Q20

[patrice.antoine]

I wanted to know how to tackle this question. I was only able to guess an answer.

[maryadkins]

This is a labor intensive question. Here's how I solved it.

We have to have a green ball in a box lower than any box that contains a red ball, and our options for red balls (since we have to have more than white) are 2, 3 or 4 (not 5, because there wouldn't be room for the GW block). In terms of numbers, there are 4 combinations we can have given that we have to have 2, 3 or 4 reds. (See attached diagram.)

Since there are multiple reds and they are constrained above G, I found it easiest to start testing where they had to go by placing them in their various combinations as low as possible in the stack.

When I listed 4 possible combinations placing the Rs as low as they could go given that G has to be lower than any of them (and filling in the rest with could-be-trues, not necessarily must-be-trues, for G and W). (See other attached diagram for these.)

Testing these against the answer choices, we can eliminate (A), (C), and (D), because boxes 2, 4, and 5 don't have repeats in every possible order. We're left with (B) and (E). Reworking the first combination into the possibility that the W in box 6 could be moved to box 1, I eliminated (B):

R
R
R
G
W
W

Through process of elimination, I identified (E) as correct.

This question likely would have been dealt with faster by saving it for last and using previous work.

[farhadshekib]

In my opinion, the best say to answer this question is to leave it for last.

I was able to use the hypo's that I made for the other questions to eliminate the 4 incorrect answer choices in 15 seconds.

[tianfeng102]

This question took me quite some time to solve:

The #6 ball could be any of the three colors: G, W, R
If G, it must have another twin to fulfill the requirement of R-R-G.
If W, it must have another twin to fulfill the requirement of GW block.
If R, well, we have at least two R's.

Therefore, #6 must have a twin at another position.

[Acing LSAT]

I also basically left it for last.

I knocked of A B and C with the work from Q 19. To do 19 I quickly wrote out the option for the question which are

6 G
5 G
4 W/ R
3 R / R
2 R / G
1 G / W

That proves 2, 3 and 4 can have only a ball that appears only once

Than after Q 23 (below) I saw that 5 could have W occurring once. that knocked off D which left me with E

Q 23 could be
6 G / R
5 W/R
4 R/G
3 R/G
2 R/G
1 G/W

[kmcaple]

I think the approach should be be which color is likely to have one, which is white. Where could white not go. I position 6 bc it had to be directly below grew.

[CharlesS800]

I struggled with this game, beginning to work on it and then leaving it until last, which ultimately proved very helpful. Using hypotheticals from other questions, I eliminated answer choices A, B, and D.

I then decided to test answer choice C. I figured that if I could only use one element, it would be better be W, so that I could satisfy the W/G chunk rule. From there I haphazardly filled in the rest of my diagram, settling on:

R
G
W
R
R
G

I checked over the rules again and this diagram seemed to obey them all, so I eliminated C as an answer choice and selected E.

[JulianG99]

I dont even understand the question.

[ohthatpatrick]

Choice (A) is saying that "Box 2 must be the same color as at least one of the other balls".

On any "must be true" question, we're seeking counterexamples in order to eliminate an answer.

So we would evaluate choice (A) by asking ourselves, "DOES the ball in box 2 have to be the same color as at least one other ball .... OR is it possible that the ball in box 2 does NOT have the same color as any other ball (i.e. is it possible that the ball in box 2 is a color that only appears once, in box 2)?"

First we would look at all our previous work (I do all the "if" questions before I tackle any unconditional questions, so we could see our work from 18, 19, 22, and 23).

If we have any scenarios where the color in box 2 is ONLY appearing in box 2, then we can eliminate (A).

If we don't have any such scenarios, then we're trying to write a scenario where box 2 has a unique color. How about
G W G R R R

That works! We can eliminate (A) since "box 2 does not HAVE to have the same color as at least one other ball." We just proved that box 2 could have a unique color.

Make sense?