Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is a Grouping: Defined-Moving, Balanced, Numerical Distribution game.

This opening game on the October 1992 LSAT presented a difficult start for test takers. Upon reading that there are 14 variables in play, you have to expect that the test makers will limit the number of options involving the 14 variables because otherwise the game would be too difficult. Not surprisingly, the first rule limits the number of numerical options, and the next two rules effectively transforms the 14 variables into 7 variable “pairs.” The game then becomes a “7 into 4” grouping game, which is considerably more manageable. The rules also create the following inferences:

• 1. Because at least one gerbil must always be with a hamster, and cages Y and Z cannot contain a gerbil, cages W and X must contain all of the gerbils and hamsters.
  
  2. Because at least one lizard must always be with a snake, and cages W and X cannot contain a lizard, cages Y and Z must contain all of the lizards and snakes.
  
  3. Because no cage can be empty, cages W and X must each contain at least one gerbil-hamster pair, and cages Y and Z must each contain at least one lizard-snake pair. This deduction is the key to the game.
  
  4. Placing the pairs in inference 3 into each cage places 8 of the 14 animals. Only one gerbil and one hamster remain, and they act as a block that must be placed in W or X. Three snakes and one lizard remain, and they act as two separate blocks: one LS block and one SS block.
  
  5. With inferences 3 and 4 above, only two numerical distributions exist for W and X: 4-2 and 2-4, and only three numerical distributions exist for Y and Z: 6-2, 4-4, and 2-6.

Combining all of the information above leads to the following setup:

O92_Game_#1_setup_diagram 1.png

[saranash1]

The answer key shows the possible set ups for w & x and Y & Z. However, it doesn't know the possibility that each letter could have a minimum of 2 animals a piece. Is this not possible?

[Dave Killoran]

Hi Sara,

Yes, each cage (shown as letter W, X, Y, and Z) must have at least two animals. This point is referenced by the discussion of the numerical distributions in the game, none of which show less than 2 animals for any of the cages.

Thanks!

[saranash1]

Right but on the answer Possible Numerical Distributions for W & X and Y & Z are layed out but neither of them show the 2-2 numerical distribution

[Dave Killoran]

Hmm, maybe I misunderstood your question? You seemed to be asking if a cage could contain less than 2 animals, to which the answer is no.

As far as 2-2, both W and X cannot concurrently be 2-2; they are either 4-2 or 2-4. For Y and Z, the only possibilities are 6-2, 4-4, and 2-6; they also cannot concurrently be 2-2. So, individually, any of the cages could have 2 animals, which the distributions reveal. At most, however, two cages could have 2 animals: one of W and X, and one of Y and Z. At least one cage always has two animals: W or X.

Please let me know if that helps. Thanks!

[saranash1]

how do you determine that only possibilities are 6-2, 4-4, and 2-6?

[Dave Killoran]

Hi Sara,

You find the distribution by combining the rules. The first rule states that, "Each cage contains exactly two, four, or six animals." Thus, you automatically know you are limited to combinations of those numbers for the cages, but that each solution must add to 14 because that is the total number of animals kept in the four cages.

Next, because W and X must contain the gerbils and hamsters, those two cages must hold 6 animals total. And thus, with a minimum of 2 animals per cage and only even numbers, we get 2-4 or 4-2.

Because Y and Z must contain the lizards and snakes, those two cages must hold 8 animals total. Thus, with a minimum of 2 animals per cage and only even numbers, we get 6-2, 4-4, and 2-6.

Please let me know if that helps. Thanks!

[saranash1]

that helps!
Thanks!

[Martina]

hi,

You state that the first rule limits the numerical options and the next two rules effectively transforms the 14 variables into 7 variable “pairs. However, I don't understand how you inferred this.

[James Finch]

Hi Martina,

The way to get to this inference is to begin with the limited possible numerical distributions: either the cages have 2-2-4-6 or 2-4-4-4 as distributions, requiring pairs of animals. Rules 2 through 4 tell us what those pairs are: 3 G-Hs, 3 S-Ls, and an S-S. Because of the odd number of each animal type, only those pairs end up being possible, as we can't have odd numbers of animals in the cages.

Hope this clears things up!