#16 - Global, Must Be True, Maximum

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13729)

The correct answer choice is (B)

Note that Maximum questions are automatically Must Be True questions (because the maximum number is an absolute). Because there are only five cities, the maximum number of theoretical connections is ten (the first city connects to the other four, the next city connects to the remaining three, the following city connects to the last two, and the second to last city connects to the last city; 4 + 3 + 2 + 1 = 10).

• However, the rules limit the number of actual connections:
  
  
  The first rule eliminates three possible connections.
  The second rule eliminates one connection.
  The fourth rule eliminates one connection.

Thus, the rules eliminate five of the possible ten connections, leaving only five actual connections that could be made. Answer choice (B) is correct.

[xishao3]

Hi PowerScore,

For Problem 16 of Game #10, it mentions in the explanation that Maximum questions are Must Be True questions and so, for that specific game there are ten possible theoretical connections [5 variables x grouping of pairs]. I wanted to ask if this method of taking the number of variables x the grouping relationship presented -- is applicable to any grouping game that asks a Maximum question.

Many Thanks,
Amy

[Eric Ockert]

Hi Amy!

Thanks for the question. No, that approach would not be generally applicable to maximum questions. This is really just unique to this particular game involving connections between variables.

And here, we aren't so much taking 5 x 2, as we are adding 4 + 3 + 2 + 1 = 10. This is because if you take any one variable, there are 4 other variables to connect with. Once you move to the next variable, there are also 4 connections, but one of those was already counted in the first number, so there are only 3 additional connections.

To illustrate, imagine we have 5 variables A, B, C, D, E. How many connections do we have? If we start with A, we have four connections:

AB
AC
AD
AE

Moving to B, we also have four connections, but one of them is BA, so that has already been counted in the first list. So, the remaining three connections would be:

BC
BD
BE

Same thing happens with C. CA and CB have already been counted, so now we have two additional connections:

CD
CE

And with D, we already have DA, DB, & DC. So all we have left is one connection:

DE

Hope that helps!

[nutcracker]

Hello! I would appreciate some insight to this question. It took me quite a while to try out different possibilities and I wasn't sure that I got the right answer in the end (although I did). Is there a more systematic way to deal with this maximum question? Where should I start? Thank you very much!

[James Finch]

Hi Nutcracker,

In this game, the maximum connections are limited by the rules and the necessity of using one of two grouping chains formed out of them:

M-H

H-T

H T, T H

H-P P-T P-V

and

P-V P-T P-H

This allows for two templates to be created (one of which will be used for the next question):

P-V, M-V/T/P, H-V, H-T, T-V

and

H-P, H-V, M-V/T/P, P-T, P-V

both of which only allow for a max of 5 connections.

These templates aren't necessary to draw out until question 16, but are still useful to set up on the front end when creating a master diagram, as it makes the preceding questions easier to answer correctly. Whenever there are two clear distinct possibilities set up by the rules, or even four, it's a good idea to draw out the templates, and, if time allows, test min/max when the numbers in groups are undefined.

Hope this helps!

[Blueballoon5%]

Hi! I have a question (not related to the OP's question). I am having difficulties understanding what the question is asking. Is it asking for the maximum number of pairs possible, or or the maximum number of pairs at one given time?

I answered this question by following the first option. First, I wrote down all the pairs:

HM
HP
HT
HV
MP
MT
MV
PT
PV
TV

I then eliminated HM because of the third rule (if HM existed, then there would have to be TM. But this is impossible because M only connects with one other city).

I then eliminated HT because of the second rule.

I ended up with this:

HM
HP
HT
HV
MP
MT
MV
PT
PV
TV

So, I selected answer choice E. However, the correct answer is answer choice B.

After reading the explanation in the back of the homework section, I tried the second option (of the max number of possible pairs at one given time).

Like before, I wrote down all the possible pairs (without considering the rules)

HM
HP
HT
HV
MP
MT
MV
PT
PV
TV

I eliminated HM and HT, like before. Then I eliminated, two of the remaining "M" pairs. I eliminated PV because of the fourth rule. I ended up with these remaining:

HM
HP
HT
HV
MP
MT
MV
PT
PV
TV

This seems to be the right answer, corresponding with the correct answer choice B. However, I am still confused how the question was asking for this method rather than what I had originally thought?

Most of these kinds of questions (maximum/minimum) questions seems to take on the approach of the second method (the max at one given time). If I have trouble understanding the question, should I try to just think that they are asking at one given time? Does the LSAT ever ask the first method (all possible pairs, at any given time)?

[Adam Tyson]

Phew, that was a lot of work for just one question, blueballoon! I would hope that your original setup for this game would have saved you a lot of that time and trouble, maybe with a couple hypothetical setups. Otherwise, to attack this question in this way, with so much labor suggests that you are doing each question in isolation and not using the work on prior questions to inform the answers on later ones, and that means a lot of wasted time and effort.

The question is asking about a single solution to the game, not about every possible pairing that could ever occur. The evidence for that interpretation is in the language of the question - "at most how many pairs" indicates a singular situation, one which maximizes the results. Asking "what's the most you can have" means "at one time". If they wanted to ask about all the possible combinations, they would have instead said something like "how many different pairs of cities could be connected?" or "the cities could be connected to each other in how many different pairs?", where "at most" isn't used as a limiter. Think about the flip side, where they might have asked "what the least number of pairs of cities that could be connected?" and you would have to interpret that as being "in a given solution".

Sometimes we get questions that ask about what could be true in a given solution, and other times we get asked broader questions about all possible solutions. They have many ways of posing those questions, but the use of the language will leave no room for subjectivity. Interpret in the way that makes the most sense within the context, and you should have no trouble. Good luck!

[g_lawyered]

Hi P.S.,
I have a question from James Finch explanation posted on 1/25/2018. I'm not sure if I'm reading at typo or not, but based on the limited solution set he posted I only see a max of 4 connections. I don't see 5 connections like answer choice B states.
In his explanation, on the 2nd Limited Solution set I see he posted: P/V and P/T, I'm sure that has to be a typo since that breaks rule 4. I think the solution set would be: P/T and V/H instead.
Can someone please clarify? Why is there max of 5 connections?
Thanks in advance!

[Robert Carroll]

G,

I think I have two things to offer: first, look at the first post of this thread. I've explained this question in private sessions before in a much more complicated way, so I'm glad the first post simplifies!

The second thing I have to offer is a specific set of connections that proves that 5 connections are possible:

PM
PH
PT
VT
VH

The first post shows why more than 5 connections are just not possible, given the rules; my hypothetical shows why 5 connections are possible.

Robert Carroll