## Help with PT 17 Game 2 Question 7

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evolution

Posts: 57
Joined: Wed Jan 30, 2013 11:04 pm

### Help with PT 17 Game 2 Question 7

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

sam62188

Posts: 32
Joined: Sat Dec 18, 2010 4:14 am

### Re: Help with PT 17 Game 2 Question 7

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:

Posts: 580
Joined: Tue Jan 08, 2013 7:50 pm

### Re: Help with PT 17 Game 2 Question 7

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.

Reframe

Posts: 78
Joined: Wed Jun 12, 2013 5:45 am

### Re: Help with PT 17 Game 2 Question 7

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.

evolution

Posts: 57
Joined: Wed Jan 30, 2013 11:04 pm

### Re: Help with PT 17 Game 2 Question 7

I see now, thanks so much for the replies!

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