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# PT34.S4.G4 - each of exactly six doctors

Member
edited September 2017 150 karma
Hi All,

I've been pretty good with setting up and reading conditional chains for in-and-out games but for some reason I'm really confused with the conditional chain for PT 34 S4 G4 (if you recall, it's about splitting six doctors to either the Souderton or Randsborough clinic).

Here is the simplified version of the conditional chain for this game:
~N --> ~O --> J --> ~K --> P

When I look at this, the minimum number of doctors that needs to be in the in group (Souderton) is two since we have two separate OR pairs (~O --> J and ~K --> P) as was asked in question 21 (and the correct answer was "two").

So this is where I get really confused. If this is the logic we are using to interpret a minimum number of doctors, why can't the correct answer for question 19 (which asks for a complete and accurate list of doctors at Souderton) be E (N and P)? Why can't the ~N --> J pair be treated the same as ~O --> J? If we only have N and P, aren't we still good since we have at least one of N and J and one of K and P?

I guess I am really confused as to how to correctly account for the OR pairs when there so many overlap between them in a particular chain like you see above? Thanks for your help in advance!

https://7sage.com/lsat_explanations/lsat-34-section-4-game-4/

• Alum Member
235 karma
I'll have a look at this game when I make my way to my computer, but based on your chain how can it be N and P? /N seems to result in P.
• Alum Member
edited October 2016 916 karma
@CrushLSAT said:
question 19 (which asks for a complete and accurate list of doctors at Souderton) be E (N and P)? Why can't the ~N --> J pair be treated the same as ~O --> J? If we only have N and P, aren't we still good since we have at least one of N and J and one of K and P?
~N --> ~O --> J --> ~K --> P

I didn't go back to look at the game but off the top of my head, and based on the conditional chain you provided, N and P can't be a *complete* list of doctors in the "in" group because you're ignoring the "or" relationship between /O and J. You would need to have one or the other of O or J and if you only have N and P, you have neither O nor J. If you had neither, it would violate that "or" rule.

Also, you can treat the /N --> J pair the same as the /O --> J pair but you can't forget that the /O and J pair still exists.
• Alum Member
235 karma
I just looked the game over and the poster above me got it right; E violates the "OR" rule by having neither O nor J. This can't happen because if one is at R then the other MUST be at S. Missing one of the entities forces the other into S.
• Member
1171 karma
@draj0623 explained it perfectly. N and P being complete clearly means ~O which means J is in and thus, we have a contradiction.
• Alum Member
23929 karma
@draj0623 nailed it!
• Member
150 karma
@draj0623 @Tinyosi1 @desire2learn Thanks for the help
• Inactive ⭐
2086 karma
@desire2learn This is spot on.
This is one very important thing to remember with in-out games. When they ask you for a complete and accurate list, any variable that is NOT mentioned in the provided list is a variable that is "out." Since O was not mentioned, O is "out," hence, J must be in.
• Member
38 karma

Can anyone help me with explaining question 21 in this game?? This is my post on the original game, here in 7sage. i will post it below.

"Can someone please explain a different/better explanation for question 21? Its asking what the MINIMUM number of doctors could be at Souderton, so the IN group.

I understand that either O/J have to be, and also, K/J.

What’s confusing me is – why couldn’t we do the exact same thing with other relationships in the chain?? for example: N/J, or N/P… either one of those could also be ones that have to be in, right? because you just read further down the logic chain and identify other OR pairings…

Am i reading the logic chain wrong? Can you not identify OR pairings unless they are right beside each other in the chain?

Because when i read it the way I am, then identifying the OR relationships (inclusive, meaning one of the pairs must be IN) makes this question come out to a lot of different possibilities, and makes it more confusing – it certainly does not clearly lead me to AC (C) – a minimum of 2.

The other way to look at it, is to read your chain outright and see that J and P could both be in while the other three in the chain are out. But what if you put only P in? then you’d have the minimum of 1 IN. BUT, then again you could say what i said above – either N/J and O/J have to be, which brings the minimum back up to 3…

I may be reading into the chains incorrectly, but any explanation or correcting of my thinking would be appreciated!"

• Member 🍌🍌
9366 karma

Hi @taranjot ,

I think the quickest way is to test.

/N —> /O —> J —> /K —> P
/P --> K --> /J --> O --> N

/P triggers three people to be in. /J triggers two people to be in. So if you put P and J in the IN group (Souderton), then the "either or" rule goes away, and you can put others in the OUT group.

either N/J and O/J have to be, which brings the minimum back up to 3…

You can have J in and both N and O out.

I hope this clarifies