#25 - Most serious students are happy students, and most

[Administrator]

Complete Question Explanation


Must Be True—Formal Logic. The correct answer choice is (B)

Here is the diagram for the stimulus, which does not contain a conclusion:

          SS = serious students
          H = happy students
          GS = go to graduate school
          O = overworked

LR1.PNG

LR2.PNG

The first inference is simply an application of the Most Train from SS to O. The second and third
inferences rely on the double-most inference discussed earlier in this chapter on page 455. The
presence of two most relationships emanating from a single group (in this case, SS) allows a some
inference to be drawn. In this case, the inference H GS can be made.

When an inference this tricky is present, you should expect to be tested on it. With that in mind,
spend some time learning to recognize the special double-most relationship and the some inference
that follows. The makers of the LSAT have had a recent emphasis on testing this relationship,
and questions appeared on the test in 2000, 2002, and 2003 (this question) that feature the special
double-most configuration presented in this problem.

Answer choice (A): Although some overworked students are happy (see the third inference), we do
not know that most overworked students are happy.

Answer choice (B): This is the correct answer. The inference in this answer is identical to the third
inference drawn from the diagram. This is a classic separation problem. Students who are properly
prepared will be able to quickly diagram the problem, make inferences, and identify the correct
answer. Other students will struggle to diagram the problem and have difficulty with the challenging
inferences, resulting in lost time and possibly an incorrect answer choice. This is the essence of
LSAT preparation: by learning the patterns used by the test makers you give yourself a tremendous
advantage over unprepared students.

Answer choice (C): At best, we can infer that some overworked students are serious students (this is
the inherent inference present in the first inference). Thus, this answer is too strong.

Answer choice (D): No inference can be made about unhappy students.

Answer choice (E): The stimulus allows for the inference that most serious students are overworked.
Although it may be possible that all serious students are overworked, this is not certain, and
therefore this answer choice is incorrect.

[JKing]

October 2003 LR section 3 # 25

SSmostHS
SSmostGS
GS-->OW

SSmostGS-->OW

SSsomeOW

I picked B. Is it SSmostHS-->GS-->OW??

A) incorrect reversal?

[Jon Denning]

Hey J - thanks for the question. The shorthand method you've used to ask this makes it a little tricky to figure out exactly what you're asking (and what gave you trouble), but I'll do my best with it.

Your initial three diagrams look correct, however I'm not sure what the two lines below those first three diagrams are meant to be...? Inferences?

In any case, there are a few inferences that can be made here, based on some rules of formal logic:

1. SS --most--> OW
2. GS <-some-> H
3. OW <-some-> H [this is answer choice B]

Answer choice A isn't a reversal per se, but rather just too strong (most) for what we know (only some). That is, we can know that since the majority of SS is both GS and H, that some H are GS (and vice versa). And if H <-some-> GS --> OW, then H <-some-> OW, which is the correct answer. But we can only get to some, not most. Note too that if A were thought to be the right answer, it would also make B correct, so A can be dismissed immediately for that reason as well.

I hope that clears things up!

[jailenea]

I am confused.

I diagrammed this differently for ease of visuals/working off of the chain: HS <--most-- SS G O

I started with SS G O and got the first additive inference of SS O. This was the easy part. From there, it got a bit messy.

I followed the double most special rule for the left half up to G (no O).
Here's the work I showed:

HS <--most-- SS G O
HS SS G O (using the inherent inference of the most to the some at the beginning).
The book then shows that the proper additive inference is HS G. I don't understand this. You've got some and you've got most. These two groups, to my understanding, don't necessarily have to overlap. So how is it that we come up with the inference? I would understand getting this inference if it was an "all" -- as in HS SS G. But how do we end up with the same thing from "most?"

Even though I remain confused on getting the second additive inference, I then recycled it combined with the last part of the chain to make HS G O. This produces the third (and correct answer) additive inference of HS O.

Please help, thank you!

[Rachael Wilkenfeld]

Hi jailenea,

Sometimes when I work with these terms, I like to use numbers to help me understand the different relationships.

Let's start by imagining we have 100 serious students. If most are happy, then we have a minimum of 51 happy students. If most serious students go to graduate school, then at least 51 students go to graduate school. If there are only 100 total students, that must be we have one happy student in graduate school. At least one can be translated to some. That's the nature of the double most. If most people in a group have attribute A, and most have attribute B, there must be an overlap between those that have attribute A and attribute B. That's the nature of the term most.

So the idea isn't that you go from some to most, it's that you combine two most to get a some.

Hope that helps!