#12 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8548)

The correct answer choice is (B)

This is not a Suspension question because the rule is simply added on to the given information. In this way it acts like a normal Local question. The extra condition stipulates that S and H cannot both be in the forest. This affects both J and M because, when either J or M is in the forest, then H is in the forest. Thus, neither J and S nor M and S can be in the forest at the same time. This information is sufficient to reject answer choices (A) and (C). Answer choice (D) can also be rejected since, when J is not in the forest, then S must be in the forest, and when S is in the forest, then H is not in the forest, and when H is not in the forest, then neither J nor M is in the forest. Finally, answer choice (E) can be disproven since if H, for example, is in the forest, then G, S, and W are not in the forest. It follows that answer choice (B) is correct.

Despite the large number of inferences in this particular game, the typical student eventually comes to find this type of game completely reasonable, and certainly doable in the allotted time. As you develop the ability to make inferences more quickly, you will begin to see a game like this one as an opportunity to make up time, especially if you know how to handle the last rule.

[hunterama1]

Surprisingly, I couldn't find any prior posts to this question so here I am again.

I worked this game three times and still was certain with question 12.

Question twelve speciously puts forward: "Suppose the condition is added that if shrikes are in the forest, then harriers are not. If all other conditions remain in effect, then which one of the following could be true?"

This changed condition converts inference 6 from its negated sufficient condition s H to a negated necessary condition S H.

The existing inference prior to this rule being 'simply added on to the given information' is:

S J + M H G W.

The LGB explanation indicates that the extra condition stipulates "S and H cannot both be in the forest" which is fine except that inference 6 above already indicates S and H both not in the forest but as a negative sufficient condition thus both can not be absent from the forest with this condition. But the inference has one absent (S) and the other present (H). The contrapositive of this existing chain inference gives us S present and H absent and complies with the explanation "This affects both J and M because, when either J or M is in the forest, then H is in the forest".

This contrapositive to inference 6 is:

W G H J + M S

In this inference "neither J and S nor M and S can be in the forest at the same time" as the explanation indicates (w/o addressing the issue of change from negatively sufficient to negatively necessary) however it is conditionally incorrect as the new condition is a negated necessary condition, again S H or contra H S.

I can't seem to figure out any other way to make it work. Any suggestions would be appreciated as there are no diagrammatical explanations to this answer.

Thanks again in advance.

[Dave Killoran]

Hi Hunter,

Ok, you've made certain assumptions in your explanation that are causing you problems in understanding the answer.

First, as noted in the explanation, the question stem simply adds a new condition, S H. You got that right, where you ran into an issue is when you made this statement:

This changed condition converts inference 6 from its negated sufficient condition s H to a negated necessary condition S H.

That's not correct. The added information doesn't change anything that was preexisting; that is, there is no "conversion" here to the other statement. What it does is further modify the meaning of inference #6. Based on our last conversation, I think inference #6 may not be entirely clear to you because it has a negative sufficient condition. Specifically, you said that "inference 6 above already indicates S and H both not in the forest." That's not the case. Inference 6 does allow for both to be in the forest. Let's look at it again.

• S H results in three possible outcomes:
  
  1. S not in forest, H in forest
  2. H not in forest, S in forest
  3. Both H and S in forest
  
  This third option is the one that you assumed could not occur, but it can--look at the sufficient condition of the rule. When S or H is present, does that tell you anything? No, and thus the other variable could be present.
  
  The addition of S H then means that "3. Both H and S in forest" cannot occur; the other two options can still occur. Thus, when the condition in question #12 is added, only two outcomes are possible:
  
  1. S not in forest, H in forest
  2. H not in forest, S in forest
  
  In other words, one is always in, and one is always out.

So, when you made your explanation near the end, because you changed the meaning of the original statement and assumed it was discarded, it then became confusing (at least, this is what I'm gathering from reading your explanation . Try taking a look at it again, with the view that the original statement is still in place. And look at that explanation in the text again--it uses words to describe the diagrams you are referencing.

PS Response: We've had instructors before in Baton Rouge, but my understanding is that our last one left, and with the standards we employ for hiring, it's not always the easiest thing to obtain an instructor in any given city. Finding instructors with a sufficient score (top 1%) is tough enough, and then lots of those people don't qualify because they don't pass the teaching or personality tests. I'll check with HR and Tutoring to see what the status is not only with Daniel, but with hiring there in general.

Thanks!

[hunterama1]

Very good. Thank you.

I re-did the problem successfully.

The following diagrams emerged:

Question Condition:

W G H (chain tilts down terminating at J) J

S H and

W G (chain tilts up terminating at 'M') M

Contra:

J (chain tilts up from 'J'):arrow: H G W

or H S

M (chain tilts down from 'M') G W


This answered the question simply and clearly for me. It's even consistent with your LGB explanation!

Feedback welcome, of course.

Perhaps my '94 BS from LSU in biochem and the sequencings of organic are making a come-back.

Let's get me in the top 1% and I'll teach for you in Baton Rouge.

Thanks again and cheers.

[Dave Killoran]

Alright, glad that worked out! If you have that science background, not doubt this stuff will ultimately make a lot of sense to you because it is very systematic. I think once you get comfortable with all the relationships, you'll dominate conditional diagramming

Let's get me in the top 1% and I'll teach for you in Baton Rouge.

Sounds good to me!

[splitterfromhell]

I am having a hard time understanding why answer choice E on question 12 is incorrect. The explanation the book gives is one scenario, but the question is a could be true question. How can I be sure that there isn't a scenario that could allow E to be true?

[Emily Haney-Caron]

Thanks for your question! In order for E to be a correct answer, there need to be 4 types of birds that can all be in the forest together without violating any rules.

If harriers are in, under the rules there are 3 (G, S, W) that cannot be in. So there is no possible answer with H that could have 4 types in.

If G is in, then H, J, and M cannot be in. So there is no possible answer with G that could have 4 types in.

If J is in, then G, W, and S cannot be in. So there is no possible answer with J that could have 4 types in.

Because there would only be 3 possible bird types left, I can know that, without trying out solutions with those types, there is no way to have FOUR types in under these rules.

I hope that helps!

[lathlee]

Hi. Actually, I don't know how to interpret the meaning of e) only two of the six kinds of bids are not in the forest. as in every scenario of when S are in the forest and harriers are not?

[Francis O'Rourke]

Hi Lathlee,

Answer choice (E) asks us if - taking the local rule into account - it could be true that four birds are in the forest and two birds are not in the forest.

I am not sure what you mean when you say "as in every scenario of when S are in the forest and harriers are not?" I'd love to help you out with that question, but I am not entirely sure what you are asking.

[LSAT2018]

Is there a biconditional being used here?

S → Not H → Not J
J → H → Not S

Not J → S
Not S → J

So given the newly added rule, S and J cannot be chosen together, and the original rule, that at least one of S or J are chosen, the inference is that either S or J, but not both, must be in?