#20 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15078)

The correct answer choice is (E)

This is an extremely challenging question if you do not understand the distribution of lots in this game. Remember, Anastasia can only park in lot X a single time.

If lot Z is the $12 lot, then from the second rule we can infer that lot X is the $15 lot. With the cost of X and Z established, we can infer that lot Y must be the $10 lot.

Because X is the $15 lot, Anastasia must park in lot X on Thursday. And, because Anastasia can only park in lot X a single time, she cannot park in X on any of the other days. This fact has a direct impact on Wednesday and Friday. Wednesday’s lot, which must cost either $12 or $15, cannot cost $15 because that would mean X was Wednesday’s lot. Consequently, Anastasia must park in lot Z on Wednesday. And, since according to the third rule the Wednesday lot costs more than the Friday lot, we can conclude that Anastasia must park in lot Y, the $10 lot, on Friday. Hence, answer choice (E) must be correct.

Answer choices (A) and (B): Neither of the days listed in these answer choices must be a day on which Anastasia parks in lot Y, and hence both of these answers are incorrect.

Answer choices (C) and (D): Neither of these answers could be a day on which Anastasia parks in lot Y, and hence both of these answers are incorrect.

Answer choice (E): This is the correct answer choice.

[k100]

Hello,

I'm confused about the wording of this question. Your explanation states that "If lot Z is the $12 lot, then from the second rule we can infer that lot X is the $15 lot." However, couldn't Z occupy both the $12 lot (Wednesday) and the $10 lot (Friday), leaving the possibility of X/Y for Thursday?

If this is true, I don't see how Y must be parked anywhere, because it could be on Monday, Tuesday, or Thursday.

I'm also not sure why X must be parked on Thursday. Even though X can only be parked once, couldn't it be either Wednesday or Thursday, as long as Z is parked on Friday for $12?

I must be missing a crucial inference somewhere. Thanks for your help!

[Luke Haqq]

Hi k100!

Happy to walk through this question.

First, we're asked, "If lot Z is the $12 lot, then on which one of the following days must Anastasia park in lot Y?" We can therefore begin to fill in what we know about the lots:

X: ?
Y: ?
Z: $12

Next, we can use the second rule of the game, which tells us that "Lot X costs more than lot Z." This can be represented as:

X$ > Z$

Since we already know that Z = $12, the only other options are $10 or $15, but we know that X must cost more than Z, we can infer that X must cost $15. This leaves $10 as the cost of Y:

X: $15
Y: $10
Z: $12

Third, we're told in the first rule of the game that the $15 lot is on Thursday. We've already fixed the prices to lots above for this question, including X being the $15 lot. Since the original rules tell us that the $15 lot is on Thursday, we therefore know in this question that X is $15 and is visited on Thursday.

Fourth, at this point you might think about diagramming two different possibilities, based on the third rule, "The lot Anastasia parks in on Wednesday costs more than the one she parks in on Friday." This can be diagrammed as:

W$ > F$

This lets us infer that the lot on Wednesday must either be $12 or 15 and the Friday lot must be either $12 or $10. There are only two possibilities, in other words. If the $12 lot is on Wednesday, this means in this question that it is the $12 lot as determined above, and that Y is the $10 lot and must appear on Friday. Another possibility that one might begin is supposing that Wednesday is instead the $15 lot. This, however, wouldn't work, since we already know that x is the $15 lot; because of the rule that the number of Zs outnumber Xs we can actually infer a global distribution (X, Y, Z, Z, Y/Z) at the outset of this game. On this particular question, Wednesday couldn't be the $15 lot, because this would require two Xs, which would require three Zs, and this would leave no space for Y.

Therefore, the only possibility that works when "lot Z is the $12 lot" involves Y being the $10 lot and being visited on Friday. It's possible that Y might also be visited on Monday or Tuesday, but that much information is sufficient to arrive at answer choice (E).