#11- Local, Could be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=351&t=3899)

The correct answer choice is (A)

The condition in this question stem stipulates that no color is used twice, and thus each light must be a different color. From a direct application of the rules, this automatically affects the second and third rules:

Because the second rule stipulates that when light 2 is green then light 1 must also be green, which would violate the condition in this question stem, we can determine that light 2 can never be green. This immediately eliminates answer choice (D) from consideration.

In the third rule, when light 3 is purple, then light 2 must also be purple. Accordingly, in this question light 3 can never be purple, and this information can be used to eliminate answer choice (E).

However, the other two incorrect answer choices can also be eliminated by examining the third rule more closely:

In the third rule, when light 3 is yellow, then light 2 must be purple. Answer choice (C), which states that light 1 is purple and light 3 is yellow, would thus cause purple to be used twice (in lights 1 and 2), a violation of the condition in the question stem. Thus, answer choice (C) cannot be true and is incorrect.

From the contrapositive of the third rule, when light 2 is yellow or purple, then light 3 must be green. Answer choice (B), which states that light 1 is green and light 2 is yellow, would thus cause green to be used twice (in lights 1 and 3), a violation of the condition in the question stem. Thus, answer choice (B) cannot be true and is incorrect.

Therefore, answer choice (A) is the correct answer, and is possible under the following solution: G-P-Y.

[ellenb]

For this question, it does not make sense the answer because:

I have the rule that says:according to the contrapositive of the third rule,

when light 2 is yellow or purple than light 3 must be green and for Answer choice A ,

we have the second light bulb purple which means that the 3 light bulb must be green and we have the first light bulb green. which would cause green to be used twice in the first and last light bulb. Please clarify?

Thanks

Ellen

[David Boyle]

For this question, it does not make sense the answer because:

I have the rule that says:according to the contrapositive of the third rule,

when light 2 is yellow or purple than light 3 must be green and for Answer choice A ,

we have the second light bulb purple which means that the 3 light bulb must be green and we have the first light bulb green. which would cause green to be used twice in the first and last light bulb. Please clarify?

Thanks

Ellen

Hello Ellen,

Is this question related to the one at http://forum.powerscore.com/lsat/viewto ... 460#p10460 ? Thanks,

David P.S. I also answered a question of yours about the "leave of absence" problem

[ellenb]

Yep it is the same one, and thanks !

[ellenb]

however it is a different question, and it not answered yet. Dear Powerscore, please answer it when you get a chance.

thanks

Ellen

[Dave Killoran]

Hi Ellen,

I think maybe there was confusion because you did not post a question number with this, but I see you are asking about question #11.

Please take a look at that third rule again--the contrapositive will be that if Light 2 is Yellow or Green then Light 3 is Green. But, in (A), Light 2 is Purple, so all bets are off.

Thanks!

[ellenb]

Thanks, funny how I thought I thought it was a different color. Thanks Dave!!!