- Sat Oct 21, 2017 10:54 am
#40811
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13302)
The correct answer choice is (A)
This question may seem similar to question #20 in terms of the unusual nature of this inference, but the correct answer is easier to identify.
In the distributions, there are always two, three, or four red balls. From the second rule, in the distributions with 3 or 4 red balls, there would always have to be a green ball in a box lower than box 4. In the distribution with two red balls (3-2-1), there could be a green ball in box 4, but because that distribution contains three green balls, there would also have to be a green ball in a box lower than box 4. Because there is always a green ball in a box that is lower than box 4, answer choice (A) is proven correct.
Note that the hypothetical shown in question #18 disproves answer choices (B) and (C). The hypotheticals shown in question #19 disprove answer choices (D) (both solutions disprove this answer) and (E) (solution #2 disproves this answer).
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13302)
The correct answer choice is (A)
This question may seem similar to question #20 in terms of the unusual nature of this inference, but the correct answer is easier to identify.
In the distributions, there are always two, three, or four red balls. From the second rule, in the distributions with 3 or 4 red balls, there would always have to be a green ball in a box lower than box 4. In the distribution with two red balls (3-2-1), there could be a green ball in box 4, but because that distribution contains three green balls, there would also have to be a green ball in a box lower than box 4. Because there is always a green ball in a box that is lower than box 4, answer choice (A) is proven correct.
Note that the hypothetical shown in question #18 disproves answer choices (B) and (C). The hypotheticals shown in question #19 disprove answer choices (D) (both solutions disprove this answer) and (E) (solution #2 disproves this answer).