This is a tricky question. I skipped it the first time I did it. I was then able to use scenarios from later questions to eliminate some answers.
Note: This question is asking if a ball cannot be the only one of its kind, and in the listed space. So if a ball can be the only one of its kind, in the listed space, that answer is wrong. For example: if there is a scenario with only one white ball, and that white ball is in spot 3, that scenario eliminates B.
This scenario from question 19 eliminates C:
This scenario from question 23 eliminates D:
In question 18, we saw a scenario with two white balls, three red balls and one green ball. We were able to place Green 2nd or 3rd, and it was the only green. This eliminates A and B:
(There are no other greens. That’s why I haven’t bothered to fill out the rest of the scenarios. In a timed situation, I just looked back at the diagram I drew for question 18 and said “ok, this eliminates A, there is only one green, etc.”)
Sometimes it makes sense to do questions out of order. Thanks to elimination from other questions, we disproved all the wrong answers.
E is CORRECT. Process of elimination is enough, but I’ll explain why it’s right using logic. If the sixth spot is red, it isn’t unique. There are always multiple reds.
If the sixth spot is white, it isn’t unique. There must be another white before a green, somewhere (rule 3). If the sixth spot it green, it isn’t unique. There must be a green before the reds too (rule 2).
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Ari says
I think there might be an easier way to think about it:
1. We know that all three colors are used from the rules. (one WG block and more reds than whites). We also know minimum there has to be two reds (more than whites). We also know that at least one green has to occur before the first red.
2. So looking through the choices you can see that if we look at box six, if it is red it is guaranteed there is another red (minimum two reds). If it is green, there has to be another green somewhere that comes before a red. If it is white, it cannot be a member of the WG block because there is no space for G to go after, guaranteeing another W in that WG block somewhere down. No matter which color you use, that color has to be used further down. I looked at the sixth position first because it gave the most chances (5) for repeat colors to occur.
TutorRosalie (LSATHacks) says
That works as well.
John Michelli says
Rule 1: R>W
Rule 2: GW
Can deduce that there are at least 2 Rs, then. Means G can never be in Boxes 5 & 6. So Box 6 is either R or W. If 6 is R, there must still be one more R. If 6 is W, must be one more W per Rule 2.
TutorLucas (LSAT Hacks) says
Just wanted to clarify one of your points about G can’t be in 5 & 6. While G can be in both 5 & 6, if there is just one G in the game it could not be in 6. Here is a scenario where G can go in both 5 & 6: (from the top) GGWRRG.
-Reply from LSAT Tutor, Morgan Barrett