Here's the setup for this game. You can use the constraints to establish two frames which can serve as a base that will speed you up through the questions. The key is to see how the first two constraints relate to each other.
If you have questions on this game, feel free to ask!
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- ManhattanPrepLSAT1
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Atticus Finch
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Diagram
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- uchechi8
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Vinny Gambini
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Re: Diagram
I'm still having trouble with this diagram?How would you answer question 2??
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- timmydoeslsat
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Atticus Finch
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Re: Diagram
I think Matt has a great diagram posted. I will attempt to go about this in a different manner in case you want a different perspective.
This is an in/out game.
We have 6 variables:
3 men: FGH
3 women: JKL
The first 2 rules are biconditional rules, which means that each condition is sufficient and necessary for the other. This means that the presence of one variable will guarantee the presence of the other. It also means that its absence will guarantee the absence of the other variable.
J <---> ~L
This biconditional rule was formed from this:
J ---> ~L
~L ---> J
We see that the arrow really goes both ways. So notice that the first rule is saying that at least one of J and L must be out.
The second rule is saying that at least one of J and L must be in.
So this means that 1 of J/L will be in and 1 of J/L will be out. What better way to frame a game than this? Do one frame with J in, do the other frame with L in.
We also have the rule that says if J is in, F is in. We also know that if any one of JKL is in, G is in. So we know already that one of J/L will always be in, which means G must always be in.
The J in frame:
J G F ........... L
The L in frame:
L G ........... J
In the first frame, we have two random variables left of H and K. They can go anywhere. The same holds true of these people in the second frame. We also have the freedom to put F anywhere as well. F was discussed as a necessary condition of J being in, and in this frame J is out, so this rule will never trigger.
So with this in mind, question #2 is a must be false question. Answer choice (D) sticks out immediately. We know that one of J/L will always be on stage, which means it is impossible that K will be the only woman on stage.
This is an in/out game.
We have 6 variables:
3 men: FGH
3 women: JKL
The first 2 rules are biconditional rules, which means that each condition is sufficient and necessary for the other. This means that the presence of one variable will guarantee the presence of the other. It also means that its absence will guarantee the absence of the other variable.
J <---> ~L
This biconditional rule was formed from this:
J ---> ~L
~L ---> J
We see that the arrow really goes both ways. So notice that the first rule is saying that at least one of J and L must be out.
The second rule is saying that at least one of J and L must be in.
So this means that 1 of J/L will be in and 1 of J/L will be out. What better way to frame a game than this? Do one frame with J in, do the other frame with L in.
We also have the rule that says if J is in, F is in. We also know that if any one of JKL is in, G is in. So we know already that one of J/L will always be in, which means G must always be in.
The J in frame:
J G F ........... L
The L in frame:
L G ........... J
In the first frame, we have two random variables left of H and K. They can go anywhere. The same holds true of these people in the second frame. We also have the freedom to put F anywhere as well. F was discussed as a necessary condition of J being in, and in this frame J is out, so this rule will never trigger.
So with this in mind, question #2 is a must be false question. Answer choice (D) sticks out immediately. We know that one of J/L will always be on stage, which means it is impossible that K will be the only woman on stage.
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