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- Wed Sep 27, 2017 4:40 pm
#40160
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8560)
The correct answer choice is (A)
If X is at 3 PM, and only one speech can be given at 3 PM, our local diagram should look like this:
According to the first rule, M and L must be given in the same room, and M must be earlier than L. With the 3 PM slot occupied, M must given at 1 PM. Since Z must also be at 1 PM (main inference), we can establish a Dual Option for M and Z in each of the two rooms:
The remaining two variables—L and Y—must therefore be given at 2 PM:
Although the setup above does not take into account the implications of the last rule, it is sufficient to prove that L and Y cannot be given in the same room. This validates answer choice (A) as the correct answer to this Cannot Be True question.
A somewhat more thorough—but less efficient—approach would be to setup each solution individually. To do that, we first need to make two local diagrams for X and E, since X can be at 3 PM in either room:
From the contrapositive of the last rule, we know that if X is in the Gold Room, then L must be in the Rose Room, with M earlier than L (first rule). Thus, the first local setup (above) is fully determined:
In the second setup, however, the placement of X at 3 PM in the Rose Room does not represent a similar point of restriction (to conclude otherwise would be a Mistaken Reversal of the last rule) . Consequently, in the second setup M and L can be in either room:
With these three setups in place, it would be exceptionally easy to prove which four answer choices could be true. The process of elimination, however, is not the most efficient way of approaching a Cannot Be True question. As discussed earlier, the best way to handle it would be to realize that L and Y cannot be in the same room, using a single local diagram.
Answer choice (A) is the correct answer choice. See discussion above.
Answer choice (B) is incorrect, because M and X could both be in the Rose Room (2B).
Answer choice (C) is incorrect, because X and Y could both be in either room (1) and (2A).
Answer choice (D) is incorrect, because X and Z could both be in either room (1) and (2A).
Answer choice (E) is incorrect, because Y and Z could both be in either room (1) and (2A).
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8560)
The correct answer choice is (A)
If X is at 3 PM, and only one speech can be given at 3 PM, our local diagram should look like this:
According to the first rule, M and L must be given in the same room, and M must be earlier than L. With the 3 PM slot occupied, M must given at 1 PM. Since Z must also be at 1 PM (main inference), we can establish a Dual Option for M and Z in each of the two rooms:
The remaining two variables—L and Y—must therefore be given at 2 PM:
Although the setup above does not take into account the implications of the last rule, it is sufficient to prove that L and Y cannot be given in the same room. This validates answer choice (A) as the correct answer to this Cannot Be True question.
A somewhat more thorough—but less efficient—approach would be to setup each solution individually. To do that, we first need to make two local diagrams for X and E, since X can be at 3 PM in either room:
From the contrapositive of the last rule, we know that if X is in the Gold Room, then L must be in the Rose Room, with M earlier than L (first rule). Thus, the first local setup (above) is fully determined:
In the second setup, however, the placement of X at 3 PM in the Rose Room does not represent a similar point of restriction (to conclude otherwise would be a Mistaken Reversal of the last rule) . Consequently, in the second setup M and L can be in either room:
With these three setups in place, it would be exceptionally easy to prove which four answer choices could be true. The process of elimination, however, is not the most efficient way of approaching a Cannot Be True question. As discussed earlier, the best way to handle it would be to realize that L and Y cannot be in the same room, using a single local diagram.
Answer choice (A) is the correct answer choice. See discussion above.
Answer choice (B) is incorrect, because M and X could both be in the Rose Room (2B).
Answer choice (C) is incorrect, because X and Y could both be in either room (1) and (2A).
Answer choice (D) is incorrect, because X and Z could both be in either room (1) and (2A).
Answer choice (E) is incorrect, because Y and Z could both be in either room (1) and (2A).
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