#12 - Global, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8627)

The correct answer choice is (E)

This global question may come as somewhat of a surprise: apparently there is an area of expenditure that is always reduced. Yet this inference did not appear in our initial diagram! To have had this inference would have made the game easier, but it goes to show that there are times where you can miss an inference and still complete the game successfully.

The best way to attack this question is to use previous work. As in question #10, start with the hypotheticals created by questions #6 and #7. G-L-M-N-W, the hypothetical from question #6, shows that P does not have to be reduced, and therefore answer choice (D) can eliminated. W-M-P-R-S, the hypothetical from question #7, shows that G, L, and N do not have to be reduced, and therefore answer choices (A), (B), and (C) can be eliminated. Thus, with little or no work, it can be determined that answer choice (E) is correct. Consider how much faster and easier using this mode of attack is than the alternative of working out several independent solutions.

Abstractly, answer choice (E) is correct because ultimately no valid solution to the game can be created unless that solution contains W. If W is not reduced, then G, P and N must be reduced (see the GSW rule and the group of 3), but when G, P, and N are reduced you cannot reduce R or L (see the second and third rules), and thus the maximum number of reductions that could be made would be four: G, P, N, and M. Because there must be exactly five reductions and eliminating W does not allow for five reductions, it follows that W must be reduced.

One final point must be made about the restriction inherent in the L, M, and R rule. Since only three basic scenarios result under that rule—LM, LR, and MR—one entirely different approach to this game involves creating three templates based on each of those options. Although we feel this approach can be quite effective, its usefulness is dependent on making the inference regarding the three reductions from G, N, S, P, and W. As many students fail to make this inference, the value of the template approach is diminished here. In a later section the Identify the Templates approach will be revisited, and we will see that at times there is no better way to attack a game.

[opubo Idoniboye]

In your explanation to question 12 of the LSAT October 1996 (Logic Games) (p.227 in the LSAT Logic Games Bible, 2002-2011), you stated in the third paragraph, line 2, that "If W is not reduced, then G, P and N must be reduced...." This is confusing to me, since the rule (that is, rule 1) states that "If both G and S are reduced, W is also reduced."
G
+ ->W
S
In light of this, is it wrong to say that the contrapositive of the preceding rule is as follows?: "If W is not reduced, then neither G nor S is reduced." Furthermore, am I wrong to say that the cotrapositive of "AND" is "NOT." Finally,is it wrong to say that "NOT" could be translated as "NEITHER"? If my analysis were correct, G must not be reduced together with "P and N."
Accordingly, please explain to me why "If W is not reduced, then G, P and N must be reduced. Thanks

[Dave Killoran]

Hi Opubo,

Thanks for the question! Let's start by taking a look at the contrapositive of that rule, because I can see that it is giving you some problems, and that is then causing confusion in the question itself.

Let's start by isolating just the first rule of the game, and consider it without considering any of the other rules or scenario information. I want to do this so we can clarify the proper contrapositive of the rule.

As you correctly note, the rule diagram for the first rule of that game is:


G
+ W
S


Now, when you take a contrapositive that contains "and," what occurs is that both terms are negated, and the "and" changes to an "or." So, the contrapositive becomes:


G
W or
S


Thus, according to the contrapositive of the rule, if W is not reduced, then G or S is not reduced (and, possibly, both are not reduced--again, that is known from just the rule itself; other information in the game affects this possibility, ultimately ruling it out, but that isn't known from just the contrapositive here).

To summarize, the contrapositive in play here can be reduced to two steps:

• 1. The "and" becomes "or" (similarly, if the original term was "or," it would convert to "and").
  
  2. Each term is negated.

This is a bit different from how you had described it, and, importantly, the "and" does not turn into "neither."

Ok, with the above covered, let's go back to question #12. It probably now makes sense how G could be reduced even when W is not reduced. But, the explanation in that paragraph is part of a discussion of an outcome that cannot occur (and which ultimately leads to the inference that W must be reduced in every solution of the game). The essential problem is that the pool of G, N/S, P, or W must fill exactly three spaces. When you take away W, and via the contrapositive also take away G or S, you ultimately aren't left with enough variables to fill those three spaces and then fill the other two as well (and thus, you can conclude that W must always be reduced).

Continuing on, in the discussion itself, you'd choose G over S because then you could still use N, giving you three variables from the group. If you chose S, you could only get two variables:

• If W and G: only P and S could be reduced (N is out due to N S). This fails because you need three variables from this group.
  
  If W and S: G, P, and N could be reduced. While this works to fill the three spaces, as explained in the rest of the paragraph this selection group does not allow a viable 5 variable group to be formed.

Please let me know if that helps. Thanks!

[opubo Idoniboye]

Hi Dave:
Thank you so much for educating me on the contrapositive/negation of "and" and "or;" as well as explaining the question and the answer to me in a very professional manner.
Again, thanks.

Sincerely,

OTI

[pacer]

Game 1 Page 117 Q12 (from October 1996 LSAT)

I was able to narrow this down to A or E. But since I made a mistake on question 7, I did not have the correct hypothetical to work with here. The solutions use the work done in Q7 to answer this question.

Is there any other way to do this question?

I thought that since G and S need to be reduced for W to be reduced, then W is not necessarily true and eliminated it because this is a must be true question.

