#19 - Global, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=16041)

The correct answer choice is (C)

This is a tricky question as well, and the question stem asks for you to identify the pair of variables where at least one of the two is always on the committee. Thus, if a scenario can be found where neither variable is on the committee, that answer choice is incorrect. In questions such as this, prior work is invaluable, so make sure to check the work you produced in solving the other questions in the game.

The hypothetical produced by the correct answer to question #12—J, K, M, Q, S—eliminates answer choice (A) because neither F nor P appears in the hypothetical. Other work in the game can be used to attack some of the other answers (#15 for answer choice (B), for example), but because many of those questions produced only partial hypotheticals, this can be challenging.

Let’s take a different attack, and look at each answer choice from a more abstract standpoint, one that assesses the conditional value of each variable:


Answer choice (A): contains two sufficient conditions

Answer choice (B): contains two sufficient conditions (one is also a necessary condition)

Answer choice (C): contains two necessary conditions

Answer choice (D): contains two sufficient conditions (one is also a necessary condition)

Answer choice (E): contains two randoms

In a game such as this one, where there is an Overloaded set of variables for the group being formed, variables that are necessary conditions are extremely powerful because when they are removed, another variable is typically also removed. Thus, answer choice (C) looks the most promising, followed by (B) and (D), then answer choice (A). answer choice (E) is the least likely to be correct as randoms are not likely to have to appear in every viable committee.

Given that (C) is the most promising answer, let’s start our analysis there. Answer choice (C) contains two representatives who are necessary for other representatives to be selected. So, eliminating both K and Q would also eliminate F and G from the committee. That leaves only J, M, P, R, and S for the committee. However, from the sixth rule, M and P cannot be on the same committee, so removing K and Q does not leave enough representatives to form a viable committee. Thus, either K or Q must be on every viable committee, and answer choice (C) is correct.

[ifstudythenkill]

Can anyone the last Q of the set for me? I find 2 choices that seem correct.

"The committee must include atleast one representative from which of the following pairs?"

[BethRibet]

Hi,

The correct answer is C. The reason why C is correct is as follows:

If K & Q are both excluded, then both F & G are also excluded, because K & Q are necessary conditions for their inclusion. Without F, G & K, the only variables remaining from the tenants group are J & M. When M is included, P is excluded, so that only leaves R & S from the homeowners group, which makes a total committee of 4 people, and the minimum required is five.

Remaining answers:
A) If F & P are excluded, G, K, J & M are still available. GK, JM, or KJM can all be selected in combination, in compliance with the rules. There are also no obstacles to using Q, R & S, so any combination that adds up to 5 works.
B) If G & J are excluded, M is also excluded, but F and K remain available. F mandates inclusion of Q, and P, R & S are all available to take up the remaining 2 slots.
D) If M & P are excluded, J is also excluded, however, F, G, K, Q, R & S are available, and several combinations will comply with the results and add up to 5 (e.g. FGKQR)
E) If R & S are excluded, P & Q must be included to ensure that 2 homeowners are included, so M (M and P can not both be used), and therefore J are excluded (J and M go together). However, F, G, and K can make up the remaining 3 slots.

Hope this helps!
Beth

[OriginalJane]

Hello Nikki and/or David,

I'm having trouble solving Global, Must Be True and Cannot be True questions. I remember that one of the best ways to approaching these kinds of questions is to first look at negative grouping rules. I've tried to do that for q #19; however, it's not helping me see the important inferences. I'm not sure what I'm missing here. Can you suggest how to approach this question and those similar to it?

Thank you,

Amanda

[David Boyle]

Hello Nikki and/or David,

I'm having trouble solving Global, Must Be True and Cannot be True questions. I remember that one of the best ways to approaching these kinds of questions is to first look at negative grouping rules. I've tried to do that for q #19; however, it's not helping me see the important inferences. I'm not sure what I'm missing here. Can you suggest how to approach this question and those similar to it?

Thank you,

Amanda

Hello Amanda,

There may be no one perfect approach for questions like this; but one inference that people should make in this particular question is that j and p don't like either too much. This is because j likes m (and vice versa), but m doesn't like p, so, j must not like p either.

And this fits well into choosing the answers. First off, since you need 5 representatives, then as for "negative grouping", so to speak, answer C might jump out at people as a prima facie good answer. That is, if you don't have k or q (which are the "necessaries" for g and f respectively), then you can't have their "sufficient" counterparts, g and f. So you have only 5 left, i.e., j, m, p, r, and s.
But we know that j hates p! from the inference we made above, and of course that m hates p. So p has to drop out (or j and m do), so there's only 4 left, not enough to fill the spaces.

One lesson, then, for questions like this is to look for necessaries that are knocked out, as answer C does, since the sufficients get dragged away too, and other rules might then come into play to send things into a red zone, as with the "j and m hate p" thing mentioned above.

Hope this helps,
David

[ashpine17]

so i had to write out a board for C i couldn't just intuit from looking at it that it was the correct answer....is that ok? i also tried out B because it took out three pieces

[ashpine17]

i mean, i didn't just look at the variables and go bingo on c even though i knew it kicked out four variables because four pieces could be out. i had to go down the list?

[Adam Tyson]

A better approach may have been to refer back to your local diagrams from the earlier questions, because all of the wrong answers are proven wrong by that earlier work. You don't need to intuit the right answer! You can instead eliminate the wrong answers and then select whatever is left.

For example, the list question (Q12) gave us a solution of JKMQS. That answer proves that we do not need one of F or P, so answer A is eliminated.

And Q15 gave us a partial solution of FK as the tenants and Q as one of the homeowners. The other homeowners could have been any two out of PRS, and that proves that we do not need one of M or P, eliminating answer D. It also proves we do not need one of G or J, eliminating answer B.

Finally, Q18, in which the chairperson is a homeowner, allows us to choose any two homeowners, such as PQ, and that proves that we do not need one of R or S, eliminating answer E.

Don't test answers until after you have sorted them out into losers and contenders, and if you can eliminate 4 losers based on your prior work, just pick the one that's left without wasting time testing it out.