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 Administrator
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#40570
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8568)

The correct answer choice is (D)

As determined during the setup, S and W cannot be on the research team together, and thus answer choice (D) cannot occur and is correct.

Incidentally, another way to view this inference is to realize that S and W have P and M as (partial) necessary conditions, respectively, and P and M cannot be chosen together per the first rule. This is ultimately the reasoning behind the inference discussed during the setup.
 lsatstudier
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#31368
Hi,

I was wondering if someone could explain how to get to the right answer for Question #9. I was trying to figure this out using an in/out group. However, I just kept losing track of all the variables and my prephrase never matched the answer choices. The S->P and T rule is also confusing me in this question. When finding the contrapositive of this, does that mean that P could still be paired with S in the "in group"?

I hope that makes sense. Thank you so much for all of your help!!
 Adam Tyson
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#31382
Hey there studier, thanks for the question.

Dealing with just that rule about S, P and T, that looks like this:

S :arrow: P & T

The contrapositive reverses the order, negates everything, and (very important here) switches that "and" to an "or":

P or T :arrow: S

So, if either P or T is missing, S must also be missing. If S and P are both in, T must also be in, because if T was out, S would be out.

All three rules in this game come into play on this question, and some prephrased relationships sure would help. Here's that first rule:

M :arrow: O & P

Another way to look at that rule is that M can never be in any solution that has either O or P in it. For that, we like to use the "double-not arrow", and it looks like this (broken into two distinct rules):

M :dblline: O
M :dblline: P

Now, think of the relationship between the first and second rules. They both have P in them, right? So if S is in, P is in, but if P is in, M cannot be in. That means S and M are having a fight about P - one says "he's in" and the other says "he's out". The inference you can draw based on that conflict is that S and M can never be together. That's this:

S :dblline: M

If that pair was one of our answer choices, we would be done. Sadly, it's not, so let's keep going (and all of this is work you should probably do as part of the original diagram, rather than waiting until you get to this question).

W requires M and Y, so that's:

W :arrow: M & Y

Now there's M again, a bit of a troublemaker because he doesn't get along with P, or O, or S. If W forces M in, and M forces those other three out, then we can infer a few new relationships for W. W can't go with those other three either!

W :dblline: P
W :dblline: O
W :dblline: S

One of these disagreeable pairs shows up in our answer choices, and that's answer D, the correct answer. There is no way that W and S can both be in, because W requires M to be there but S requires M to not be there (because they fight about P - see above).

All of this should show up in one long conditional chain, where W forces M and Y in, M forces O, P, and S out, and we don't know anything about T or Z. That would be my main diagram for this game, along with, perhaps, another chain based on S (forcing P and T in, M and W out, and O, Y and Z all become randoms that can come or go in any combination).

Draw that out and see if this question, and the rest of them for this game, doesn't get much easier for you. That's the key to these games where all the rules are conditional. Create the chain, make the inferences about who cannot go together, and then head into the questions.

Good luck, keep pounding!
 Jon Denning
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#31383
Well I typed up something as Adam was working on this as well, so I figure I'll leave my reply, too :-D Between the two of us I'm hopeful you'll figure things out!

By combining all three rules, what you find is that W causes things to happen for a lot of variables. Specifically, W brings with it M and Y (rule 3), and M gets rid of O and P (rule 1). And because S needs P (rule 2), then one of the big inferences here is that W and S cannot go together: W brings M, M removes P, no P means no S (since S needs P)...so ultimately W and S cannot go together.

And that's answer choice D!

A somewhat involved inference, but this is precisely the sort of thing the test makers love with LG and conditional rules, so it's critical that you can track these types of connections.

I hope that helps! Please let me know if it's still unclear!
 demk26
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#77543
Hi PS,

I made the inference: S :dblline: W, but can you please explain why Answer (B) is not feasible?

I had diagrammed:

W :arrow: M + Y :arrow: O + P

via the contrapositive, wouldn't O :arrow: Y be correct?

Thank you!
 Adam Tyson
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#77566
That's not a good contrapositive, demk26, because it wasn't Y that forced O out in that chain. It was only M that had any impact on O, because of the first rule. Y never has any impact on anything - it is a necessary condition for W, but is not sufficient for anything else to be in or out. It's a little challenging to draw that in this forum, but if you had the M and Y stacked vertically to the right of the arrow in your chain, then the arrow pointing towards O and P being out would be coming exclusively from the M, and not from the Y.

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