#9 - Global, Must Be True, Maximum

[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 (C)

In this question you must select the variables in such a way as to maximize the number of birds in the forest. This means that birds that tend to knock out several other birds must be removed. An examination of the list of inferences indicates that W must be removed since, when W is in the forest, then H, J, and M cannot be in the forest. The other bird that must be removed is G since, when G is in the forest, then H cannot be in the forest, and when H is not in the forest, then neither J nor M can be in the forest. If G and W are removed from consideration, then the remaining four birds are J, H, M, and S. Ultimately, four is the maximum number of birds in the forest, and answer choice (C) is correct.

Another approach to this question involves referring to work done on other questions. When this approach is used, hypotheticals are examined to eliminate certain answer choices. For example, in question #7 we were able to determine that four birds could be in the forest: M, H, S, and J. This hypothetical eliminates answer choices (A) and (B). To effectively use this approach, however, it would be best to skip this question and return after completing all other questions in order to have as many hypotheticals as possible.

A third approach involves considering the negative grouping rules in the setup. For example, since the first rule establishes that H and G cannot both be in the forest, answer choice (E) can be rejected. And since at least one of H and G cannot be in the forest and at least one of W and H (or W and J for that matter) cannot be in the forest, a case can also be made against answer choice (D). When you examine these rules, it is important to consider negative grouping rules that contain entirely different sets of variables. You cannot simply count all the negative rules and arrive at an answer because some of the rules will revolve around the same variables, and when those variable are removed the other variables can be selected.

Finally, let’s take a moment to examine a mistake that is made by a number of test takers. Some students, upon encountering the final rule, make the classic error of assuming that the jays and the shrikes cannot be in the forest together. As discussed in the Avoiding False Inferences section, this is a false inference, and thus it is possible for J and S to be in the forest together. Let us take a moment to review:

Most LSAT Logic Game conditional rules place the “not” on the necessary condition, but in this rule, the “not” is on the sufficient condition. While that difference may seem minor, in effect it completely changes the meaning of the rule.

According to the rule, if J is not in the forest, then S must be in the forest, and via the contrapositive if S is not in the forest, then J must be in the forest. Essentially, this means that if either bird (S or J) is absent (not in the forest), then the other bird cannot also be absent (they cannot both be out of the forest at the same time; at least one must always be in the forest). In other words, all we know from the fourth rule is that they will never both be out of the forest, because as soon as one is out the other must be in. They could, however, both be in the forest at the same time. The rules never tell us anything about what happens when either J or S is in the forest, only what happens when J or S is not in the forest.

Since it is possible for both J and S to be in the forest at the same time, that possibility impacts the choices you have in questions #9 and #10. In #9, most people assume that one of J or S is always out of the forest. Obviously, that is not the case, and when arriving at the maximum number of birds in the forest you should include both J and S. In question #10, many people select answer choice (B) because they misinterpret the fourth rule. As explained above, J and S—the pair in answer choice (B) of question #10—can both be in the forest at the same time.

[Xantippe]

My approach to answering Q9 was the following: First, I looked at the second rule that seemed to allow for the greatest number of birds; If jays, martins or both are in the forest, then so are harriers. I then applied the third rule; If jays are not in the forest then shrikes are. That gave me a total of four birds. The contrapositive not g then not w of the third rule supplied me with the negative pairings for m h j (s can be with j) for a total of four birds.

Was my process correct?

Thanks in advance.

[James Finch]

Hi Xantippe,

The key here is to make two conditional chains at the setup, one the contrapositive of the other, as all the rules connect. You can then use those chains, similar to what you did, to see which would allow for the most birds, and then simply count them up. The only tricky part is to remember that S is only a necessary condition, and S only a sufficient one, so S could be in in the J/M/H chain.

Hope this helps!