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- Wed Jan 11, 2017 6:18 pm
#32018
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15203)
The correct answer choice is (D)
If R is alone in her group, we know we have a 3-2-1 distribution, with R as the single, and group M as the double. We also know that R’s group cannot be M for the same reason. Thus, R must be assigned to either L or P. This may once again warrant two separate local diagrams, as shown below.
In either solution, we need to ensure that Y and V are kept in separate groups. Consequently, either Y or V must be assigned to M, and thus the ST block cannot be assigned to M:
It is now time to consider the implications of the last rule: Z and Y must both be assigned to P if either one of them is. Placing either Y or Z into P would be impossible in either solution, as it would exceed the number of variables allowed for P (Local Setup 1, where P would have four people) or with R (Local Setup 2, since R must be alone in this question).
Accordingly, neither Z nor Y can be assigned to P, forcing Z to always be assigned to M:
To answer this Must Be True question, focus on the consistency between the two solutions: in both of them Z is assigned to M, confirming that answer choice (D) is correct.
Answer choice (A): This answer choice is incorrect, because the ST block can also be assigned to P.
Answer choice (B): This answer choice is incorrect, because V can be assigned to any of the three groups.
Answer choice (C): This answer choice is incorrect, because Y can also be assigned to L.
Answer choice (D): This is the correct answer choice.
Answer choice (E): This answer choice is incorrect, because the ST block can be assigned to either L or P.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15203)
The correct answer choice is (D)
If R is alone in her group, we know we have a 3-2-1 distribution, with R as the single, and group M as the double. We also know that R’s group cannot be M for the same reason. Thus, R must be assigned to either L or P. This may once again warrant two separate local diagrams, as shown below.
In either solution, we need to ensure that Y and V are kept in separate groups. Consequently, either Y or V must be assigned to M, and thus the ST block cannot be assigned to M:
It is now time to consider the implications of the last rule: Z and Y must both be assigned to P if either one of them is. Placing either Y or Z into P would be impossible in either solution, as it would exceed the number of variables allowed for P (Local Setup 1, where P would have four people) or with R (Local Setup 2, since R must be alone in this question).
Accordingly, neither Z nor Y can be assigned to P, forcing Z to always be assigned to M:
To answer this Must Be True question, focus on the consistency between the two solutions: in both of them Z is assigned to M, confirming that answer choice (D) is correct.
Answer choice (A): This answer choice is incorrect, because the ST block can also be assigned to P.
Answer choice (B): This answer choice is incorrect, because V can be assigned to any of the three groups.
Answer choice (C): This answer choice is incorrect, because Y can also be assigned to L.
Answer choice (D): This is the correct answer choice.
Answer choice (E): This answer choice is incorrect, because the ST block can be assigned to either L or P.
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