- Sat Oct 21, 2017 12:52 pm
#40815
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13302)
The correct answer choice is (B)
If there are exactly two white balls, then the 1-3-2 distribution is in effect. From the second rule, the green ball must be in a lower box than all of the red balls, and thus immediately we can infer that the green ball can never be in box 4, 5, or 6 (otherwise one of the three red balls would be forced to be lower than the green ball). This eliminates answer choices (C), (D), and (E).
Because there is only one green ball, and from the third rule a white ball is always in a lower box than a green ball, the green ball can never be in box 1 in this scenario. That eliminates answer choice (A). Thus, answer choice (B) is proven correct by process of elimination.
Answer choice (B) can be correct under the following solution:
This is not the only solution to the question, but every solution involves a white ball in box 1 and a green ball in box 2 or 3.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13302)
The correct answer choice is (B)
If there are exactly two white balls, then the 1-3-2 distribution is in effect. From the second rule, the green ball must be in a lower box than all of the red balls, and thus immediately we can infer that the green ball can never be in box 4, 5, or 6 (otherwise one of the three red balls would be forced to be lower than the green ball). This eliminates answer choices (C), (D), and (E).
Because there is only one green ball, and from the third rule a white ball is always in a lower box than a green ball, the green ball can never be in box 1 in this scenario. That eliminates answer choice (A). Thus, answer choice (B) is proven correct by process of elimination.
Answer choice (B) can be correct under the following solution:
This is not the only solution to the question, but every solution involves a white ball in box 1 and a green ball in box 2 or 3.
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