Diagram

[aileenann]

Diagram and solution to problem 3 attached. Let me know if there are other questions on this game you'd like to talk about! I'm all ears.

PT 53, S2, G1 -Talent Agencies - Manhattan LSAT.pdf

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[sge4]

I encountered something bizarre when I diagrammed this puzzle. Instead of FPS on the bottom, I put T W X Y Z on the bottom with slots above each, where I put an f, p, or s. This led me to split the puzzle into two frames. For some reason I can't get the f, p, and s's to stay put but in frame I we have the base of T W X Y Z with f, p, and p over X, Y, and Z, respectively. In frame II we have f, s, and s over X, Y, and Z, respectively.

In frame I if S goes in T, S goes in W. Fine. But the contrapositive isn't true -- If S is not in W then S is not in T! That means P/F must be in W and P/F must be in T. But that would leave no room for an S. So do we have to discount the contrapositive somehow?

The opposite happens in frame II. If S goes in T, S can't also go in W because then there are 4 S's and no P's. In this case, though, the contrapositive could apply (P/F in T and W).

This wasn't that big a deal on this puzzle but I've never seen a case where the contrapositive or the actual rule has to be OVERRULED by another rule. Am I missing something here, or else how do you know if a rule applies or not. Specifically, in frame, would S --> S rule still hold, but not the contrapositive. In frame II, would the contrapositive of S --> S still hold?

Thanks!

[noah]

In frame I if S goes in T, S goes in W. Fine. But the contrapositive isn't true -- If S is not in W then S is not in T! That means P/F must be in W and P/F must be in T. But that would leave no room for an S. So do we have to discount the contrapositive somehow?

It's not that you "discount" the contrapositive, it's that you simply realize you can't apply it here. By the way, you could also have S assigned to W but F or P to T.

The opposite happens in frame II. If S goes in T, S can't also go in W because then there are 4 S's and no P's. In this case, though, the contrapositive could apply (P/F in T and W).

Right, you know that S can't go in T in that frame.

This wasn't that big a deal on this puzzle but I've never seen a case where the contrapositive or the actual rule has to be OVERRULED by another rule. Am I missing something here, or else how do you know if a rule applies or not. Specifically, in frame, would S --> S rule still hold, but not the contrapositive. In frame II, would the contrapositive of S --> S still hold?
Thanks!

There are plenty of examples where you can't apply a conditional statement. It's an interesting question, but I think you're over-thinking this. If I tell you that Q is either first or last, and I also tell you that if H is third, Z is last, then you know that if Q isn't first, H can't be third. You can't have the condition triggered.

I hope that helps!

I tried your frames, BTW, and I found it easiest to write an "S" in a cloud above T and W in frame 1 (I need at least one S), and in frame 2 I put a "P" in a cloud above T and W (and you could put not S underneath T as well).

[tzyc]

For unconditional questions like Q2 and Q4,
we have no way but test each answer choice and find which is MF...right?
It took a lot of time so I wonder whether I forget something...

[noah]

For unconditional questions like Q2 and Q4,
we have no way but test each answer choice and find which is MF...right?
It took a lot of time so I wonder whether I forget something...

You're right that there's no inference chain to follow in either, however both can be handled pretty quickly.

Take a look at my explanations for each:

2: q2-t6836.html

4: q4-t6837.html

[PaulaJ949]

AileenAnn, was there a specific reason you didn't actually put X in F on your diagram? Just want to make sure I'm nailing these setups!

Thanks!

[ohthatpatrick]

Yikes, no that was definitely just a mistaken omission. Honestly, that pdf is a bit of a travesty. I'm sorry.

For what it's worth, my setup would look something like this:

RULES:
X = Fame
~(XY)
(ZY)
Ts -> Ws

.. Fame ....... Premier ..... Star ...
.......X............__+.........___+

Since there's a chunk, (ZY), I would frame its options.
Since Y and X can't be together, ZY can only go into P or S.

.. Fame ... Premier ... Star ...
.......X............Z Y.........___+..........frame 1
.......X............__+.........Z Y...........frame 2

In frame 1, we can further infer that W will definitely be in Star.
If T is there, then W is there.
If T isn't there, then W is the only one left and someone needs to go there.

.. Fame ... Premier ... Star ...
.......X............Z Y.........W..........frame 1
.......X............__+.........Z Y...........frame 2