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Getting mixed up with failing the sufficient and the necessary.

cole.w.murdochcole.w.murdoch Alum Member
edited June 2014 in Logic Games 228 karma
Hey everyone, if someone could give me a quick explanation on failing either the sufficient or necessary I would greatly appreciate it.
In LSAT Prep Test 15 (June 1995) - Section 4 - Logic Game 3 JY does exactly that during question 19. Here is a link:

There I asked, "Can I get clarification on how rule 5 becomes irrelevant during question 19? I usually understand satisfying the necessary but this mixed me up. If F is on 4 and 6, then J has to be on 3; that I understand. Because of this, G does not have to be on 1, correct? However, it still can be. When the necessary condition is satisfied through the question stem (as that is what happens in 19) then sufficient condition (G) becomes a floater?"

Furthermore, JY also shows a similar scenario when explaining rule 3 here:



  • CFC152436CFC152436 Alum Member
    edited June 2014 284 karma
    A ---> B ---> C

    Deny (fail) sufficient:
    We're told that A is in the out group. This means the first part of the rule falls away and we're left with B ---> C as our only rule.

    Deny (fail) necessary:
    We're told that C is in the out group. This means we can run the contrapositive back through our chained-up conditional statement. So, if C is out, then we know B is out and A is out.

    Bonus round:
    We're told that B is out. When this happens we're failing the necessary condition for A ---> B, but failing the sufficient condition for B---> C. What does this mean? Well, it means A and B are out, and C is a floater.

    Try and do something similar for this:
    /A ---> B ---> /C

  • cole.w.murdochcole.w.murdoch Alum Member
    228 karma
    Just for clarification, when you say the "out group" do you mean a question stem that has placed A (for example) into a slot that immediately goes against the intial rules set out? So, if the rule is: A1 ---> B4, and the question tells us A3, B would then be free to float anywhere? Further, if the question told us B3, then A cannot be in one as the contrapositive is /B4 ---> /A1?
  • ikethelsatikethelsat Alum Member
    193 karma
    Cole - that's exactly correct.
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