# Reply To: Diagramming Logic Games With LSAT Trainer.

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.