PT 60 third Game landscaper

 Posts: 34
 Joined: Thu Oct 07, 2010 9:34 pm
PT 60 third Game landscaper
I had no concrete method on this game. I just wrote out as much possibilities as I could think of. Is there a better way to approach the game?
 ocean
 Posts: 27
 Joined: Wed May 05, 2010 1:58 pm
Re: PT 60 third Game landscaper
shemori wrote:I had no concrete method on this game. I just wrote out as much possibilities as I could think of. Is there a better way to approach the game?
No, that's exactly the way to do it.
 rso11
 Posts: 126
 Joined: Fri Sep 10, 2010 9:25 pm
Re: PT 60 third Game landscaper
That's what I did to. It sucks.
 incompetentia
 Posts: 2307
 Joined: Thu Sep 30, 2010 2:57 pm
Re: PT 60 third Game landscaper
I can see why this game was difficult for people, but you don't have to do just possibilities.
7 periods of time...3 have to be mulch. If we think about this NOT as filling in 7 blanks, but rather 3 digits where we mulch instead of stone, I think it's much easier to think about. It's much easier to think about and refer to a case as "145" rather than MSSMMSS. Note that every case we have is going to be a "5xx" case, that is, 5 will always be mulch.
If we look at inferences we can make about flipping
I. At least two of those datapoints have to be consecutive. Why? If not, you'd have a minimum of 4 switches. Cases like 135 and 257 prove this.
II. If all 3 are consecutive (345, 456, or 567), the case is valid. Why? Maximum of 2 switches.
III. If two are consecutive, but not all three (such as 235, 156, 457), one or both of the separate groups has to include an endpoint (1 or 7). Why? This one is the trickiest, but will help the most. If you think about flipping a switch 3 times, if you start with it on, you will have two periods of it being on, but it'll have to end with it off. MMssMss (125), for example: 3 cleanings, starts with M and ends with S. On the other hand, sMssMMs (256) has 4 cleanings, since M is neither on the beginning or the ending. (There are no cases where BOTH endpoints are taken, but mulch is still on 5.)
Note that with those three principles, we've covered all possible configurations of M already.
Q13.
A fits into category I, so that's definitely FALSE.
B, C, and D fit into category II, so we can't rule in either direction.
E fits into category III, so that's definitely TRUE. You don't even have to look at the other three.
Q14.
D is pretty much the reverse argument of Category I, where we assume:
If 3 Separate > Not True
Here we have:
True > Not 3 Separate
Q15.
3rd load mulch means that the case is going to be "35x"  which means that category III is automatically out. (You can't have something consecutive with either 3 or 5, AND have it be either 1 or 7.)
That means category II is the only option, and the only possible way to get 3 consecutive is the 345 case.
Run through the choices, and E is the only answer which fits the 345 case.
Q16
Exactly two cleanings once again rules out category III, in this case because having mulch on 5 always means category III will require 3 cleanings:
Two separate groups of mulch with stone in between only will have 2 cleanings when all 4 stone are in between.
However, you can't have mulch on 5 when there are 4 stone in a row unless they are 14. That's the 567 case, which is category II to begin with.
So if we know that the case is either 345, 456, or 567, we know for a fact that A is true in all three cases.
Q17.
Without the possibility of 3 loads consecutively, category II is out. This means we'll have a group of 1 mulch, and a group of 2 mulch.
If you want to calculate all the possibilities for category II, we have 145, 156, 125, and 457. However, 156 and 457 have 3 stone in a row, so those are out.
We're left with 145 and 125 as our cases: this leaves B as the only thing that's not 100% false.
If this makes no sense, lte me know and I'll try to revise what I'm saying. Sometimes I kind of suck at explaining my thinking with LGs...
7 periods of time...3 have to be mulch. If we think about this NOT as filling in 7 blanks, but rather 3 digits where we mulch instead of stone, I think it's much easier to think about. It's much easier to think about and refer to a case as "145" rather than MSSMMSS. Note that every case we have is going to be a "5xx" case, that is, 5 will always be mulch.
If we look at inferences we can make about flipping
I. At least two of those datapoints have to be consecutive. Why? If not, you'd have a minimum of 4 switches. Cases like 135 and 257 prove this.
II. If all 3 are consecutive (345, 456, or 567), the case is valid. Why? Maximum of 2 switches.
III. If two are consecutive, but not all three (such as 235, 156, 457), one or both of the separate groups has to include an endpoint (1 or 7). Why? This one is the trickiest, but will help the most. If you think about flipping a switch 3 times, if you start with it on, you will have two periods of it being on, but it'll have to end with it off. MMssMss (125), for example: 3 cleanings, starts with M and ends with S. On the other hand, sMssMMs (256) has 4 cleanings, since M is neither on the beginning or the ending. (There are no cases where BOTH endpoints are taken, but mulch is still on 5.)
Note that with those three principles, we've covered all possible configurations of M already.
Q13.
A fits into category I, so that's definitely FALSE.
B, C, and D fit into category II, so we can't rule in either direction.
E fits into category III, so that's definitely TRUE. You don't even have to look at the other three.
Q14.
D is pretty much the reverse argument of Category I, where we assume:
If 3 Separate > Not True
Here we have:
True > Not 3 Separate
Q15.
3rd load mulch means that the case is going to be "35x"  which means that category III is automatically out. (You can't have something consecutive with either 3 or 5, AND have it be either 1 or 7.)
That means category II is the only option, and the only possible way to get 3 consecutive is the 345 case.
Run through the choices, and E is the only answer which fits the 345 case.
Q16
Exactly two cleanings once again rules out category III, in this case because having mulch on 5 always means category III will require 3 cleanings:
Two separate groups of mulch with stone in between only will have 2 cleanings when all 4 stone are in between.
However, you can't have mulch on 5 when there are 4 stone in a row unless they are 14. That's the 567 case, which is category II to begin with.
So if we know that the case is either 345, 456, or 567, we know for a fact that A is true in all three cases.
Q17.
Without the possibility of 3 loads consecutively, category II is out. This means we'll have a group of 1 mulch, and a group of 2 mulch.
If you want to calculate all the possibilities for category II, we have 145, 156, 125, and 457. However, 156 and 457 have 3 stone in a row, so those are out.
We're left with 145 and 125 as our cases: this leaves B as the only thing that's not 100% false.
If this makes no sense, lte me know and I'll try to revise what I'm saying. Sometimes I kind of suck at explaining my thinking with LGs...
 gdane
 Posts: 11910
 Joined: Sat Sep 26, 2009 2:41 pm
Re: PT 60 third Game landscaper
This game took me 6 minutes and all I did was plot out all the possibilities. There arent that many. It was much easier than people on here made it out to be. Now the 4th game with the Interns was a little tougher...
 rso11
 Posts: 126
 Joined: Fri Sep 10, 2010 9:25 pm
Re: PT 60 third Game landscaper
I actually thought the intern game was easy. Whatever, I guess it just depends.
 LSAT Blog
 Posts: 1262
 Joined: Mon Dec 07, 2009 9:24 pm
Re: PT 60 third Game landscaper
Here is a complete explanation of the setup and all questions for PT60, G3 (mulch/stones). Explanations for other LGs in that exam are linked within that post.

