#16 - Local, Must Be True, Maximum

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=7106)

The correct answer choice is (D)

If 5, 6, and 7 are cleaned in numerical order, the following setup is produced:

Because there are two solutions when 2 is cleaned on Thursday, and two solutions when 2 is cleaned on Friday, there are four total solutions, and answer choice (D) is correct.

[rameday]

I don't understand how you get to the answer D?

I see 3 different schedules.

5,6,7 respectively at Tuesday, Wed, Thur
5,6,7 respectively at mond, tues, thur
5,6,7 respectively at Mon,wed, thur

Where is the 4th schedule?

A

[Jon Denning]

Hey rameday,

Thanks for the question. There is actually only ONE possible order for 5, 6, and 7, since you must have 8 cleaned on Monday afternoon (to have it be on an afternoon and before 4, which is cleaned Tuesday morning). That means that 5, 6, and 7 must go on Tuesday PM, Wednesday AM, and Thursday AM (always for 7), respectively.

The question isn't about the possible orders for just those three however. It asks for the total number of schedules possible for all eight streets, so we need to consider what's still uncertain at this point.

With 8, 4, and 7 always fixed, and now 5 and 6 fixed as well, the only streets left are 1, 2, and 3, and the open days are Monday AM, Thursday PM, and Friday PM. Since 2 is the only variable there with any restrictions--must be an afternoon--let's start with it.

1. 2 could be cleaned Thursday PM. That means Monday AM and Friday PM are open for 1 and 3, meaning two possible orders (1 Monday 3 Friday, and vice versa).

2. 2 could be cleaned Friday PM. That means Monday AM and Thursday PM are open for 1 and 3, meaning two possible orders (1 Monday 3 Thursday, and vice versa).

That gives us a total of four possible schedules, making D the correct answer.

I hope that helps!

Jon

[rameday]

oh wow this makes so much more sense now. thanks

[hasanya3]

In reference to the concept of Q16 p. 4-43 of the homework, what is the easiest way to determine the answer? I understand that (2,1/3) are in the Thurs/Fri afternoon spots but the answer explanation only addresses the possibility of 2 being on Thursday or Friday but does not seem to account for the event that streets 1/3 could each alternate between Monday, Thursday and Friday as well.

My main concern is figuring out how to narrow does the different possible sequences in questions like this one.

[hasanya3]

Thank you for the thorough explanation! but his would you go about arriving at this answer quickly? is there a certain shortcut strategy for this?

[James Finch]

Hi Hasanya,

My recommended approach would be to first diagram it out, as with any local question, then look at what is left uncertain. Here, the only unknowns are whether 6th street is cleaned on Wednesday or Thursday morning, and 1st or 3rd are cleaned Monday morning (whichever isn't in that spot goes to whichever spot 6th didn't take). So that leaves two possibilities for 1st and 3rd, and two possibilities for 6th. 2 X 2 = 4, so four possibilities. If you're still felling unsure, it only takes a couple seconds to game out the possibilities:

1--3-6
1--6-3
3--1-6
3--6-1

Hope this helps!

[ray57]

Hello,

For this question I put C because I thought there were only 3 maximum places for 1,2,3. When 2 is on Thursday, 1 can be Friday or Monday, and when 2 is on Friday, 1 can be Thursday or Monday, so, 1 can either be on Monday, Thursday or Friday (3 total). That goes for 3 as well. The question asks what is the max number of schedules for any ONE of the crewmembers, not a combination of all three. I don't understand why 4 is the answer here even after reading the administrator post above. Thanks for your help.

Ray

[Rachael Wilkenfeld]

Hi Ray,

Let's take a look at the four possible solutions.

We know that on Monday morning, we either need to clean First or Third street. Second street can't be Monday morning, because we know it's has to be in the afternoon.

Let's say that First street is cleaned Monday morning. Second street could be cleaned Thursday and Third on Friday. Alternately, Third street could be cleaned on Thursday, and Second on Friday. That gives us two possible solutions when First Street is cleaned on Monday.

Now, let's say that Third street is cleaned on Monday. Second street could be cleaned Thursday and First street on Friday. Or, First street could be cleaned on Thursday and Second street on Friday. That's two more possible solutions.

That gives us a total of 4 possible solutions.

Hope that helps!
Rachael

[akreimerman1]

Hello,

In the explanation of this game, I am just trying to figure out what this notation means when diagramming different scenarios: ( 2 , 3/1 ).

Thank you.