Diagramming Logic Games With LSAT Trainer.

    • June 15, 2016 at 10:50 am #2039
      kylsatter
      Participant

      Hi I’m having some trouble with diagramming on the Lesson 14 drills in the LSAT Trainer (and pretty much all Logic Games for that matter)

      Game 1 I don’t understand because of the intensely fast inferences made through the process of the diagram. I’m lost as to how to find and test right and wrong answers in the diagram here. When I started diagramming diagonally, I started to notice an even bigger problem which is that there is no way to distinguish an order between the top and bottom lines in a diagram when you have a situation like in Step 2 of the Sometimes Two… section.

      Step 3 on the prior page points to my main problem about testing even more. Section 1 offers no advice on properly plugging elements into diagrams to test for right or wrong answers.

      On Game 2, I’m completely lost as to how Mike set up the diagrams here. He starts off with four slots on the bottom shelf of two shelves in the first diagram which I don’t understand he got.

      On diagram 2, there are now three levels and the second shelf has 3 slots. Whaa? I’m stumped on there Logic Games questions. 200 pages into the trainer I feel like I’ve barely learned anything.

      I haven’t even gotten to game 3 yet. Thinking about completely starting over again even though I’m 200 pages into this book.

    • June 15, 2016 at 11:05 am #2040
      kylsatter
      Participant

      And here were are on Game 3. Mike draws two diagrams and then says in the 1st answer. “If W is first we know we’re dealing with the link on the left of our diagram.”

      How are we supposed to know to use the 1st diagram when W is technically in the first spot on both diagrams? If anything, I’d be more likely to pick Diagram 2 since W stands out more as the first element. Again, completely confused as to how he got here and how to plug the elements in to test right and wrong answers.

    • June 15, 2016 at 11:32 am #2042
      kylsatter
      Participant

      Game 4: Again on the answer to the first question I’m confused as to how Mike immediately jumps to the answer G,L,N. This doesn’t appear to violate the rules of the diagram. L and N are together and either F or G can go with JL so that leaves us with G,L,N being ok. I see how it doesn’t map on to the diagram he created but that doesn’t explain how it violates the rules. Dumbfounded at these leaps in logic with no explanation.

      That basically leaves me with no foundation on which to base my answer for question 2.

      On Game 5, I do not understand the diagram on the bottom line whatsoever. I have no idea how he drew these inferences from the rules for the diagram beginning with J. H/P in second. No idea where that came from. L/O in third. Why? H/P in fifth. No idea. O/L in the last positions. Sure. No idea how he got here.

    • June 15, 2016 at 11:45 am #2043
      kylsatter
      Participant

      How did you do? Terrible. Missed all questions or at least half on every drill except for Complex “Or” Rules and Math Inferences.

    • June 15, 2016 at 6:00 pm #2053
      LSAT Dan
      Participant

      With respect to game 2, Mike has drawn two separate templates, based on the first rule. If twice as many books go on the bottom shelf as the top shelf, then there are only two possibilities for the distribution: Either the top shelf has two books and the bottom shelf has four books (that’s the diagram on the left), or the top shelf has one book, and the bottom shelf has two books (that’s the diagram on the right). Every valid layout must fit into one of those two arrangements.

      In the one on the left, where two books are on the top shelf and four on the bottom, that accounts for all six books, so there are no books on the middle shelf, and F, which we’re told is not on the top shelf (last rule), must be on the bottom shelf.

      In game 3, the second rule says that either T or W is third. On question 1, since W is first, T must be third. If T is third, we have to use the chain on the left, since in the chain on the right, W, V, and R all have to be placed before T. There’s no room for all of them to be to the left of T when T is third.

      In Game 4, on question 1, did you try to put all 9 people on three teams with one of the teams being G, L, and N? If you do, you should stumble across the rule violation. Because of the second rule, if G is with L and N, then it must be F who is with J and I. That leaves H, M, and K to make up the third team, but H and K can’t be on a team together. The rules will interact with each other in ways that are not always readily apparent, so you can’t stop your analysis at the point where you see (correctly) that F is still free to be with I and J. Before you say that the arrangement in (D) is ok, you should have a complete assignment of all 9 people that doesn’t violate any rules.

      The diagram for Game 4 is based on the rules – H and K are separated, as we’re told they are. The third team must be I, J, and either F or G. So every valid arrangement must fit into that template, or it will break at least one of the rules.

      On game 5 in the bottom layout, once J is in position 1 and M is in position 4, ask yourself what positions L could go in. Why L? Because L is affected by multiple rules, and variables affected by multiple rules tend to be the ones where inferences are available. For instance, L cannot be in position 2, because there would nowhere to put K (first rule; remember, M is in position 4). L cannot be in position 5, because there’s nowhere to put O (second rule). If you go down the line and look for potential places for L that don’t break the first two rules, you’ll see that it’s only possible to place L in position 3 or position 7. Those two variations give you the basis for placing K and O, with H and P (unconstrained) getting whatever’s left.

      I’m not sure where exactly Game One gave you the most trouble, but notice the effect of that first rule – If N has to be after L or O, but can’t be after L *and* O, then N must be in between L and O. That gives rise to the two possible chains:

      N is after L (but not O): L – N – O

      N is after O (but not L): O – N – L

      Other than that, the only addition to the chain is to tack P onto the back of O, because of the second rule.

      Hope this helps.

    • June 15, 2016 at 7:29 pm #2054
      kylsatter
      Participant

      Wow. Thank you for the response, Dan. I just started trying out the questions on the practice exams a few minutes ago so I’m not giving up hope yet. Got a bit stressed earlier so I apologize for spamming.

      Definitely going to review this again tomorrow before Lesson 15. You’ve no idea how much I appreciate the detailed response on this. Hope it wasn’t too much trouble.

    • June 15, 2016 at 7:43 pm #2055
      LSAT Dan
      Participant

      No worries…you’re welcome. Feel free to follow up if you have further questions after you’ve reviewed them.

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