[Nikki Siclunov]

Hi pacer,

This is an exceptionally difficult question to solve if you don't already have the inference that W must be selected. The LGB outlines one possible approach. Here are two more:

Templates:

• 1. Given the last rule, every solution to the game must contain two of these three variables: L, M, and R. So:
  
  (a) L, M
  (b) L, R
  (c) M, R
  
  2. There are only five variables remaining to compete for the remaining three spaces: G, N, P, S, and W. Remember that N and S cannot be reduced at the same time (second rule). After applying this inference along with the remaining rules, the first two templates can be solved as follows:
  
  (a) L, M, G, W, N/S
  (b) L, R, G, S, W
  (c) M, R, (3 of G, P, S, W)

Due to the first rule, Template (c) above prevents us from reducing G, P, and S (you'd need W as well, for which you have no room). So, Template (c) contains the following solutions:

• (c. 1): M, R, G, P, W
  (c. 2): M, R, G, S, W
  (c. 3): M, R, P, S, W

As you can see, W is reduced in every single solution to our game. This is one way to prove that W must be reduced. What's the alternative?

Contrapositive of first rule:

Let's see what happens if W is not reduced. Thanks to the contrapositive of the first rule, if W is not reduced, then either G or S cannot be reduced:

• Option 1: W and G are not reduced. This would violate the requirement that you reduce 3 of these five areas: G, N, P, S, and W (see my second point above), because N and S cannot be reduced simultaneously. So, if W is not reduced, it is impossible that G is not reduced as well.
  
  Option 2: W and S are not reduced. In that case, we are left with G, N, and P. Fine, but in that case R cannot be reduced (second rule), and neither can L (third rule). So, we cannot reduce exactly two of L, M, and R.

Clearly, neither of the two options above is possible, providing that it is impossible not to reduce W. W must be reduced.

So, this gives you two more approaches to Question 12 No matter what, it's a really difficult question.

Good luck!

[leslie7]

Hi Opubo,

Thanks for the question! Let's start by taking a look at the contrapositive of that rule, because I can see that it is giving you some problems, and that is then causing confusion in the question itself.

Let's start by isolating just the first rule of the game, and consider it without considering any of the other rules or scenario information. I want to do this so we can clarify the proper contrapositive of the rule.

As you correctly note, the rule diagram for the first rule of that game is:


G
+ W
S


Now, when you take a contrapositive that contains "and," what occurs is that both terms are negated, and the "and" changes to an "or." So, the contrapositive becomes:


G
W or
S


Thus, according to the contrapositive of the rule, if W is not reduced, then G or S is not reduced (and, possibly, both are not reduced--again, that is known from just the rule itself; other information in the game affects this possibility, ultimately ruling it out, but that isn't known from just the contrapositive here).

To summarize, the contrapositive in play here can be reduced to two steps:

• 1. The "and" becomes "or" (similarly, if the original term was "or," it would convert to "and").
  
  2. Each term is negated.

This is a bit different from how you had described it, and, importantly, the "and" does not turn into "neither."

Ok, with the above covered, let's go back to question #12. It probably now makes sense how G could be reduced even when W is not reduced. But, the explanation in that paragraph is part of a discussion of an outcome that cannot occur (and which ultimately leads to the inference that W must be reduced in every solution of the game). The essential problem is that the pool of G, N/S, P, or W must fill exactly three spaces. When you take away W, and via the contrapositive also take away G or S, you ultimately aren't left with enough variables to fill those three spaces and then fill the other two as well (and thus, you can conclude that W must always be reduced).

Continuing on, in the discussion itself, you'd choose G over S because then you could still use N, giving you three variables from the group. If you chose S, you could only get two variables:

• If W and G: only P and S could be reduced (N is out due to N S). This fails because you need three variables from this group.
  
  If W and S: G, P, and N could be reduced. While this works to fill the three spaces, as explained in the rest of the paragraph this selection group does not allow a viable 5 variable group to be formed.

Please let me know if that helps. Thanks!

Just curious ( I missed the option that there is a possibility for both not to be selected in the contrapositive - is that true for all contrapositives of this form?)

G
& --> W
S

/W --> / G or /S

So there are three possible outcomes
Not G OR
Not S OR
Not Both?

Just generally speaking how useful do you think this contrapositive is when trying to make inferences? (I guess it depends on the question and its constraints/rules?) But since so many options are plausible, is it safe to say that it's less of a "powerful" rule if we were to place generally some of these rules on a scale of "powerfulness" so to say?

I'm just thinking that if I see something like this again whether its worth scrutinizing for too long unless a question asks me something of it directly?

Also,

I noticed in this question there were a number of hypotheticals made which is great and fascinating for studying but during a timed test is it viable to do that? I think my inclination would be just to get the basic diagram out and then move on to the questions (because I think that by the time I got to figuring out those hypotheticals on my own I would have definitely run out of time to try and answer the questions)

Let me know your thoughts generally or anyone else who can answer

Thanks so much !

[leslie7]

Also just to be sure

G and S must occur together to trigger W correct?

Even though we can separate these as

G--> w?
S-->w?

but we need BOTH G and S to trigger W?

Additionally I loved that we separated the N S R rule into 2 separate diagrams that would have made it so much easier to understand /do the questions quickly

How often should we be separating these? should we make it practice to do it all the time ?

e.g.

should I immediately separate G&S --> w to 2 diagrams?

this is more of a general question for methodology

[Dave Killoran]

Even though we can separate these as

G--> w?
S-->w?

No, you can't separate them! See these two articles for more info on separating conditional statements:

https://blog.powerscore.com/lsat/bid-26 ... Condition/

https://blog.powerscore.com/lsat/bid-26 ... condition/

I think that will help sort this out as that is when compound conditionals can be separated