If there are two white balls, then there must be three red balls (rule 1) and one green ball – you always need a green ball.
Since there’s only one green ball, there must be a white ball directly in front of it (rule 2). And this WG pair must be in front of all red balls (rule 3):
There is a second white ball. This is the only ball that can go in front of the WG pair. So green can go second or third:
B is CORRECT.
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S says
I got this answer by realizing that the only spot in the answer choices that was missing was box 2, meaning that’s probably where there would be ambiguity, so I guessed B because I had already eliminated A due to the second rule…
is this a valid approach?
Zoe says
The official answer is actually A…
TutorLucas (LSAT Hacks) says
I looked into it, and it seems the official answer is indeed B.
Here’s one common reason that your answer looks different: occasionally, other prep test companies provide practice tests that have an additional section in them, i.e. the tests will have 5 sections rather than the official 4 that LSAC releases so that you can practice with a full 5-section test. So, you might be looking at the answer key for the unofficial section.