LG Game 2 Question 9 Explanation, by LSATHacks
I skipped this question my first time through the game. I eventually came back and solved it by brute force. I drew diagrams disproving all the wrong answers.
This isn’t the best approach, but it works perfectly fine if you have time for it. I did have time on this question since I finished the first game quite quickly.
The short way to solve this question is to have made the deduction that Zimmerman must go at 1pm.
If you know that, then it’s obvious that B is CORRECT. This is because Long always go at 2pm or later (rule 1).
Here are diagrams proving the other answers are possible.
This diagram proves that A is possible:
This diagram proves that C is possible:
This diagram proves that D is possible:
This diagram proves that E is possible:
You don’t need to draw these brute force diagrams. I’m only drawing them in case you thought one of the wrong answers wasn’t possible.
I interpreted rule 2 incorrectly apparently. I thought it meant that X and Y could not even be in the gold room at 2pm or 3pm even if Z was at 1pm in the rose room. Why was this wrong? Because technically x and y are “before” z, no?
G | R
Y Z
M X
L
In this situation, isn’t Speech Y given a day”before” Speech Z?
An easier way than brute force is knowing that Z must be at 1pm. And since M has to be before L, that means M has to be either 1pm or 2pm. That means L has to be 2pm or 3pm. Though this knowledge, one can realize Z and L cannot ever be together, no further deduction made.
And I like that you pointed that out. Realizes Z must be at 1 helps with 9 and 13 and saves so much time with this game. Thanks for the explanations
For LSAT 73, Logic Game 2, Question 9 why isn’t the possibility of Zimmerman in the gold room taken into account. It is possible for example to have Z,X,L in gold and the rest in rose. This approach rules 1 and 2 do not apply.
Why would it matter? The point of the diagrams was to eliminate the wrong answers. There’s no need to draw *every* possible scenario.