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Help with PT 17 Game 2 Question 7

Posted: Wed Jun 12, 2013 10:05 pm
by evolution
I'm having trouble understanding why C is the correct answer for question 7.
I don't see how C is a MBT since when I diagram like this, it seems like no rules are violated - and would contradict answer choice C.

P: F/G/H
Q: F/L/L
S: G/M/K

Can someone tell me if I'm missing something here, because the 7sage video diagrams like this:

P: F/G/H
Q: F/K/K
S: G/M/K

Re: Help with PT 17 Game 2 Question 7

Posted: Wed Jun 12, 2013 11:07 pm
by sam62188
Basically the new rules added from question 7 restrict the groups into 2 scenarios.
Scenario 1:
P: F, G, H
Q: F, L, K
S: G, M, K
Scenario 2:
P: F, G, H
Q: G, M, K
S: F, L, K

In both cases K is locked into group Q and S, making C the correct answer.

The reason why there cannot be an F, L, L grouping is because the game groups people and unless L was cloned he cannot be in the same group twice.

BTW I am just restating what was said on this video I found when trying to look up the game:
http://www.youtube.com/watch?v=DFnA4F-61Aw

Re: Help with PT 17 Game 2 Question 7

Posted: Wed Jun 12, 2013 11:23 pm
by aboutmydaylight
evolution wrote:I'm having trouble understanding why C is the correct answer for question 7.
I don't see how C is a MBT since when I diagram like this, it seems like no rules are violated - and would contradict answer choice C.

P: F/G/H
Q: F/L/L
S: G/M/K

Can someone tell me if I'm missing something here, because the 7sage video diagrams like this:

P: F/G/H
Q: F/K/K
S: G/M/K
You can't have the same person on the same committee twice, that wouldn't even make sense.

We know F can't be with M and H is already placed and can only be placed once, so we know that M must be with G. You put that block on q or s, and L has to be on the other.

Re: Help with PT 17 Game 2 Question 7

Posted: Wed Jun 12, 2013 11:35 pm
by Reframe
It's often helpful to remember that you can think about who must be out instead of thinking about who must be in. For groups Q and S, we know h must be out (because h's one group is P), and we know that at least one of each pair f/M and g/L must be out, since those pairs can't be in a group together. That means there's no space left in either Q or S for K to be out. That gives us (C): K must be in for both groups.

Re: Help with PT 17 Game 2 Question 7

Posted: Wed Jun 12, 2013 11:41 pm
by evolution
I see now, thanks so much for the replies!