PT13.G1.Q6 Forum

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glucose101

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PT13.G1.Q6

Post by glucose101 » Sun Jul 11, 2010 4:26 pm

I think I've been reviewing this question so much that perhaps I'm missing my own diagramming, but I don't see the correct answer. I got it down to B and D, but don't know why it's D. I know it involves Rule 4. Thanks in advance!

NYCLSATTutor

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Re: PT13.G1.Q6

Post by NYCLSATTutor » Sun Jul 11, 2010 4:36 pm

glucose101 wrote:I think I've been reviewing this question so much that perhaps I'm missing my own diagramming, but I don't see the correct answer. I got it down to B and D, but don't know why it's D. I know it involves Rule 4. Thanks in advance!
The question premises that L is in group 2. If L is in group 2 then via the contrapositive of H1->L1, H must also be in group 2. If L and H are both in group 2, that means that FJ must be in group 1. After all, if it was L,H,F,J all in group 2, then G & M would be together in group 1. That violates a rule, so now we have L &H in group 2 and F&J in group 1.

Now, D puts K &M in group 1. That would mean that F,J,K,M are in group 1 and L,H,G,N are in group 2. We have a rule that says N2->G1. Therefore, having both G & N in group 2 is impossible. Since that won't work, D is the correct answer.

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glucose101

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Re: PT13.G1.Q6

Post by glucose101 » Sun Jul 11, 2010 4:45 pm

I guess I wasn't using the choices, but trying to do it based on the rules. N2-->G1/contra: G2-->N1.

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blackwater88

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Re: PT13.G1.Q6

Post by blackwater88 » Sun Jul 11, 2010 4:55 pm

NYCLSATTutor wrote:
glucose101 wrote:I think I've been reviewing this question so much that perhaps I'm missing my own diagramming, but I don't see the correct answer. I got it down to B and D, but don't know why it's D. I know it involves Rule 4. Thanks in advance!
The question premises that L is in group 2. If L is in group 2 then via the contrapositive of H1->L1, H must also be in group 2. If L and H are both in group 2, that means that FJ must be in group 1. After all, if it was L,H,F,J all in group 2, then G & M would be together in group 1. That violates a rule, so now we have L &H in group 2 and F&J in group 1.

Now, D puts K &M in group 1. That would mean that F,J,K,M are in group 1 and L,H,G,N are in group 2. We have a rule that says N2->G1. Therefore, having both G & N in group 2 is impossible. Since that won't work, D is the correct answer.
Dammit, you beat me to it :D

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