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- Sat May 05, 2018 12:27 pm
#45465
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4080)
The correct answer choice is (C)
If there are twice as many lizards in Z as in Y, then Z contains two lizards and Y has one. So there must be either two or four snakes in Z to make an even number (a total of four or six animals), and there must be either one or three snakes in Y (a total of two or four animals).
Answer choice (A) is incorrect because Y contains one lizard.
Answer choice (B) is incorrect because Y contains either one or three snakes.
Answer choice (C) is the correct answer choice.
Answer choice (D) is incorrect because Z contains either two or four snakes.
Answer choice (E) is incorrect because Z contains a total of either four or six animals.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4080)
The correct answer choice is (C)
If there are twice as many lizards in Z as in Y, then Z contains two lizards and Y has one. So there must be either two or four snakes in Z to make an even number (a total of four or six animals), and there must be either one or three snakes in Y (a total of two or four animals).
Answer choice (A) is incorrect because Y contains one lizard.
Answer choice (B) is incorrect because Y contains either one or three snakes.
Answer choice (C) is the correct answer choice.
Answer choice (D) is incorrect because Z contains either two or four snakes.
Answer choice (E) is incorrect because Z contains a total of either four or six animals.
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/