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 Dave Killoran
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#60228
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=6469)

The correct answer choice is (B)

The question stem asks for the number of other friends who must appear when U and Z appear together in a photograph.

As we have already established, Z is irrelevant in this type of question because Z is a random. Thus, the question revolves around U. From the first two rules, we know that when U appears in a photograph then S and W must also appear in that photograph. The appearance of W in the photograph (and S and U), forces Y to appear in the photograph. None of the other friends must appear, and thus the correct answer is three: S, W, and Y.

Answer choices (A), (C), (D), and (E): These answer choices are incorrect because they do not contain a number that equals the number of other friends who must appear in the photograph.
 CEF
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#10738
I had some difficulties on this game. After further review, I now see that common misinterpretation of the third rule which answered many of my questions but I still believe I am missing a piece of the puzzle.

On question 16, why is it three friends that must appear in the photograph with Umiko and Zach and not two?

I made the connection that if U then S, and if S then W. So, S and W must also be present. Is the third variable that either R or Y must also be present? I understand that it could be R and not Y, or Y and not R, or both. I just want to clarify that the third element is that you must always have Y or R.

With that said, is this the same concept that is applied to question 17 as well? That if S and Z, then W must also be in, but you must also have Y or R and therefore you cannot have exactly three?

This was a tricky game! It really put into perspective how the misinterpretation of one rule can throw off the entire game.
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 Dave Killoran
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#10758
Hi CEF,

You are correct that the third rule is critical here, and playing a role in both questions. For other readers, I discuss the third rule in some detail over at http://forum.powerscore.com/lsat/viewto ... 6243#p6248.

In question #16, when U appears, then S and W appear. But, when W appears, then R does not appear, which forces Y to appear. So, it's not the choice of R or Y, but rather that Y has to be the choice. Thus, the three other friends that must appear are S, R, and Y.

In question #17, it's the same logic at work, that when you have S and ZS, you also have W, and because of that, ultimately Y.

There's no question it is very tricky, mainly because of the way they present the rules! This is one where you need to be rock solid on conditional statements, otherwise the game will seem quite difficult.

Please let me know if that helps. Thanks!
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 SGD2021
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#93120
For question 16, why do we interpret it as though we can determine the number of people who go in the same photograph? I thought we could really only determine how many people go in photographs together in general (since the intro suggests there are multiple photos) and how many go alone, but not the exact number of people together in one photograph. Are we able to answer question 16 (how many must appear in that same photo as U and Z) because we assume that the question is asking us about one hypothetical situation, and one hypothetical situation represents all those together in one photo vs all those not in that photo (i.e. alone)?
 Robert Carroll
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#93250
SGD2021,

We can determine the maximum and minimum number of people appearing in a single photograph, and, in fact, that's about all the rules talk about - single photographs. In this game, we have no clue what's happening in the entire album. The rules are all about things happening in single photographs. So, could everyone be together in the album, just in different photographs? Sure, that's easy enough to do - any people who can't be together would just be in different photographs. So there's really not much point asking about the album as a whole, and the questions don't really talk about that. This question is talking about a single photograph - if U and Z are in that one, then how many must be in that photograph? What's happening with other photographs in that album, which may or may not exist, has nothing to do with what happens to the single photograph, which has at least U and Z, that we're talking about in this question.

Robert Carroll

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