Diagram

[dj_grey]

Can someone go through this problem?

[ManhattanPrepLSAT1]

Attached is a set of frames that will make it clear that K must be lower than J. A key second level inference is that if J is not 4th, then J will have to be 6th lowest.



This makes it certain that K must be lower than J.

Hope this helps!

[noah]

I was just preparing this game for a student and came up with a different diagram. I did not put in the initial notation of the rules, but instead only have put the two frames that I came up with and a demo of how to approach Q23.

PT54, S3, G4 – Six Contract Bids - ManhattanLSAT.pdf

(56.3 KiB) Downloaded 1109 times

[brandblum]

Either r or s is the fifth lowest.

lowest -------> highest

___ ____ ___ ___ ___ ___
1 2 3 4 5 6

why wouldnt the fifth lowest mean slot 2?
wouldnt the sixth lowest mean slot 1, and so on.

[ManhattanPrepLSAT1]

If it goes

Lowest --> Highest

then

1 would be the first lowest
2 would be the second lowest
etc...

The test-writer does a lot to confuse you, but they would never refer to the 5th lowest, and expect you to count backwards.

[brandblum]

Thanks.

[spfelker]

OK, this is probably something simple that I am overlooking. I was looking at Noah's diagram and wondering if J is NOT fourth, how do you know to put J sixth?

[noah]

OK, this is probably something simple that I am overlooking. I was looking at Noah's diagram and wondering if J is NOT fourth, how do you know to put J sixth?

Thanks for the question. If J is not fourth, we know it has to be higher than S and T. Since J is already higher than H, it would be higher than at least three other elements, requiring it to go in slot 4 or higher. But, the whole condition is based on J NOT being fourth, so that leaves slot 5 or higher; since slot 5 is taken by R/S, that leaves only slot 6.

Tell me if that doesn't make sense.

[ShariS937]

Hello. I see the Q has already been addressed and mostly answered, for this problem, but I need to understand how to apply the logic to similar situations.

Qs:
1. How do I know that 1 2 3 4 5 6 lists least $ bids to most $$$ bids? Am I assuming least $ bids are first because the company wants to choose for cost?
2. I had $$$ - 1 2 3 4 5 6 - $ and to choose the third least costly, I'd count back from 6 to get to 4. Is there any reason that this strategy is invalid logic? LSAT would never ask me to count backward seems - scary, as a rule to apply across all games. Unless it really is a bedrock rule?

I get tripped up by "common sense" designations like this, in the stimulus, quite often. Should I just try to understand the normative position? (What is normative for LSAT?) Or are there solid rules? If there are rules - please help.

Thanks!
s

[Laura Damone]

Sorry for the delayed response here. We had a technical difficulty that caused your post to be mistakenly categorized and it slipped through the cracks.

Yes, backwards counting is scary, and no, I can't imagine the test writers would ever ask you to do that!

In this game, the first rule uses the phrase "second or third lowest in cost." That tells me I should organize my diagram from lowest on the left to highest on the right, so that slot 2 represents "second lowest in cost." That phrase in the rules is the only reason to organize the diagram in that way. You don 't need to assume anything about the company and its selection in order to make that call