#12 - Global, Must Be True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8568)

The correct answer choice is (D)

This question asks you to identify the pair of variables from which at least one is always on the research team. Thus, the correct answer is a pair of variables that cannot both be eliminated. Historically, questions of this type have been difficult, and this question is no exception.

In a question of this type, the variables fall into two categories: problem variables that are unlikely to be part of the answer, and desirable variables that are more likely to be part of the answer. Let’s look at both groups.

• Problem variables: As discussed during the setup of the game, M and W both alienate three other employees. Thus, it is unlikely that either would be part of a pair where one is required to be in every solution. This makes answer choice (B) unlikely to be correct based on a surface scan of the answers.
  
  Desirable variables: In these questions, variables that serve as necessary conditions for the presence of other variables are highly valuable because their absence eliminates more than just the variable themselves. For example, if P is not part of the team, then S cannot be part of the team either. From this perspective, P, T, M, and Y are desirable. However, since M has some undesirable characteristics, it becomes less desirable than the others, and thus P, T, and Y are the most desirable variables, and thus the most likely to appear in the correct answer choice.

Having identified P, T, and Y as the variables most likely to be in the correct answer choice, scan the answer choices to see if one or more of those variables appears in an answer choice:

Answer choice (A) contains none of the desirable variables.

Answer choice (B) contains one of the undesirable variables.

Answer choice (C) contains one of the desirable variables.

Answer choice (D) contains two of the desirable variables.

Answer choice (E) contains one of the desirable variables.

From this scan of the answer choices, answer choice (D) appears the most likely to be correct. To test whether (D) is the right answer, simply attempt to remove both T and Y from the team:

If T is eliminated, then via the contrapositive of the second rule S is eliminated. If Y is eliminated, then via the contrapositive of the third rule W is eliminated. That eliminates four variables, leaving just M, O, P, and Z to form a viable team. However, the inclusion of M with O and P violates the first rule. Thus, both T and Y can never be eliminated from the team and at least one of the two must always be on the team. Consequently, answer choice (D) is the correct answer.

For each of the incorrect answer choices, a viable solution can be found where neither of the listed variables is on the team.

[emilyfoster2013]

I am still confused about number 12. I was able to figure out the answer by process of elimination. (I got rid of A and C because they both could be not selected if W was selected) and E did not seem like a likely variable because there were not a lot of rules about Y and Z. After that I tried out B and D and after a few tries D seemed like a more likely answer but I was still not 100% sure. I was wondering if there is a faster way for me to figure this problem out?

Thank you!

[Ron Gore]

Hi Emily,

The key to this question, and a driving force for the entire game, is that you can never have more than 4 variables in the out group (i.e., not selected). You always have to reserve at least one spot in the out group for the variables in the first rule, because either M is out or both O and P are out.

This understanding can help focus you in on the variable in the list that, if not selected, force out at least one other variable. Think of the all of the variables in the other answer choices, and what they force out.

Notice that only T and Y force out a variable other than one involved in the first rule. If you have T (with S) and Y (with W) out, then there isn't sufficient room in the out group to separate the variables from the first rule.

Understanding that you always have to provide room for at least one of the variables from the first rule in the out group (or, in other words, that there are at most 3 spaces available in the out group for variables other than M, O and P) will help you focus in on answer choice (D) without having to use process of elimination.

Hope that helps,

Ron

[emilyfoster2013]

Hi Ron!

Sorry I just can't seem to wrap my head around this question!

I now understand why D works but I am also getting confused about why answer choice B is not correct:

By the Contrapositive of the first rule: O or P NOT M
and by the contrapositive of the third rule: NOT M or NOT Y NOT W

I diagrammed out a chain
O NOT M NOT W

So wouldn't it follow that one of either O or W has to be selected/ not selected?

Thanks for the help Wish I had asked this question in class I did not realize how confused I was!

[Nikki Siclunov]

Hi Emily,

You're correct in that O and W cannot both be selected. However, that doesn't mean that at least one of them must be selected. On the contrary, there is nothing stopping us from eliminating both O and W. We'd be left with:

S, P/M, T, Y, Z

If we include S, we'll end up the following setup:

S, P, T, Y or Z (or both)

If we don't include S, we have even more combinations of variables to choose from. Therefore, it is possible that both O and W are out.

However, if we exclude both T and Y, we must also exclude S and W. We're left only with P, M, O, and Z. We cannot include all 4, because M cannot be included with either P or O.

Attached is my setup for this game... this is actually all I did before moving onto the questions. It may be a somewhat unconventional approach, but as long as you can see your chain relationships and keep in mind the need to select at least 4 variables, you're good to go:



Thanks!

[emilyfoster2013]

Got it, thanks so much

[Arindom]

Nikki,

Totally dig the diagram! Will try to incorporate this method going forward. One question though - here Z is the random. So, looking at this chain and if we have similar chains where T and Y, although part of an equation, are really by themselves like Z, can we always conclude that either T and Y will have to be in the In group? I spent quite sometime on this. I first chose ans. choice E till I came up with STPO and then got D! But I spent so much time.... I want to avoid that in case this is a shortcut?

Thanks.

- Arindom

[Nikki Siclunov]

Arindom,

T and Y function as necessary conditions for other variables to be included. This fact alone does not guarantee that T or Y must be included. Be careful!

If T and/or Y must be included, this would be for one of two reasons only:

1. Either S or W is included, triggering the conditional relationship that forces either T or Y, respectively, to be included.

2. Given that at least 4 variables must be included, it is possible that either T or Y will end up included regardless, in part because of the conditional relationships pertaining to the remaining variables in the game. For instance, here P M and M O. Given that S T and W Y, the exclusion of both T and Y would necessitate the exclusion of both S and W. We'd be left with only four variables - P, M, O, and Z - which cannot all be included because of M. Thus, either T or Y must be included.

The last inference results from 1) the numerical stipulation of min. 4; and 2) the conditional relationships that govern the remaining variables, particularly M. The inference is specific to this setup! My advice would be to extrapolate the method of reasoning, not the inference resulting from this method of reasoning, when attacking similar games going forward.

Good luck!

[Arindom]

Thanks, Nikki!

- Arindom

[PamelaO]

can someone explain what this question is asking for?