# December 2015 LSAT – Game #4 – Volunteers and Committees

This is the other game from the December LSAT that some people found difficult. This game was a defined grouping game that if set up correctly was not too difficult, but the way the game starts out it might lead you into a different set up.

First we have 5 volunteers H, J, K, M and N being assigned to 3 committees, X, Y, Z and then these volunteers will each hold a position on the committee L, S, T

The basic set up for the game is:

Not:

You do not want to put the positions held as a static variable, because then you cannot add a volunteer onto a committee unless you know their position. And it is possible you won’t know there position but you will know which committee they are on so you want to leave the option open to record that.

The rules for the game are:

If N —-> N
_______L

M = exactly 1 committee

K = Y and ~Z

J = Ys and ~X and ~Z

So when you plug the original deductions into your diagram it should look like this so far:

But now that you know J and K cannot be in Z that means the remaining 3 volunteers H, M and N have to be in Z. And since N is there, N has to be the leader.

And since M can be used only once, we now know there is no M in X and Y.

This means there is no J and no M on committee X, so that means N, H and K must be on committee X, with N being the leader again.

So that leaves either H or N being on committee Y, with most of the positions remaining unknown. However, just note that if we knew N was on committee Y, since N is always the leader we would be able to deduce that K would have to be T, but since H/N could be there we can’t make any further deductions.

Your final diagram should look like:

QUESTIONS:

18. For committee Z we know it is made up of N, M, and H, with N being the secretary. So quickly scan to see if an answer has those 3 and has N being the leader. Only choice E has this. So E is correct and then we can quickly move on.

19. If K is the T for two committees that means K is the T for committee X and committee Y. If you add that into your diagram and make your deductions you will see that H has to be the S for committee X since all the other positions are now filled, And that the remaining H/N in committee Y is the L. and then committee Z is unchanged.

This means that each of the following could be true except: C – because H cannot be the treasurer of X, as you can see from our diagram H must be the S for committee X

20. Must be true except means: we should be able to figure out the answer by looking at our original deductions. By looking at the answers this leaves us with B. H does not have to be assigning to Y. It is possible H is assigned to Y, but it is also possible that spot could be filled by N. That means this B does not have to be true.

21. If K is going to be the leader for one of the committees that committee has to be Y, because the only other committee that K is on is committee X, and N is already the leader for the committee. So if K is the leader for committee Y that means the last volunteer on the committee has to now become H, because if it was N then N would need to be the leader, but the leader position is now being filled by K.

Our diagram should now look like:

So the amount of positions is fully determined for B – only committee Y, as you can see from the diagram. We still do not know the positions of the other committees.

22. To know what type of answer we need to find for determining the positions of everyone on every committee we need to think strategically. First off, we don’t know if committee Y has H or N, so we would need an answer that would force one of them on committee Y. Also, if we knew one more persons position on each committee we could figure out the third position by process of elimination and solve the whole game. So we need to see which answer does that.

A) H is the leader for one committee. This forced H to be the leader of Y and then we would know K is the T for committee Y, but we still would have no idea of the positions of the other people on committee X and Z. So this answer is out.

B) H is the secretary for two committees, well H could be the S for X and Z and then we would have no idea still if H was on committee Y or if N was. So this answer is out.

C) H is the treasurer for all three committees. This means H is on committee Y and is the T, so then K becomes the L. And since H is the T on X, that means K now becomes the S on X. And since H is the T on Z, that means M is the S on Z, so we would know everyone’s role. This is the right answer! Your diagram would look like:

D) K is the T for two committees would give us the same diagram as #19, and as we can see we still had a lot unknown. So this answer is out.

E) Nash is the leader for all three. Means N is the L on Y as well so K would be the T, but we would not know the roles of the volunteers on the other committees still so then this answer is out too.

23. This is the substitute a rule question. It tells us that we need to find a new rule that could replace the rule stating that M is only assigned to one committee. So to figure out how to replace it we need to consider how this rule affected our original diagram. Since we figured out that M was on committee Z by default, because K and H couldn’t be there, we would need a rule that would force M not to be on committee X and Y.

