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Setup and Rule Diagrams

Jon Denning
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Finally, some Grouping.

The fourth game on the June 2017 LSAT is a standard Grouping Game where four product managers—F, G, H, and I—visit three cities—M, S, and T. We're then told that each manager visits at least one city (we need to use each of F G H I) and each city is visited by two people.

So we'll end up with the three cities, M S T, as the base, with two stacks above each letter to represent that city's two visitors. The numbers here don't quite match up, and that deserves our attention straight away: in those six total spaces we need to use all four managers at least once—so that's four spaces filled—and then figure out how to fill the remaining two "empty" spots.

We have two distribution possibilities at this point then:

..... 3-1-1-1 where one manager visits all three cities, and the other three managers visit one city apiece

..... 2-2-1-1- where two managers visit two cities, and the other two managers each visit one city

Let's see if the rules can help us determine which distribution is at work (note: it's possible at this point that both distributions could be used for this game; narrowing it down to just one though, if we can, would be nice).

Here are the rules:

..... I visits exactly 2 cities. Well then. Problem solved. It's the 2-2-1-1 distribution of people to cities, with I as a 2.
..... Knowing how important numbers tend to be for Grouping I'd keep a close eye on the other three variables to
..... see if I can figure out who else is a double and who the two singles are (whether it's constant, like I, or question
..... to question). And always remember the overall distribution as you go!

..... F and H don't visit the same city. We could show this as a vertical Not Block (probably my choice), or with
..... a Double Not arrow: F :dblline: H

..... GM :arrow: HT , and the contrapositive HT :arrow: GM (I showed the cities as subscripts, but it's
..... entirely up to you—do whatever you want as long as you have the order correct)

..... Lastly, a Not Law for G under the S column.

So what inferences can we draw from this?

In games with variable sets this small (four people, three cities, two spots on each city...) it's extremely common to find inferences anywhere you can begin to eliminate options. In short, when you have few options to begin with, and begin to reduce them even further, you often find yourself in situations where things MUST occur!

So the first thing I would do is look to my most restricted city, S. We know from the last rule that G cannot go there, so that leaves just F, H, and I available. But we also know that F and H cannot visit the same city as each other...that means at most they could take just one of the two S spots. What that means then is that I must visit S! G can't, F and H both can't...I must.

So one of I's two visits is known: S. The other S spot? Either F or H. That's absolutely worth showing:

..... __ .....F/H .....__
..... __ ..... I_ ..... __
..... M ..... S ..... T

Is there more that we could discover if we keep testing hypotheticals or toying with possibilities? Perhaps. But this is the stage where I'd forego all of that and eagerly move to the seven questions. It's also the final game of the test, so if this is the last game you're attempting (i.e. you've done the other three already), how much time you choose to spend setting it up can be in part determined by how much time you have remaining for the section. Assuming time is tight though, I'd move on.
Jon Denning
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Etsevdos
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Time permitting, thoughts on setting up 3 scenarios with G in M, G in T, and G in M and T. It would establish three templates and establish I in S.
Claire Horan
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Hi Etsevdos,

It would be fine to do templates, but I would advise just two templates: one with G in Manila and one with G in Tokyo. The reason is that this is an either/or. You know that if G is not in Manila then G must be in Tokyo. The purpose of templates is to show EVERY possible solution, so I would not make a third diagram with G in both places. The reason is that we don't know if G is used twice, and you are not going to draw a template where F is used twice and another where H is used twice. It would be better to just have the two templates I described, and leave it open which of G, H, or F is used twice.

I hope that is clear!