September 2017 LSAT Section 2 Question 19

# Which one of the following could be a matching of the students to the subjects on which they present?

6 Replies

Christopher on July 28, 2018

@RJEh, this is a grouping game, so the first thing to do is to identify what your groups will be. In this case, since you can have multiple students present on each of three lectures, your groups are M, J, S. Before getting into the rules, you know that there must be at least one presentation on each subject, and there's no limit to the number of presentations, so it is theoretically possible (before getting into the rules) that all three students present on all three subjects. I'd start with this as the diagram and go from there into the rules._ _ _ _ _ _ _ _ _

M J S

Then getting into the rules:

Rule one diagrams as GJ -> RJ and not RJ -> not GJ. This rule doesn't do much to the diagram. It is still possible for J to only have one presentation from W, but it could also have two from R and G or three from R, G, and W.

Rule two diagrams as RS -> not WS and WS -> not RS. This rule means that S can have a max of 2 presentations.

Rule three diagrams as M -> S and not S -> not M. This rule piggy-backs on rule two and means M can only have a max of 2 presentations.

So the diagram is narrowed to:

_ _ _ _ _ _ _

M J S

As you start going into the questions, if you see any that have more than two presentations on M or S, you can immediately rule them out (for instance, (B) on the first question). Otherwise, you'll have to run hypotheticals through each rule.

Does that help?

on December 15 at 09:19PM

How do you know that this rule means s can have a max of 2 presentations? Same for M?"Rule two diagrams as RS -> not WS and WS -> not RS. This rule means that S can have a max of 2 presentations.

Rule three diagrams as M -> S and not S -> not M. This rule piggy-backs on rule two and means M can only have a max of 2 presentations."

on May 7 at 09:09PM

Where are the details to this games answers?Skylar on August 9 at 04:15PM

@ca_teran1@yahoo.com, happy to help!Rule #2 tells us:

R(S) -> not W(S)

W(S) -> not R(S)

This means that if R is in S, W cannot also be in S. On the other hand, if W is in S, R cannot also be in S. In short, R and W cannot both be assigned to S.

We start with a total of three options of students (R, W, and G) that can be assigned to presentation S. This rule tells us that it would be impossible to have all three students assigned to S because R and W cannot coexist there. Therefore, we could only have a maximum of two students assigned to presentation S (G and either R or W).

Rule #3 tells us:

M -> S

not S -> not M

This means that whatever is in M must also be in S. We know from Rule #2 as discussed above that we cannot have both R and W in S. Therefore, we cannot have both R and W in M (because if we did, we would have to place both R and W in S, which would violate Rule #2). This means that we could only have a maximum of two students assigned to presentation M (G and either R or W), just like we concluded with S above.

Does that make sense? Hope it helps! Please let us know if you have any other questions.

Skylar on August 9 at 04:18PM

@Bonning, there are currently no video explanations uploaded for this game, but most of the questions have walkthroughs posted in their corresponding message board threads. If you are struggling or need more details on a specific problem, please feel free to create a message board post and we'd be happy to help!Thalia Wednesday at 05:32AM

@Skylar - how can I request a video explanation for this?