#17 - Local, Not Necessarily True, False to True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4635)

The correct answer choice is (D)

The question stem indicates that Q is the only employee to attend L. This places the game into the 2-2-2-1-1 distribution, and the local condition can be added to the base diagram as follows:

D10_game #3_#17_diagram 1.png

Of course, H can only have 0 or 1 attendees, and so H is the other talk with just one attendee (which must be S or T), and F, G, and I each have two attendees:

D10_game #3_#17_diagram 2.png

With Q attending both G and L, Q cannot attend any other talks. When the fourth rule is considered, T’s first talk must be either G or L. Since T’s first talk can’t be the last talk at the conference (because T must attend two talks), we can infer that T attends G:

D10_game #3_#17_diagram 3.png

This could also be inferred by recognizing that the only candidates to attend G are Q, S, and T, and because S is already attending two other talks, Q and T are the only two employees that can attend G in this scenario.

In addition, as established during the game setup, the only employees that can attend F are R and S. Thus, because F must have two attendees, R and S both attend F in this question:

D10_game #3_#17_diagram 4.png

At this point, the attendees of F, G, and L are fixed, leaving only H and I still unresolved. H must be attended by S or T, but there is no way to determine which one attends. The remainder of S/T attends I, along with R, who has not yet been assigned to a second talk. This leads to the final diagram for this question:

D10_game #3_#17_diagram 5.png

The only uncertainty involves S and T, and so you should immediately examine the answer choices that contain S or T. Because S can attend I instead of H, answer choice (D) is not necessarily true and is therefore correct.

[bghose]

Hi,

I think that for S (Spivey), the choices for the 2 conferences are limited to F/I and H. My reasoning is as follows:

1) Q is the only one attending L. Hence R has to attend F and I, since L is ruled out for R. R's choice of I would limit Q to choose G.
2) Since Q and T share one common and it cannot be L, so T has to go to G.
2) Since R & S share one common, then S has to choose from one of F/I. For the 2nd conference, R cannot choose L or G anymore, so I think that it must choose H.

Looking at the answer choices, A,B,D,E look true to me based on the above reasoning. Which leaves C as my answer which is apparently Incorrect! Am I reading too much into the rules which say that R & S and Q & T have one common? Could it be that they can have both common as well? Does the answer hinge on this fact, or is there a different explanation?

Thanks,
-Bish.

[moshei24]

Hey, here's how #17 should be diagrammed.

The order is F-G-H-I-L.

Q's second talk is L, and is the only one in L, so that one is locked up.

Now, we know that there has to be a block of T's first talk with one of Q's talks, and now we know that it has to be Q's first talk that is in a block with T's first talk. That has to be in G, because Q can't be in F or H because of the rules and Q can't be in I, because then R won't have two places to go, and there won't be a place for the second T.

So we have G: Q1T1 and L: Q2.

Next, we know that S has to be in a block with R's first talk. That has to be F. It can't be G or H because of the rules, and if it's I, there's no place for R's second talk.

So we have F: S1R1 G: Q1T1 and L: Q2.

That leaves us with H and I. We know that the second R has to be in I because the rules say R can't be in H.

So we have F: S1R1 G: Q1T1 H: ? I: R and L: Q2.

There's still three open spots - 2 in H and 1 in I. We know from the rules that S and T can't go together, which leaves us with either T or S in H and then the remaining one in R. That's the uncertainty in this question. So which one could be false? The answer is (D), since S could go in H, and then T could with R in I in that situation.

Please let me know if that helped.

Thanks!

[bghose]

Thanks for the reply, I understand how you got to D) as the answer choice.

The only part of with which I had difficulty is that why couldn't I be R1 & S1, why did it have to be F only? But I accept that the answer won't change due to the choice of F or I as R1/S1. Thanks for taking the time to resolve the issue!

[bghose]

Just to complete the thought process for Q17: in my original posting, if I had accounted for the fact that R & S could share both conferences, then both of them could have gone to F & I. Which would have left G & H to be assigned to T. So it is not necessarily true in that hypothetical scenario that H is assigned to R, hence D) could be False!

In retrospect, I guess that I did read too much into the rule which indicated that R & S have one common;; nowhere did it restrict it from having two common conferences.

[moshei24]

It had to be F only because QT was in G, R can't be in H, and if they were in I, there would be no place for the second R.

Got it, now?

[bghose]

Is there anything wrong with the following hypothetical scenario for Q17:

L F F H
G I I G
-----------
Q R S T

It seems to satisfy all the rules, including that Q & T share the 1st conference, R & S share the 1st conference and that S & T don't have any common conference. If the above is correct then not only F but even I could be the 1st shared conference between R & S, no?

This is a slightly tangential point, since it has already been established and accepted that D) is the correct answer. Thanks again for that help!

[moshei24]

I think you should diagram it with the base as FGHIL. Try doing it that way. It will clear everything up for you.

[moshei24]

And btw, that scenario works for #17, but you need to realized the the order has to be FGHIL, so that should be the base without a doubt in this case.