PT87.S4.G4 - A charity is assigning volunteers to work

Adam Z. Xie

This game has virtually no rules to solve MPT questions.

During PT, I spent 15 minutes for the first 3 games, but I couldn’t finish game 4 in 20 minutes. I technically guess the last 4 answers because there are infinite possibilities.

I must did something terribly wrong. Please help.

Admin note: edited title

Adam Z. Xie

Also, is there any game like this? I have done most of not all LGs, but I cannot recall any similar game.

xenonhexafluoro

How did you set the game up?

I treated it as an in/out game with subcategories (days). For each day you have 3 slots in, 2 slots out.

Timed & upfront, the only inference I made was that L must be in on Saturday, because P is already occupying 1 of the 2 out slots, and kicking L out would kick out M too, putting 3 players in the out group.

You also have a total of 9 in slots, with 5 players, each of which can be in a maximum of 2 times (b/c none of them can be in all 3 days). So, 1 player goes in once, and the rest go in twice. If M goes in twice, then L must go in twice, with M. This distribution inference helped me out on a couple of questions (20/23).

Edit: I've definitely found myself with uncomfortably open in/out subcategory games. But if you do the set up correctly, and keep a close look on that out group, the questions shouldn't drain you of time.

Michael.Cinco

There are a couple of infrences you can make in this game.

Namely. There is a piece that MUST got on Saturday

Also if you think about distribution you soon see that this game must have a distribution by where 4 of the pieces are used twice and one is used once.

It's wide open but because of the amount of open spaces and the small number of pieces relative to the spaces you are actually quite restricted on how you distribute the pieces.

Adam Z. Xie

@xenonhexafluoro said:
How did you set the game up?

I treated it as an in/out game with subcategories (days). For each day you have 3 slots in, 2 slots out.

Timed & upfront, the only inference I made was that L must be in on Saturday, because P is already occupying 1 of the 2 out slots, and kicking L out would kick out M too, putting 3 players in the out group.

You also have a total of 9 in slots, with 5 players, each of which can be in a maximum of 2 times (b/c none of them can be in all 3 days). So, 1 player goes in once, and the rest go in twice. If M goes in twice, then L must go in twice, with M. This distribution inference helped me out on a couple of questions (20/23).

Thanks for sharing about the In/Out method. I will experiment it now.

Adam Z. Xie

@"Michael.Cinco" said:
There are a couple of infrences you can make in this game.

Namely. There is a piece that MUST got on Saturday

Also if you think about distribution you soon see that this game must have a distribution by where 4 of the pieces are used twice and one is used once.

It's wide open but because of the amount of open spaces and the small number of pieces relative to the spaces you are actually quite restricted on how you distribute the pieces.

Thanks for reply. Could you be more specific, please? Which piece must go to Saturday and why?

Edited: Ok, I got it now, with the In/Out method.

Michael.Cinco

P cannot go in Saturday. Leaving you with only 4 pieces for 3 spots. If you remove L from Saturday you also remove M leaving you with 2 pieces to fill 3 spots. So L must go on Saturday.

The distribution rule plus how restricted some of the other rules are helps you get to the right answer choices on a few of these questions relatively quickly

xenonhexafluoro

@"Adam Z. Xie" said:

@xenonhexafluoro said:
How did you set the game up?

I treated it as an in/out game with subcategories (days). For each day you have 3 slots in, 2 slots out.

Timed & upfront, the only inference I made was that L must be in on Saturday, because P is already occupying 1 of the 2 out slots, and kicking L out would kick out M too, putting 3 players in the out group.

You also have a total of 9 in slots, with 5 players, each of which can be in a maximum of 2 times (b/c none of them can be in all 3 days). So, 1 player goes in once, and the rest go in twice. If M goes in twice, then L must go in twice, with M. This distribution inference helped me out on a couple of questions (20/23).

I set the game as a standard 3x3 grouping gameboard. I understand the distribution mechanisms that at least four pieces need to repeat once - but that is not enough because every piece can repeat in theory. With respect, I disagree with respect to the inference about L.

"On any day that Morse works, Lentz also works." - only means that M -----> ML. It does not mean /L ------> /M. L can exist without M. For example, slot 1 (Thursday) can have LPQ and does not violate any rules.

Thanks for sharing about the In/Out method. I will experiment it now.

I'd diagram "On any day that Morse works, Lentz also works" as M ------> L. /L ------> /M does not mean that L needs M. It's the valid contrapositive of M ------> L. If you were to diagram L needs M, that would be L ------> M, not /L ------> /M.

