#19 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=2450)

The correct answer choice is (E)

If F drives exactly twice, but does not drive on either Tuesday or Wednesday, F must drive on Thursday and Saturday, and this forces J to drive on Wednesday:

J95_Game_#3_#19_diagram 1.png

The diagram is sufficient to eliminate answer choices (C) and (D). In addition, because F drives exactly twice, we can determine that the drivers are in a 2-2-1-1 distribution. This information can be used to eliminate answer choices (A) and (B). By process of elimination, answer choice (E) is proven correct.

Another way to eliminate answer choices (A) and (B) is to realize that they are functionally identical, and any two identical answer choices must both be wrong because each correct answer choice is unique (the Uniqueness Theory of Answer Choices).

[ccampise]

So I understand the question and why the answer is E, but I'd like to know how exactly A and B are functionally identical. I can't figure it out and I feel like it'd be a good thing to try and understand moving forward.

[Lucas Moreau]

Hello, ccampise,

This question has all to do with numerical distributions. Since Fritz is driving exactly twice this week, that means that the distribution of drivers to days must be 2-2-1-1 - two drivers driving two days, two drivers driving one day.

Answer choices A and B are therefore wrong for the exact same reason, which is that one person cannot drive three times (with Fritz driving twice, that would force you into a 3-2-1-0 distribution, impossible by the first rule), nor can three people drive one time (that would force you into a 2-2-2-1 distribution, also impossible by the first rule).

Hope that helps,
Lucas Moreau

[lfial1011]

I completely understand why answer choices A, B, C, and D are totally impossible. I also understand why E is the answer by process of elimination, of course. However, what I don't understand is how E can be the answer if we would have F driving on three consecutive days. Would not that be a violation to the second rule? Rule #2: No person drives on two consecutive days.

You would have something like this:

___, ___, J, F, F, F
M T W TH F S

[Jon Denning]

Hi lfial1011,

Thanks for the question, and welcome to the Forum! I think the issue here might come down to a simple misreading of (E), where F has been accidentally confused with J in the answer, but let's take a look at #19 and see.

19. Here we're told that F drives exactly twice, but not Tuesday nor Wednesday (nor Monday, of course, from the third rule). So that leaves days Thursday, Friday, and Saturday open.

Now, as you noted, no one can drive on two consecutive days. So how can we have two Fs on Thurs, Fri, and Sat without the Fs being consecutive? It must mean that F is Thursday and Saturday, and not Friday.

What that would also mean is that J drives Wednesday (since not Saturday), and days Monday, Tuesday, and Friday are open.

Answer choice (E) says it is possible that J drives Friday. That's totally acceptable! That would be J on two days (Wed and Fri), F on two days (Thur and Sat), and G and H would be one day each (Monday and Tuesday).

So as I said up top I suspect you may have misread (E) as saying F on Friday, which would indeed be a problem. But J on Friday is totally alright, and thus (E) is a possibility and correct.

I hope that helps!