A) H is on more committees than M. Well this would leave the possibility open that M could go on two committees since it is possible H is on all 3. This would change our game so A is out.

B) J must be on more committees than M. J can only be on one committee, so that would mean M would have to be on 0, which would change our original game as well, so B is out.

C) K must be on more committees than M. Well K must be on 2 and can only be on 2, since K cannot go on Z, so then that would force M to only go once or 0, but M cannot go 0 anyways because M gets forced onto committee Z, so this works!

D) M must be on more committees than H. This is out because H goes 2-3 times! and M can’t go more than that! M needs to go only once.

E) Nash must be assigned to more committees than M. N goes 2-3 times, so once again this would make it possible for M to go 1-2 times, which  would be too many! So this answer is out as well.

I hope that made the game seem a little easier for everyone!!!

# U.S. Law Schools – Can you get in?

I came across a useful tool for predicting whether you can get into certain U.S. Law Schools. You can check to see if your score is high enough to get into a certain school based on other people’s stats that they have posted. So if you are wondering if you have a good shot to get into certain US schools then you should definitely check it out!

People post their stats on: http://schools.lawschoolnumbers.com/ and you can also see your chance of getting in graphed on this affiliated site with exact percentage chance based on your stats: http://mylsn.info/

# 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 !!!

# The LSAT Trainer

I just wanted to write up a quick post to let everyone know about a new site at http://www.lsatters.com — Mike Kim the author of the LSAT Trainer books is starting a new website that will connect tutors with students, as well as, a forum for people studying for the LSAT to go on and participate in if they have any LSAT or law school related questions.

It will be a good source to connect future law students/LSAT takers. If you ever have any LSAT questions you can always e-mail me directly, but feel free to take advantage of his forum as well. I, along with other experienced LSAT tutors, will be answering any questions you may have on his site as well.

# Canadian Law Schools that Accept the February LSAT

Most law schools recommend writing the LSAT in December or earlier, but most Canadian law schools do accept the February LSAT. There is a rolling admissions at most law schools so your chance of acceptance may go down if you choose to write later. However, if you don’t have a high enough score already or you are scoring poorly on your prep tests then I would recommend waiting until the February sitting. Also, the more time you spend studying for the LSAT your score WILL go up! (As long as you know how to attack questions correctly)

Law Schools that accept the February LSAT:

1. University of Toronto
2. Osgoode
3. Queens
4. University of Western Ontario
5. University of Ottawa
7. University of British Columbia
8. University of Victoria
9. Thompson Rivers University
11. Mcgill
12. Dalhousie

Law Schools that accept December as their last LSAT:

1. University of Alberta
2. University of Calgary
3. Windsor
4. University of Manitoba

# Prep Test 68, Section 2 Question 10

This is a question about people who have allergies to cats. We are asked to find the statement that is most strongly supported from the information above. Which means we have an Inference Question.

Analysis:

(1) People are allergic to the proteins found in the skin and saliva of cats.

(2) The particular protein responsible for the allergies varies from person to person

(3) All cats can provoke an allergic reaction

(4) However, some cats can cause an allergy in one person, but not in another person who is also allergic to cats.

(A) We can’t say that every person is allergic to some breeds but not others. Some people may be allergic to all breeds for all we know.

(B) One cat could cause an allergic reaction in all allergy suffers. Nothing in the stimulus forbids that. Part (4) of the analysis doesn’t mean that all cats affect some but not all people.

(C) Correct! “Not all cats have identical proteins in their skin and saliva” — The reason this has to be true is — if all cats had identical  proteins then it wouldn’t matter which protein a given person was allergic too, because all cats have the same proteins (lets call them X, Y, Z). Then it wouldn’t matter if person A is allergic to X and person B is only allergic to Y, they would both be always allergic to the same/all cats. This would violate the rule set out in Analysis (4).

Therefore, if cats have different proteins then this could explain why a person may be allergic to Cat C but not allergic to Cat D.

(D) We do not know anything about the intensity of the people with allergies…

(E) May sound tempting… but the stimulus tells us nothing about how hard or easy it may be to predict which cats will cause an allergic reaction in which person.

