#20 - Global, Must Be True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13302)

The correct answer choice is (E)

This can be a challenging question, and contains an inference that seems to come out of left field. This is certainly not the type of inference you would be expected to draw during the setup (it’s simply far too specific for such a broad game), but you can solve this question efficiently even if you do not initially see the answer. How? By referring to the local diagrams that you have already created, and by skipping part of this question and returning to it after you have created more solutions.

Let us assume that when you approach this question, you realize that the Global nature of the question lends itself well to using prior work. You can then refer back to questions #18 and #19:

• From the solution to question #18, we can determine that the ball in box 3 does not have to be the same color as a ball in another box. Thus, answer choice (B) can be eliminated.
  
  From the solutions to question #19, we can determine that the balls in boxes 1 (solution #1) and 4 (solution #2) do not have to be the same color as balls in another box. Thus, boxes 1 and 4 can be eliminated. Box 1 does not appear among the answers, but box 4 is referenced in answer choice (C), and thus answer choice (C) can be eliminated.

This leaves only answer choices (A), (D), and (E) in contention. At this point, you could choose to solve the question by either considering the remaining answer choices abstractly or by using hypotheticals. However, given that there are still three more questions left in this game, if you must use hypotheticals, then the best choice would actually be to stop working on this question, move on to the remaining questions, and then to return to this question with whatever new knowledge you have gained. Using this approach, the solutions to questions #22 and #23 can be used to eliminate answer choice (D). At that point, only answer choices (A) and (E) would be left, and your work to solve this question would be as minimal as possible.

In analyzing the correct answer, why must it be that the ball in box 6 is the same color as a ball in another box? Let us look at each color option and see why:

• If the ball in box 6 is G: If the ball in box 6 is green, then to accord with the second rule, there must also be another green ball lower in the stack.
  
  If the ball in box 6 is R: According to the game distributions, there are always at least two red balls in every solution to this game. Thus, if the ball in box 6 is red, there will be a red ball in another box.
  
  If the ball in box 6 is W: If the ball in box 6 is white, then to accord with the third rule, there must also be another white ball lower in the stack (directly below a green ball).

Thus, regardless of the color of the ball in box 6, there will always be another box containing a ball of the same color. Hence, answer choice (E) is correct.

[smile22]

Hello - I ran into trouble with this question. I was able to do this game, and determined the appropriate numerical distributions associated with the color of the balls. However, I was unable to answer this question. I ran out of time on this section and was unable to go back and answer this question. Any assistance that you could provide would be greatly appreciated.

[Robert Carroll]

smile,

It's helpful to examine this question and phrase it in a different way. If there were some color such that only one ball were of that color, then that ball, being the only one of its color, would not be the same color as any other. In the context of this question, we're being asked, "If there's only one ball of some color, which box can't contain that ball?"

Now, because the third rule tells us that at least one white ball exists, and the first rule tells us that there are more red balls than white balls, there must be at least two red balls! So even if there is some color with only one ball representing it, that ball could never be red.

We know that there has to be a green ball, because the second and third rules give us conditions under which we can place green ball(s). We also know that there are at least two red balls. By the second rule, if there is only one green ball, it would have to be under at least the two red balls, so a lone green ball certainly couldn't be in box 6 or box 5.

So far, we know that box 5 and box 6 can't contain the stack's only red ball (because the stack always contains more than one red ball, so there is never an "only red ball") and they can't contain the stack's only green ball (because, if there is only one green, it must be the one below all the reds, which would push it below 5). We also know that there has to be a white ball right below a green ball somewhere in the stack. So if box 6 contained a white ball, the green-then-white would have to occur somewhere else, so box 6 couldn't be the only white ball! Thus, if a ball without a color match exists in the stack, there's no way it can be box 6, because none of the colors works!

Robert Carroll

[smile22]

Thank you very much for the explanation.

[htngo12]

I have been rereading your response on this question over and over again to comprehend your view on how reinterpret the question.

However, I'm still processing the first part of your response.

When I reread the question, I was able to interpret it as asking utilizing the information gathered based on rules, I need to prove that 1 box -> the same color at least one of the balls. So meaning which box has a at least a color match with another ball.

Could I interpret the contrapositive: not at least one same color of other ball -> not 1 box? translates to none of the same color then not that 1 box. (if not the same color, then not in that box)

I was tied between D and E, but picked D. I was able to prove that was wrong later in question # 22.

[Kristina Moen]

Hi ht,

I think I understand your question, but if I don't answer it satisfactorily, please ask again!

This is a Must Be True question, so you can certainly solve it by proving that an answer can be false and thus eliminating it. In this case, the answer choice must be one that EVERY TIME will contain a ball that is the same color as a ball in another box.

So if you go to answer choice (A), you could disprove that answer by creating a scenario where Box 2 has an independent color that is found in no other boxes. Or you could look at some of your previous scenarios from other questions. But if you can find NO scenario in which Box 2 has its own color, then that means answer choice (A) is correct. You can do that with all of the answer choices. Unfortunately, that would be time-consuming with this question, since the correct answer choice is (E)!

[htngo12]

I think it get it now. I start to confuse myself when I read too many possible solutions instead of staying in line with my own methods. So it basically ask which box must have at least another match in the sequence. For answer D is not possible because in Q22 I was able to get a hypothetical (balls 6-1) G W G R R G. Leaving me with E) Box 6. Box 6 must have a color that matches another box.

[blaisebayno]

Hi,

I would greatly appreciate if someone could visually represent the situation in which box 4 is Not the same color as any other of the boxes.
Thank you!

[Adam Tyson]

Sure! How about this, from the top (6) down to the bottom (1):

G
G
W
R
R
G

We have more reds than whites (first rule);

We have a green below all the reds (second rule);

We have a GW block (third rule).

We're good! And box 4 has the only white ball!