 Posts: 34
 Joined: Thu Oct 07, 2010 9:34 pm
Re: PT 60 third Game landscaper
I didn't think it was extremely difficult just thought there might be a more efficient way to approach...The long explanation above is a little too convolute for me. I would think of contrapositives unless i were doing an in/out game. I've seen the explanation on the LSAT BLOG and it amounts to just writing out the possible outcomes.
Interns Game wasnt too bad for me.
Interns Game wasnt too bad for me.

 Posts: 767
 Joined: Tue May 11, 2010 2:17 am
Re: PT 60 third Game landscaper
Wow there were like literally two possibilities for this, a lot of over thinking going on here.
 KevinP
 Posts: 1324
 Joined: Sat Sep 26, 2009 8:56 pm
Re: PT 60 third Game landscaper
I did the mulch game in about 34 minutes. Which was very nice because it gave me the extra time I needed to finish the other games as I stalled on them too long.
As someone above said, the intern game was much harder for me but I generally do 01 LG anyway.
As someone above said, the intern game was much harder for me but I generally do 01 LG anyway.
 Lasers
 Posts: 1576
 Joined: Sat Jul 10, 2010 6:46 pm
Re: PT 60 third Game landscaper
i thought this game was easy and the 4th game hard.
in the 4th game, i just couldn't see the big inference. this game really was just about trying stuff out.
in the 4th game, i just couldn't see the big inference. this game really was just about trying stuff out.

 Posts: 2526
 Joined: Sat Jul 18, 2009 12:12 am
Re: PT 60 third Game landscaper
the 4th game had 1 maybe 2 inferences you HAD to get before you started or it made it a lot harder than it really was (GL block and F/K) I had 56 mins left for the game and completely was baffled by the fact that 6 interns had to be either P or Ws... lol
 Dany
 Posts: 11580
 Joined: Mon Sep 28, 2009 3:00 pm
Re: PT 60 third Game landscaper
incompetentia wrote:I can see why this game was difficult for people, but you don't have to do just possibilities.
7 periods of time...3 have to be mulch. If we think about this NOT as filling in 7 blanks, but rather 3 digits where we mulch instead of stone, I think it's much easier to think about. It's much easier to think about and refer to a case as "145" rather than MSSMMSS. Note that every case we have is going to be a "5xx" case, that is, 5 will always be mulch.
If we look at inferences we can make about flipping
I. At least two of those datapoints have to be consecutive. Why? If not, you'd have a minimum of 4 switches. Cases like 135 and 257 prove this.
II. If all 3 are consecutive (345, 456, or 567), the case is valid. Why? Maximum of 2 switches.
III. If two are consecutive, but not all three (such as 235, 156, 457), one or both of the separate groups has to include an endpoint (1 or 7). Why? This one is the trickiest, but will help the most. If you think about flipping a switch 3 times, if you start with it on, you will have two periods of it being on, but it'll have to end with it off. MMssMss (125), for example: 3 cleanings, starts with M and ends with S. On the other hand, sMssMMs (256) has 4 cleanings, since M is neither on the beginning or the ending. (There are no cases where BOTH endpoints are taken, but mulch is still on 5.)
Note that with those three principles, we've covered all possible configurations of M already.
If there's a way to make this game difficult, this is probably it.
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