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 gweatherall
  • Posts: 39
  • Joined: Jun 29, 2017
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#39710
Gahhh noooo are you kidding, LSAC?? I've never come across a game like this before, where the rules are simply contradictory. How do we know, in this instance, that we can safely just cross off L(S) and move on?
 Luke Haqq
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#39738
Hi gweatherall!

Two responses to your question,
How do we know, in this instance, that we can safely just cross off L(S) and move on?
First, in terms of how you approach the rules and inferences on logic games, to me L(S) was a pretty restrictive variable, meaning that a lot would follow if L(S) were true. Thus you might "know," or at least have a good idea, simply from looking at the rules that there might not be a lot of possibilities if L(S) were true.

Second, in terms of "knowing" as a proof that we can safely cross of L(S) and move on, then I'd recommend just assuming L(S) were true, and testing it out in your diagram. Again, a reason for testing out L(S), given the limited amount of time you have, is that it is a restrictive variable.

If L(S) is true, this immediately gives us further information because of the 3rd rule:

L(S) :arrow: N(R) & P(R)

So thus far, if we separated doctors into Souderton and Randsborough, we'd have:

Souderton: L
Randsborough: N, P

On this game, it's important to note that these options are binary, and all doctors are used--if the doctor's not at one location, then the doctor's at the other location. We know from the 4th rule that,

N(R) :arrow: O(R)

So we'd then have:

Souderton: L
Randsborough: N, P, O

So far that looks okay. However take a look at the 2nd rule, and its contrapositive:

2nd rule: J(R) :arrow: O(S)
Contrapositive: O(S) :arrow: J(R)

Since the options are binary and all doctors are used, then "not being in S" means "being in R," and not being in R means being in S. So another way of writing the contrapositive is:

O(R) :arrow: J(S)

Remember that we started by assuming L(S) were true--and this meant that N, P, and O were in Randsborough. So since we have O(R), we can then fill in more:

Souderton: L, J
Randsborough: N, P, O

The 1st rule from the game tells us that J(S) :arrow: K(R), so we could plug in the last variable, K:

Souderton: L, J
Randsborough: N, P, O, K

And that is where the problem is, which proves why L(S) can never be true/why L(R) must be true. The 5th rule of the game was:

P(R) :arrow: K(S) & O(S)

If L(S) were true, P would have to be in R, but so would K and O. In other words, L(S) being true will always conflict with that last rule, which is why it can never be the case.

Hope that helps!
 elliotklein
  • Posts: 2
  • Joined: Oct 17, 2017
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#40609
Hello,

I was running through the following Logic Game on LSAT 34 (June 2001) and upon drawing out the rules, I noticed what appeared to be a logical fallacy.

The 3rd rule states that if doctor Longtree is at Souderton, then both Nance and Palermo are at Randsborough.
The 4th rule states that if doctor Nance is at Randsborough, then so is Onawa
The 5th rule states that if Palermo is at Randsborough, then both Kudrow and Onawa are at Souderton

Where I find there to be a complication is that rule 3 requires that both rules 4 and 5 are occurring simultaneously, yet they both have opposite outcomes relative to the doctor Onawa. Here is the diagram I have drawn:

Ls ---> Nr + Pr
Nr ---> Or
Pr ---> Os + Ks

Therefore: Ls ---> Or + Os + Ks.... but how can Onawa be in both places at once? The rules state that each doctor is only at one place at a time.

Hopefully someone can provide me some insight as to how this isn't a problem, because for me I just can't understand how two cases can contradict one another.
 James Finch
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#40645
Hi Elliot,

This is actually a major inference for the game: because placing Longtree at Souderton would force multiple other variables to be at both hospitals simultaneously, we can infer that Longtree cannot be at Souderton, and thus must be at Randsborough. While this inference isn't tested directly, it does help eliminate several wrong answers immediately within the questions. Also, it serves as a frame for the two logic strings used for the other variables:

NR :arrow: OR :arrow: JS :arrow: KR :arrow: PS

and

PR :arrow: KS :arrow: JR :arrow: OS :arrow: NS

While L is always in R.

Hope this clears it up!
 elliotklein
  • Posts: 2
  • Joined: Oct 17, 2017
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#40649
That was my assumption, but I was just so confused because I've never seen them put in a rule that is directly avoided - it's always either a secondary deduction that has to be made.

Does this happen often? Wherein a rule that is stated is rendered impossible by following rules, and its negation becomes an important deduction?
 Francis O'Rourke
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#40694
There are not many games that involve this type of inference. This game is also one that stands out rather prominently in the history of the logic games.

Someone else can comment on exactly how frequently this occurs, but do not worry too much. It is not very likely that you will see it again. Don't ignore what you learn from this completely, since it is possible that the test makers will decide to insert this inference into the upcoming logic games section. It is also possible - though unlikely - that a logical reasoning stimulus will reason via this sort of reductio ad absurdum.
 powerscoreQasker
  • Posts: 23
  • Joined: Nov 24, 2020
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#84785
Hi, I have a question about writing out the logic chain. I originally wrote out the chain as shown in initial drawing.jpg.

I included the parts in red in my original diagram. I notice that the diagrams other people have made seem to exclude these "smaller branches," so to speak. That is, their diagrams show only the chains running from Nr to Ps or Pr to Ns, as shown in 2nd drawing.jpg (attached).

Is there any use in showing the parts I wrote in red in the first picture? What about for other games? When I construct a chain of conditional statements for an in-out or 2-value system game, should I initially include the smaller branches and then discard them? How should I approach this issue?
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 Ryan Twomey
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#84910
Everyone does this differently, but I would personally keep your drawings in red. I like everything to be in one place and I connect everything in the conditional games with connecting statements so I am sure I am not missing anything.
 powerscoreQasker
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  • Joined: Nov 24, 2020
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#85706
Gotcha. That's sort of been my impulse as well. Thanks!
 lsatstudying11
  • Posts: 54
  • Joined: Jul 30, 2020
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#87099
Hello,

I had a question about the templates that I did. Based on rule 1, I wrote down that at least one of J/K must be in R. Based on rule 2, I wrote down that at least one of J/O must be in S. Then I mixed and matched to make my templates.

Base:
S - O/J
R - J/K

Templates:
S - O
R - J

S - O
R - K

S - J
R - J
(^ fails because J cannot be in both)

S - J
R - K

Then, I was able to place some more of the doctors based on the rules. My question is whether this is a sensible way to make templates? Thanks in advance!

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