# Question of the Day! PrepTest 55, S.1, Q.15

A lot of people studying for the LSAT have troubles with inference questions. One question that was giving some of my students trouble is from Prep Test 55, Section 1, Question 15.

This question is about Zach’s Coffee Shop. (I cannot legally post the question unless I give licensing fees to LSAC). However, I am going to explain why each answer choice is wrong or right.

Analysis of the Question:  Zach’s coffeeshop has poetry readings almost every Wednesday. And then the question tells us that Every day that holds a poetry reading has 1/2 priced coffee.

We are looking for an inference – which is a statement that must be true based on a sentence or combination of sentences.

Therefore, something that MUST be true based on the stimulus is that – on the Wednesday’s with a poetry reading, the coffeeshop must offer 1/2 priced coffee. Let’s take a look at the answer choices to see if something matches our prediction.

(A) The stimulus does not tell us that Wednesday is the most common day to offer 1/2 priced coffee, since the stimulus does not tell us that they ONLY offer 1/2 priced coffee when there are free poetry readings. For all we know the coffeeshop offers 1/2 priced coffee everyday.

(B) Like A, the stimulus does not tell us that the coffeeshop only has poetry readings on Wednesday’s. Perhaps they have free poetry readings every single Friday.

(C) We know that: if there is a free poetry reading —> then there is 1/2 priced coffee. But, we cannot assume the reverse. We don’t know that every time there is 1/2 priced coffee there is also a free poetry reading. Be careful not to mix up necessary and sufficient conditions!

(D) CORRECT – We know that: Almost every Wednesday –> free poetry reading —> 1/2 priced coffee! Which is what this question is saying. If you were confused by the “most, if not all” part of the question, just note that — they could still offer 1/2 priced coffee on the Wednesday’s without poetry readings. They don’t have to have 1/2 priced coffee on those days, but they could offer it. That is why it is possible that they could offer 1/2 priced coffee on all Wednesdays. (“Almost every” means 51-99%, but since the 1% of days could still have 1/2 priced coffee for all we know – this answer choice is correct)

(E) Since we now understand why the coffeeshop could offer 1/2 priced coffee every Wednesday. This answer choice must be false, since we do not need to have a day without 1/2 priced coffee. (and “some” means – at least one)

# Formal Logic – Necessary & Sufficient?

One of the major concepts to understand on the LSAT is the use of formal logic. Formal logic is found predominantly in the Analytical Reasoning section of the LSAT, but it is also found in the Logical Reasoning section in inference questions and assumption questions.

Formal logic is founded on the understanding of necessary and sufficient assumptions. A necessary assumption is something that MUST be true, whereas a sufficient assumption is something that COULD be true. Formal logic is the use of rules to make deductions.

In the Analytical Reasoning section of the LSAT you want to diagram formal logic and necessary/sufficient problems. For If—then statements you want to diagram them as:

If A then B:         A —–> B

The left side of the arrow (A) is the sufficient condition, and then right side of the arrow is the necessary condition (B). Thus, it is possible for A to happen (although it doesn’t have to happen), but if A does happen then B MUST happen as well.

When you have a conditional statement you can also write the if—–then statement’s contra-positive. A contra-positive is just a further deduction that can be made from a conditional statement. To do this you will negate both terms and then flip them to the other side of the arrow.

If A then B:      ~B —-> ~A

Other Examples:

1) If not X then Y:      ~X —–> Y     OR     ~Y —–> X

2) if not S then not T:    ~S —-> ~T    OR    T —-> S

Another common type of formal logic seen on the LSAT is the “only if” statement.  For example, A only if B. Only if means that if A does happen then B must also happen. That means that:

A only if B can be rewritten as: if A then B:     A—–> B

Once again the contra-positive would be ~B —-> ~A

Another common type of formal logic seen on the LSAT is the “if and only if” and “if but only if” statements. “If and only If” is a bi-conditional logical connective between statements. This means that the truth of one of these statements requires the reverse to also be true. For example:

A if but only if B can be written as: if A then B AND if B then A:      A <—–> B

The contra-positive would be: ~A <—–> ~B

Also, the phrase “if and only if” means the same thing as “if but only if” in terms of formal logic and would be written the same way as the above example.

***”~” means NOT ***