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- ManhattanPrepLSAT1
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Atticus Finch
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- shaun_davis5
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Vinny Gambini
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Re: Diagram
On question 13-18, would it be possible to get breakdown of the questions and answers. During my timed section practice, I missed 17 and 18, but I would like to know "why" on the correct answers as well as the wrong answers.
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- ManhattanPrepLSAT1
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Thanks Received: 1909
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Atticus Finch
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Re: Diagram
Solutions for each question are up!
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- AubreyH4
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Vinny Gambini
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Re: Diagram
Solution is not up for Q12
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- SamanthaW170
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Vinny Gambini
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Re: Diagram
One of the rules for this game is that if L is not on Saturday, then L comes before F. The contrapositive of the statement is: If L is not before F (in this game, F-L, since F has to go in the beginning of the week, where it's not possible for two librarians to go on the same day), then L is on Saturday.
From the tree diagram, we see that there are four elements which have to go later in the week than F (and each has to go later in the week the one proceeding it on the chart): K-Z and M-G. This led to me conclude that F can only go on Monday or Tuesday, since K-Z and M-G need to go one each on Wednesday-Saturday.
From the diagram, we see that the only two librarians who could be on duty on Monday are F and H, since all other elements have at least one other element proceeding them.
I was thinking about these inferences, since both involve F, when I realized that if I am correct, is it ever possible for L to go before F? In other words, if F always has to go before L, then isn't L always on Saturdays?
Since the only option for the Monday librarian is F or H, if we put H first, F has to go on Tuesday, since it cannot go on Wednesday-Saturday. There would be no room for L before F.
It seems highly unlikely that a conditional rule could only play out one way in a game on the LSAT - where am I going wrong?
From the tree diagram, we see that there are four elements which have to go later in the week than F (and each has to go later in the week the one proceeding it on the chart): K-Z and M-G. This led to me conclude that F can only go on Monday or Tuesday, since K-Z and M-G need to go one each on Wednesday-Saturday.
From the diagram, we see that the only two librarians who could be on duty on Monday are F and H, since all other elements have at least one other element proceeding them.
I was thinking about these inferences, since both involve F, when I realized that if I am correct, is it ever possible for L to go before F? In other words, if F always has to go before L, then isn't L always on Saturdays?
Since the only option for the Monday librarian is F or H, if we put H first, F has to go on Tuesday, since it cannot go on Wednesday-Saturday. There would be no room for L before F.
It seems highly unlikely that a conditional rule could only play out one way in a game on the LSAT - where am I going wrong?
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- ohthatpatrick
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Atticus Finch
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Re: Diagram
I think you're right that it will RARELY be the case that L - F, so it will USUALLY be the case that L is on Saturday.
But we can make L - F, given that tree, as long as we start with H.
Here's one such scenario:
H L F M G K Z
In any such scenario, we'd know that it starts
H L F
So it would have been prudent to frame this game based on whether L - F or F - L
Frame 1:
H L F M/K __ __ G/Z
Frame 2:
H/F __ __ __ __ __ L
Hope this helps.
But we can make L - F, given that tree, as long as we start with H.
Here's one such scenario:
H L F M G K Z
In any such scenario, we'd know that it starts
H L F
So it would have been prudent to frame this game based on whether L - F or F - L
Frame 1:
H L F M/K __ __ G/Z
Frame 2:
H/F __ __ __ __ __ L
Hope this helps.
6 posts Page 1 of 1