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- Sat May 05, 2018 12:27 pm
#45464
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4080)
The correct answer choice is (D)
If the number of animals in W equals the number of animals in Z, the cages could either both have two animals or both have four animals, and these are the only two distributions possible:
Answer choices (A) and (B) can be immediately eliminated because they are each numerically impossible.
Answer choice (C) can be eliminated as there are not enough lizards to match with a single snake in order to meet the numerical requirements above.
The correct answer choice, (D), is the only answer that allows these distributions to be possible: If there are three snakes in Y then there are two snakes in Z, which means there are two lizards in Z for a total of four animals in Z. This works with the distribution possibilities so answer choice (D) could be true, and is thus correct.
Answer choice (E) is incorrect because if there were four snakes in Z, then there would have to two lizards as well, combining for a total of six animals, which is more than either distribution allows for.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4080)
The correct answer choice is (D)
If the number of animals in W equals the number of animals in Z, the cages could either both have two animals or both have four animals, and these are the only two distributions possible:
Answer choices (A) and (B) can be immediately eliminated because they are each numerically impossible.
Answer choice (C) can be eliminated as there are not enough lizards to match with a single snake in order to meet the numerical requirements above.
The correct answer choice, (D), is the only answer that allows these distributions to be possible: If there are three snakes in Y then there are two snakes in Z, which means there are two lizards in Z for a total of four animals in Z. This works with the distribution possibilities so answer choice (D) could be true, and is thus correct.
Answer choice (E) is incorrect because if there were four snakes in Z, then there would have to two lizards as well, combining for a total of six animals, which is more than either distribution allows for.
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Dave Killoran
PowerScore Test Preparation
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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/