Setup and Rule Diagrams

[Dave Killoran]

This is a Basic Linear: Unbalanced: Overloaded, Numerical Distribution game.

The game scenario indicates that nine students will attend a total of six classes, creating an
Overloaded Linear situation:

Capture29.PNG

The scenario further states that some classes will have multiple students and that four will have
exactly one student. This creates a classic Numerical Distribution, and we will consider the
distributions after analyzing the rules.

The first rule establishes that I and L form a single class:

Capture30.PNG

This affects the distribution by establishing that one of the multiple-person classes is made up of two
people, I and L.

The second rule indicates that the other multi-person class consists of P and two other students:

Capture31.PNG

When combined with the game scenario, the first two rules establish that there is a single unfixed
Numerical Distribution:

Capture32.PNG

Of course, I, L, and P can be assigned to certain groups:

Capture33.PNG

We will revisit the distribution after considering the remaining rules.

The third rule indicates that K is the first female—but not the first student—to attend a class:

                                   KF 1st F

Because K is not the first student to attend a class, K cannot attend the first class. Because K must
be the first female to attend a class, we can infer that G, H, and I (the other three females) cannot
attend the first class (and thus K > IL block, G, and H). And, since K cannot attend the first class, the
earliest class that K can attend is the second, we can infer that G, H, and I cannot attend the second
class. In addition, because I and L form a block, we can infer that L cannot attend the first or second
class, leading to the following:

Capture34.PNG

The fourth and fifth rules can be combined with the first and third rules to produce a powerful chain
sequence:

Capture35.PNG

Without consideration of the other requirements of the game, the sequence indicates that O and P cannot
be in the first, second, or third class, that I and L cannot be in the fifth or sixth class, and that G cannot be
in the sixth class. Consequently, seven of the nine students are eliminated from attending the first class,
and therefore only N or S—but not both—can attend the first class. The N/S inference is one of the keys
to the game.

When the chain sequence is considered in conjunction with the other rules of the game and the
numerical distribution, a number of additional inferences can be made:

          Either H or O or both must be in the group of three.
          Either N or S or both must be alone.
          Because K’s class is ahead of P’s class, K must attend class alone.
          Because G’s class is ahead of P’s class, G must attend class alone.
          The group of three must attend either the fifth or sixth class.
          The group of two must attend either the third or fourth class.
          The sole attendee at the first class is male.

All of this information culminates in the final setup to the game:

Capture36.PNG

Overall, this is a very complicated sequence of inferences, and a tremendous amount of information
to juggle during a setup.

[eober]

Hi,

I am not sure whether I have an accurate set up for this game. I thought it could be either:

3 student class - 2 student class - 1 student - 1 student - 1student - 1 student

or
2 student class - 3 student class 1 student - 1 student - 1student - 1 student

Is this the correct order for the classes?

Also, Kate is the first female but not the first student, what restriction do we infer from this rule? Would it be correct to say that the first class (with 3 or 2 students) has no females?

[David Boyle]

Hi,

I am not sure whether I have an accurate set up for this game. I thought it could be either:

3 student class - 2 student class - 1 student - 1 student - 1student - 1 student

or
2 student class - 3 student class 1 student - 1 student - 1student - 1 student

Is this the correct order for the classes?

Also, Kate is the first female but not the first student, what restriction do we infer from this rule? Would it be correct to say that the first class (with 3 or 2 students) has no females?

Hello,

The first distribution, 3-2-1-1-1-1, is usually the way we show it, with the largest number first.
It does seem Kate cannot be in the first class, nor can any other female. Also, if Kate is the first female, that seems to mean that no other female can be with her, including Iyanna (who is with Leung). So Kate can't be earlier than second, and Iyanna can't be earlier than third...so that Gimena can't be earlier than fourth, or later than fifth, since Pedro and Oscar have to be after her (though they could both be in the same class). So answer A is correct.

David

[sdesousa]

Hello!

I just finished working through Game #7 in the LSAT Game Type Training II (October 1999, Questions 20-24) and I had a lot of trouble setting up my main diagram.

I was able to link the rules together to make one linkage; however, I had difficulty determining the positions of random variables such as N, S and H as well as determining where certain variable could not be placed. I'm hoping someone can discuss the optimal setup for this game and the inferences that can/should be made.

Thank you so much for all your help.

Sandy

[Jon Denning]

Hey Sandy,

Thanks for the question! This is a tricky game for a lot of people, but hopefully I can help clear things up a bit.

Numbers first. With 9 students into 6 positions, the game at first feels unbalanced, but the test makers then give you the groups: 3 block with P, 2 block with IL, and 4 singles for the other 4 spaces. We don't know the exact placement of the blocks yet of course, but at least we know the definitive group sizes.

To me the real key to this one is noticing the long chain that forms with K, IL, G, and both O and the P block. That's really powerful, especially once you see that K can't be first (first position must be a Male), since it pushes K into 2 or 3, IL block into 3 or 4, G into 4 or 5, and the P block (which could contain O, so be careful) into 5 or 6.

Consider:

First position is single Male since K is first female but not first student, and since L, P, and O are all pushed further down by our chain then either N or S must be first. That answers question 20 and really helps on 22.

Starting at position 2 the "K > IL > G > Pblock and O" sequence takes over. That answers question 21 and is tested directly in question 23, where K is forced into 2, IL is 3, and G is 4.

The final question puts O between K and G, meaning K is 2, IL and O are 3 and 4 (not sure of order), G is 5, and the P block is 6. Since only the P block has an opening for H, then H must be 6.

And that's it! You could certainly go through and do a bunch of Not Laws, but really just realizing how limited that chain is crushes this game.

I hope that helps!

[sdesousa]

Hi Jon,

Your breakdown of the game really helped clear things up for me. Thanks so much for all your help! It's very much appreciated.

[srcline@noctrl.edu]

Hello Jon

I understand the chain that K> IL> G >P forms but I dont see the PO box. MAybe I am not understanding the second rule but wouldnt it be P-- as diagramed becasue its saying that Pedro and exactly 2 others together constitute one class, so wouldnt that be three people?

Thankyou
Sarah

[Adam Tyson]

Sarah, check Jon's explanation again and you will see that you are in complete agreement with him. P is in a block with two others, and O may or may not be one of those two. He refers to "both O and the P block" (not to "the O and P block", if that's what you thought he meant) and then later to "P block and O". He is not suggesting that there is a block of P and O only. Chalk that up to a misunderstanding?

[srcline@noctrl.edu]

Hello Adam

Still confused on this. My chain was K> IL>G>P, If there is a PO box what would be the third variable grouped with P, would it be N, because the hypothetical that S is first? So then H would be last?
I apologize for the confusion, Thankyou
Sarah

[Robert Carroll]

Sarah,

Any of the randoms could be the third variable with P and O, if P and O indeed are together. They can be, but don't have to be. So while N could be, there is no reason it has to be. Is there a specific question you have in mind that is setting a hypothetical?

Robert Carroll