P (Y must be used in one of the windows and P must also be used when Y is used), and the second rule involving 2 R’s must also be accounted for. However, there is no need to continue to create the diagram when the initial inference allows you to eliminate each of the four incorrect answers.[/quote]

Just so I understand, if we are following this outline of what could be in the third window, it would also have to include R, correct?

Thanks!

[KelseyWoods]

Hi ncolicci11!

The 3rd window would not necessarily have to have R in this scenario. We'd have to place 2 R's but they could really be placed in any combination: 1 & 2, 1 & 3, 2 & 3.

Hope this helps!

Best,
Kelsey