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- Thu Nov 30, 2017 3:12 pm
#42028
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8341)
The correct answer choice is (E)
Due to the action of the first rule, a maximum of three stores on a side could possibly be green (this is the case without considering the other rules). Thus, to meet the condition in the question stem, three stores on one side must be decorated in green and two stores on the other side must be decorated in green.
If there are five stores decorated with green lights, and stores 3 and 6 are already green, then three stores remain to be decorated in green. On the north side of the street, store 3 is already decorated in green, and stores 1 and 5 are other colors. Only stores 7 and 9 are not fully determined, but, due to the first rule, only one of stores 7 and 9 can be green. Thus, the north side must be the side with two stores decorated in green. Those stores are store 3 and either store 7 or 9.
On the south side, store 6 is green, and, to meet the condition in the first rule, the other two stores decorated in green must be store 2 and store 10. With store 10 decorated in green, store 9 cannot be green and must be red; store 7 is therefore green. Store 8 must be yellow because store 8 is the only store remaining on the south side that is able to fulfill the stipulation in the third rule that exactly one store on each side is decorated in yellow. This information results in the following setup:
Answer choice (E) is therefore correct.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8341)
The correct answer choice is (E)
Due to the action of the first rule, a maximum of three stores on a side could possibly be green (this is the case without considering the other rules). Thus, to meet the condition in the question stem, three stores on one side must be decorated in green and two stores on the other side must be decorated in green.
If there are five stores decorated with green lights, and stores 3 and 6 are already green, then three stores remain to be decorated in green. On the north side of the street, store 3 is already decorated in green, and stores 1 and 5 are other colors. Only stores 7 and 9 are not fully determined, but, due to the first rule, only one of stores 7 and 9 can be green. Thus, the north side must be the side with two stores decorated in green. Those stores are store 3 and either store 7 or 9.
On the south side, store 6 is green, and, to meet the condition in the first rule, the other two stores decorated in green must be store 2 and store 10. With store 10 decorated in green, store 9 cannot be green and must be red; store 7 is therefore green. Store 8 must be yellow because store 8 is the only store remaining on the south side that is able to fulfill the stipulation in the third rule that exactly one store on each side is decorated in yellow. This information results in the following setup:
Answer choice (E) is therefore correct.
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Dave Killoran
PowerScore Test Preparation
Follow me on Twitter at http://twitter.com/DaveKilloran
My LSAT Articles: http://blog.powerscore.com/lsat/author/dave-killoran
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PowerScore Test Preparation
Follow me on Twitter at http://twitter.com/DaveKilloran
My LSAT Articles: http://blog.powerscore.com/lsat/author/dave-killoran
PowerScore PodCast: http://www.powerscore.com/lsat/podcast/