Q17

[monicaiannacone]

I understand the diagram and that if R is out, then Y must be in since R or Y must be in.

This made me start thinking:
So, for this game, is it true that both W and R can be out?
In the same sense, is it true that both W and T can be out?

This makes sense to me when drawing the In/out boxes. But is there another way to realize this quicker using the logic chain?

Thanks!

[timmydoeslsat]

I understand the diagram and that if R is out, then Y must be in since R or Y must be in.

This made me start thinking:
So, for this game, is it true that both W and R can be out?
In the same sense, is it true that both W and T can be out?

This makes sense to me when drawing the In/out boxes. But is there another way to realize this quicker using the logic chain?

Thanks!

It is true that in this game both W and R....and....W and T, even both T and R can be out when W is out.

We know what happens when W is in. We have no information as to what happens when W is out.

I do not use the logic chain, but the in/out set up.

~A ---> B .....This means at least one in.

A ---> ~B .....This means at least one out.



As for this question, #17, we know that 3 are in, 4 are out.

We know that one of those slots of 3 will be for one of Y/R. We could have both of those variables, but at least one will be part of the three.

So we have two slots left. Since this is a must be false question, I know the chain of U ---> S ---> W will be tested.

No way could U be in. No way that S could be in with somebody other than W. That is what (A) does. Puts S and Z together. So we would now have one of Y/R, S, Z, and W. (The W due to the conditional of S ---> W.

Z was a random/floater/free agent in this game.

[monicaiannacone]

Thanks! That makes sense, I understand that concept better and that hopefully means I can get my logic chain setup and answers quicker.