# LSAT December 2015 – Logic Game #3 – Employees & Offices

The LSAT results were released today and I’ve had a chance to go over the Logic Game section of the test myself already. None of the games were too different then the typical games seen on the LSAT other than Game #3. Game #1 was an ordering game (or linear game) and same thing with game number #2, except it was an over-funded game so there was a “out” section as well. Game #4 was a straight forward defined 3 group game.

That brings us to game #3. This game was a Pattern game. Pattern games are more rare on the LSAT but they do exist and they have appeared on two recent LSAT tests now, because they also appeared on the June 2014 LSAT. These games at first glance appear to take on the form of a linear or grouping game but then you will quickly realize that your typical game setup won’t work. I always go through pattern games with my clients because I do not want them to have the shock of seeing a game they think has never existed before. Just like mapping games, they can happen. And the LSAT seems to be bringing back a lot more rare question types on recent tests (not only in the game section – like point of agreement or evaluate the argument on Logical Reasoning).

Game #3 appears daunting at first but if you do not get trapped into a specific diagram the game is quite manageable. You just need to see the pattern. If they have to choose their most desirable office as their first choice there are only so many options. For example if J goes first he would choose Y, then if L goes next she would choose X, then if P goes he’d choose Z, and then that means T would choose W. That was an arbitrary ordering but there are only so many options present and it’s just checking if an option is possible.

Question 13:

A) is incorrect because for J to choose W that means J = 4th choice, but if L chose Y then L= 4th choice. They can’t both choose their last choice. That would be impossible.

B) Is the right answer: if J chose Z then J =3rd, if L chose X then L=1, if P chose W then P=4, and if T chose Y then T=2

This works! Lx = 1, Ty = 2 (and T would have to choose Y since their first choice was gone) then Jz=3 (since Y and X are gone J would have to choose Z) and then that leaves Pw=4 (and since all the other options are gone P must choose their last choice)

Question 14:

This answer is C and is a “gimme” since at least one of the employees must select the office they rank first — it would be impossible to not do this if they have to choose the office that they rank the highest that isn’t already chosen.

Question 15:

It would be impossible for C, D and E since 3 employees can’t choose any option all at the same rank as each other based on the order of preference. For 3 to select their 2nd choice they would all need the same 1st choice and all have a different 2nd choice – which isn’t the case. 3 can’t choose their 3rd or 4th choice because someone must chose their first choice and their first again or at least 2nd choice!

And you can eliminate B as well since 2 can’t chose their 4th choice, because you must chose your highest ranking option still left, and you can’t skip option 3, for example.

This leave A. And this works if PY=1, Tx = 2, Jz=3, Lw= 4

Question 16:

You can eliminate C quickly because it is impossible for 3 to choose their first choice since the only first choice options are X and Y for everyone.

You can also eliminate D since if Pw, then Pw=4, and then Jx means J=2, but for J to go second someone must choose Y first, and the only person who could do that is P but P must choose W so this answer is gone.

Answer A is wrong since two can’t choose their second option  — P is going 4th, so then if J=1, then Jy, L or T next would get choice 1 of X. If L goes 1 then L gets X (so then J can’t go next for sure or they get option 1) so then it would be TY but that makes Jz and Z is 3rd choice for J. So our last hope would be T first with X, then L with Z then J gets option 1 of Y, so then choice doesn’t work

Option B is wrong because 2 can’t get their 3rd choice, max one can get their 3rd choice. Even looking at work from A.

BUT I would have tested E before A and B anyways because it is easier to test. And this is the correct answer.  if Lz, then someone must just choose X before L’s turn so the order could go: Tx=1, Jy=2, Lz=3, Pw=4

Question 17:

You can eliminate C and D right away based on past work.

The answer is E, because for P to be able to choose X then that means Y and Z must have been chosen before P’s turn. I would have started with this answer choice because X is rated first or second by everyone else so it is highly likely it would be chosen before someone could choose it 3rd. So if Jy=1, then L or T would sweep up X, so that doesn’t work. And everyone else would choose X first so then it would be impossible for P to get it 3rd.

This game was probably a lot more time consuming then the other games on the test, but if you didn’t panic and realized these games do exist and if you practiced them a few times beforehand then this game on test day probably went okay. Make sure you always leave yourself enough time to study and that you don’t skip over “rare” LSAT question types. You don’t want to leave yourself unprepared on test day for one of the most important tests of your legal career. Hope everyone’s test went well !!!