If you're feeling shaky on your conditional logic, I'd return to in CC. Also, check out the question bank and filter for logic games that are "In/Out - Subcategories" to practice.

Lucas Carter

L must go on Saturday because if it does not, M does not either. That would force the remaining pieces into Saturday, violating the rule that P does not go in Saturday. Additionally, since we know L cannot go into all 3 days, it must be out at least 1 day. This means M and L must be out on one day so there will be at least one day where it is PNQ. However, L cannot be out 2 days because this would mean Thursday and Friday would be PNQ. That would leave ML on Saturday with nobody else since no piece can be in all 3 days. Therefore we can make the inference that L must be in on Saturday and exactly 1 of Thursday and Friday, with the slot it is not in containing PNQ.

Adam Z. Xie

@"Lucas Carter" said:
L must go on Saturday because if it does not, M does not either. That would force the remaining pieces into Saturday, violating the rule that P does not go in Saturday. Additionally, since we know L cannot go into all 3 days, it must be out at least 1 day. This means M and L must be out on one day so there will be at least one day where it is PNQ. However, L cannot be out 2 days because this would mean Thursday and Friday would be PNQ. That would leave ML on Saturday with nobody else since no piece can be in all 3 days. Therefore we can make the inference that L must be in on Saturday and exactly 1 of Thursday and Friday, with the slot it is not in containing PNQ.

Fabulous explanation, thank you!

BlindReviewer

@"Michael.Cinco" brings up a huge inference that I didn't make until after the exam, that because you need 9 items but you only have 10 pieces (5 people and each can only go a max of two days), when you have a situation where one person only works one day, then everyone else must be working two days (for the last question you can then immediately say that P must go Thurs/Fri).

That inference, plus what @"Lucas Carter" mentions regarding L and how if you knock out L you also knock out M, are the two inferences key to solving this under time. Unfortunately I also just kind of stared at it and then tried to brute force rather than really step back haha

Lucas Carter

@BlindReviewer said:
@"Michael.Cinco" brings up a huge inference that I didn't make until after the exam, that because you need 9 items but you only have 10 pieces (5 people and each can only go a max of two days), when you have a situation where one person only works one day, then everyone else must be working two days (for the last question you can then immediately say that P must go Thurs/Fri).

That inference, plus what @"Lucas Carter" mentions regarding L and how if you knock out L you also knock out M, are the two inferences key to solving this under time. Unfortunately I also just kind of stared at it and then tried to brute force rather than really step back haha

Don't feel bad I think everyone relied on brute force and felt pretty paralyzed by this one. #20 and #23 are absolute killers/massive time sinks.

N3rve_333-1

Game 1 - 3.8 minutes
Game 2 - 3.7 minutes
Game 3 - 10.7 minutes
Game 4 - 16.6 minutes

I knew I was going to see ownage-tier games from LSAT writers later in the section after I finished the the first two quick lol. Made a stupid mistake in Game 3 which added 2-3 minutes to it, but damn, Game 4 had me sweating. I made the "L must be on 6" inference, and knew that M could not be on both 4 and 5, but didn't push these far enough to get to the PNQ "combo" - fuck me. This led me to having to draw out possible scenarios and cross out the ones that violated rules. All in all, could have been an easier time had I pushed myself to spend more time trying to connect the rules.

4.4m for 20 and 5.5m for 23 alone. Properly salty lol.

Adam Z. Xie

@N3rve_333 said:
Game 1 - 3.8 minutes
Game 2 - 3.7 minutes
Game 3 - 10.7 minutes
Game 4 - 16.6 minutes

I knew I was going to see ownage-tier games from LSAT writers later in the section after I finished the the first two quick lol. Made a stupid mistake in Game 3 which added 2-3 minutes to it, but damn, Game 4 had me sweating. I made the "L must be on 6" inference, and knew that M could not be on both 4 and 5, but didn't push these far enough to get to the PNQ "combo" - fuck me. This led me to having to draw out possible scenarios and cross out the ones that violated rules. All in all, could have been an easier time had I pushed myself to spend more time trying to connect the rules.

4.4m for 20 and 5.5m for 23 alone. Properly salty lol.

This game is a In/Out game disguised as a Grouping game. I have noticed that our fellow 7Sagers who tackle this game as In/Out game have found this game super easy.

I only spent 5 minutes to solve it in the second time after reading xenonhexafluoro’s advice.

The Master game board will have N in on Friday, L in and P out on Saturday. With new rules appear in each question, the game board fills up very fast.