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- Fri Jan 21, 2011 12:00 am
#80602
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8622)
The correct answer choice is (B)
In the prior question, we discussed the fact that there are still three hospitals remaining to be placed. But, in that question, the hospitals had to be separated in order to “cover” cities 1, 4, and 5, each of which did not contain an institution. This is a Global question, and there is no restriction on the hospitals being together, so it is possible that the three hospitals could all be placed in any of those three cities, and thus we can eliminate cities 1, 4, and 5 from contention. This removes answer choices (A), (C), and (D) from consideration (another way of thinking about those three answers is that according to our diagram, cities 1, 4, and 5 are identical, so none of them is likely to be the correct answer).
Only cities 2 and 6 remain in consideration. City 6 already has a hospital, and there is no restriction on adding two more hospitals (which would leave the remaining hospital to pair with the jail in city 1, 4, or 5), and so city 6 can have three hospitals, and is eliminated from consideration.
City 2 cannot contain all three remaining unplaced hospitals because city 2 cannot contain a jail, and one of the three hospitals must be with the jail that goes in city 1, 4, or 5. Because the maximum number of hospitals in city 2 is thus two, answer choice (B) is the correct answer.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8622)
The correct answer choice is (B)
In the prior question, we discussed the fact that there are still three hospitals remaining to be placed. But, in that question, the hospitals had to be separated in order to “cover” cities 1, 4, and 5, each of which did not contain an institution. This is a Global question, and there is no restriction on the hospitals being together, so it is possible that the three hospitals could all be placed in any of those three cities, and thus we can eliminate cities 1, 4, and 5 from contention. This removes answer choices (A), (C), and (D) from consideration (another way of thinking about those three answers is that according to our diagram, cities 1, 4, and 5 are identical, so none of them is likely to be the correct answer).
Only cities 2 and 6 remain in consideration. City 6 already has a hospital, and there is no restriction on adding two more hospitals (which would leave the remaining hospital to pair with the jail in city 1, 4, or 5), and so city 6 can have three hospitals, and is eliminated from consideration.
City 2 cannot contain all three remaining unplaced hospitals because city 2 cannot contain a jail, and one of the three hospitals must be with the jail that goes in city 1, 4, or 5. Because the maximum number of hospitals in city 2 is thus two, answer choice (B) is the correct answer.
Dave Killoran
PowerScore Test Preparation
Follow me on X/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 X/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/