by ManhattanPrepLSAT1 Thu Mar 25, 2010 4:55 pm
According to the way you described the constraints, I think you're on track. To compare your setup, I've attached a diagram of the way I would approach this game.
Q2: If G's last meeting is with R, then which one of the following could be true?
If G's last meeting is with R, then it cannot be true that G's sixth meeting is with S.
_ _ _ _ T S R
You simply don't have room for three F's which are nonconsecutive.
Q4: If G's meeting with R is the very next one after G's meeting with S, then which one of the following must be true?
If G's meeting with R is the very next one after G's meeting with S, then we have a chunk TSR, that must be scheduled either 234 or 456.
_ T S R _ _ _
_ _ _ T S R _
This is the only way to have room for three nonconsecutive F's. You're choice (D) could be true, but doesn't have to be true. Therefore, isn't correct. Notice above, if the F's are not going to be consecutive, F's will have to be scheduled first and last. Thus, answer choice (E) is correct.
Q6: If G's meeting with M is the very next meeting after G's meeting with R, then who must go fourth?
If G's meeting with M is the very next one after G's meeting with R, then there are two chunks. TS and RM. To get three nonconsecutive meetings with F, F must be scheduled first, fourth, and seventh.
F T S F R M F
or
F R M F T S F
So the correct answer is (A)
One way to help you see this game would be to ask yourself how many possible arrangements of F are there and where do those arrangements force your TS chunk?
- Attachments
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- PT44, S3, G1 - Seven Different Meetings - ManhattanLSAT.pdf
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