Linear set-up from powerscore homework p230

[artcast58]

Hello,
I have question from my LSAT Powerscore test preparation homework on page 335 the linear practice answer for question #8 on page 230
Why is 'o' not able to be seated in table#4?
Thanks Art Castillo

[Morgan O'Donnell]

Thanks for your question, Art!

Can you clarify what material you are working out with? Are you using the Weekend Course book or one of our Full Length Course books? If it is a Full Length Course book, we just need you to clarify what lesson!

Once we get this clarification, one of our instructors can help you out!

Thanks,
Morgan O'Donnell
PowerScore Test Preparation

[artcast58]

This is the weekend course book question 8 p 230. the answer page is 335

[Dave Killoran]

Hi Art,

I don't have the book in front of me right now (I'm on the road), but I'm betting I know which question this is—the one about the tables and the patrons, right? If so, the inference you’ve asked about is a difficult one. Please let me try to shed some light on the process:

When O is placed at Table 4, then Tables 1, 2, and 3 must each occupied by a patron. This forces the Empty table (variable “E”) to be assigned to Table 5, 6, or 7. Unfortunately, that assignment is a violation of the third rule that states that tables 5, 6, and 7 must be occupied by a patron.

Please let me know if that helps. Thanks!

[artcast58]

Hi Dave,
Sorry but still don't get it. O is the fourth patron according to the rules thus the fourth spot. I don't see the how it becomes a violation if E is only a place holder.

[Jon Denning]

Hey Art,

E isn't so much a "placeholder" as it is an indication of "no patron" (Empty, hence "E"). The trouble here is that tables 5, 6, and 7 cannot contain the E--they can't be empty according to the rules--and placing O at table 4 would force the empty table to be one of 5, 6, or 7, causing a violation. So that's why O cannot be 4.

[artcast58]

Ok I'll figure it out according to your explanation.

[Dave Killoran]

Hi Art,

Two things:

First, I separated out your second question in this post and made it into a new thread. We try to keep separate questions in separate threads so other students can find the answers easily The other question is now at http://forum.powerscore.com/lsat/viewto ... =26&t=7340.

Second, let's go back to the problem above. One of the best ways to figure out why Not Laws don't work is to place the variable in question into the position where it supposedly can't go. so, in this case, we'd place O into the Table 4. What happens then? Well, because of the other rules that connect to O, patrons would be forced into tables 1, 2, and 3. Ok, so far, that's not an issue. But, we know that one of the tables is also empty. That empty table has to go where? The only tables left are 5, 6, and 7, so it has to be one of those. But that's a violation of the rules too, and now we have a problem. Thus, the thing that started this all—O on 4—won't work.

It's a tough inference, but a classic "domino sequence" type of inference where a series of events is triggered by placing O on 4.

Please let me know if that helps clear it up. Thanks!

[artcast58]

Ok. By re-reading the question-one seat will be empty except 5,6,7 thus seat one and seat four can be empty. Right?

[Dave Killoran]

Hi Art,

I'm not sure I'm following you there. If O is in 4, then it's not empty, nor can 1 be empty. That's what forces 5, 6, or 7 to be empty (which is the violation).

If I've misunderstood you, let me know. Or, if you're asking about a different part of the problem now, also let me know. I'm going off memory at the moment since I'm travelling Thanks!