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#44127
Setup and Rule Diagram Explanation

This is a Grouping: Partially Defined, Numerical Distribution game.

The game scenario indicates that there are two cages and one exhibition for ten birds. At least two birds and at most four birds are on exhibition, and there are at most four birds in each of two cages:
F95_Game_#3_setup_diagram 1.png
This leads to some easy distributions, such as a 4-4-2 (cage-cage-exhibit) or 4-2-4 and 2-4-4, etc. However, the distributions do not play a major role in this game. Instead, it is the grouping rules and the restrictions between all of the birds that answers the questions.

These are the ten birds:
F95_Game_#3_setup_diagram 2.png
The second rule states that birds that are both of the same sex and of the same kind cannot be caged together. Therefore, only one male parakeet can be assigned to each cage. Since there are three male parakeets and only two cages, it follows that at least one male parakeet must always be exhibited, along with a corresponding female parakeet. Thus, one of Q, R, or S and one of T or W is always on exhibit.

So, at least one pair of parakeets must always be exhibited. However, one of the initial rules states that at most two pairs of birds can be exhibited at a time. Therefore, a pair of goldfinches and a pair of lovebirds can never be exhibited together. This inference is tested directly in question #17.

The second rule states that two birds of the same sex and kind cannot be caged together. Thus, J and K (two female goldfinches) cannot be caged together. Same for Q and R (two male parakeets), Q and S (two male parakeets), R and S (two male parakeets), and T and W (two female parakeets). These inferences can be diagrammed as:
F95_Game_#3_setup_diagram 3.png
While the cage inferences are all based on the second rule, there is more variation in the exhibit inferences.

First, from the third rule, S cannot be exhibited with J or W:
F95_Game_#3_setup_diagram 4.png
Of course, if S cannot be exhibited with W, then we can infer that S must be exhibited with T, the only other female parakeet:
F95_Game_#3_setup_diagram 5.png
The next few inferences derive from the stipulation that pairs of one male and one female of the same type must be exhibited. Because S can only be exhibited with T, and not with W, neither Q nor R can be exhibited with S:
F95_Game_#3_setup_diagram 6.png
Again, this occurs because Q or R would require W, and S cannot be exhibited with W.

Another inference in that same vein involves J and K, the two female goldfinches. J and K can never be exhibited together, because there is only one male goldfinch, and for both J and K to be exhibited, there would have to be two male goldfinches:
F95_Game_#3_setup_diagram 7.png
The final inference on the exhibition list reflects the inference discussed above that goldfinches and lovebirds cannot be exhibited together.
F95_Game_#3_setup_diagram 8.png
Combining all of the above rules and inferences leads to the following:
F95_Game_#3_setup_diagram 9.png
With this information in hand, we are ready to attack the questions.
 atirvine88
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#22278
Hi guys,

I had trouble diagramming this question. I made a couple of inferences concerning which birds could not be with each other, but otherwise, this question for me took a real long time. I found myself having to go back to review the rules and the variables involved. Should I have listed the variables even if they have listed them for me up at the top? How should I diagram this?
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 Dave Killoran
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#22279
Hi AT,

The key is to pull a separate set of inferences for the cage and for the exhibits. There are about 5 good inferences for each, and using those 10 inferences allows you to really push through the questions. Although there are 10 variables, with the inferences the game becomes easier than it looks at first.

There are also some easy distributions, such as a 4-4-2 (cage-cage-exhibit) or 4-2-4 and 2-4-4, etc. However, the distributions do not play a major role in this game. Instead, it is the grouping rules and the restrictions between all of the birds that answers the questions.

Please let me know if that helps. Thanks!
 florbonita
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#22280
I spent a ridiculous amount of time on this problem and completely missed the very important cage/exhibit inferences. I now understand your explanation of the numerical distribution, but I didn't even get that far in this problem. I spent a great deal of time straining to figure out what could go where from the rules, and I didn't come to any inferences on my own. Maybe I missed the entire fact that there was 1 exhibit in addition to the 2 cages. I also overlooked the importance of the number of male parakeets. I'm trying not to get discouraged when it comes to games. Is there something that I am missing in my initial review of the rules? Honestly, I was not really sure which way to go with this game. I did not even get the first question.
 Nikki Siclunov
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#22281
Hello florbonita,

This is a confusing game indeed, but you can take solace in the fact that there are very, very few games where the variable set has not one, not two, but three distinct attributes. Here, each bird has a name, a kind, and a sex.

You missed the fact that there are 3 groups here (two cages, one exhibition), suggesting that you didn't read the scenario closely enough. Slow down! You also didn't focus on the male parakeets, which was a mistake. You should have realized that the male parakeets contain 3 variables (more than any other group). Because of this, the second rule will be significantly more restrictive with respect to the male parakeets than any other type of bird: a male parakeet must be exhibited, because we can't distribute all three of them between the two cages. What's more, the last rule affects S (which is a male parakeet).

The key to a game like this is 1) close reading; 2) identify the power variables; 3) see the connections between the rules. The numerical distribution analysis was of secondary importance here.

Hope this helps!
 LustingFor!L
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#30629
Please explain the exhibition inference H,J,K double not arrow M,N. Why can you not have H J/K and M N in the exhibition?
 LustingFor!L
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#30634
I'm really struggling with the inference that one pair of goldfinches and one pair of lovebirds cannot be exhibited together. Is it because there are two female goldfinches and only one can fit in the exhibition? Why can't the female goldfinch be alone in a cage?
 Adam Tyson
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#30727
Hey there Lusting, good to see you again! Let me see if I can help.

The problem with exhibiting the goldfinches and lovebirds has nothing to do with the exhibition, and everything to do with the cages. We cannot have any two birds of the same type and sex caged together, per the 2nd rule. But we have three male parakeets, QRS, and only two cages in which to put them. If none of them are exhibited, we have to put two male parakeets together. We can't have that! The inference, then, is that among those birds exhibited there is always at least one pair of parakeets. With no more than two pairs to be exhibited, we cannot have goldfinches and lovebirds both in the exhibit because parakeets will take up one of those two slots.

If you have any doubts about that, try coming up with a solution with goldfinches and lovebirds exhibited, and see if you can follow all the rules. Testing these things out should lead to new inferences, a fuller diagram, a greater understanding of the game, and ultimately more correct answers and a higher overall score. Go for it!
 Katherinthesky
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#88097
Hello,

Can the numerical distribution of 4-3-3 (exhibited-caged-caged) work as well?

I am aware that the numerical distributions in this particular game don't really play a significant role in solving this particular set of questions, but just making sure I am not missing an inference.

Thanks in advance.
 Adam Tyson
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#88108
Yes! While the exhibited birds must be in pairs, the caged birds don't have to be, so 3 birds per cage is fine as long as no cage has two birds of the same sex and kind. For example:

Cage: HJM
Cage: KNS
Exhibit: QT, RW

or

Cage: QJW
Cage: RKH
Exhibit: ST, MN

Good eye!

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