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- Wed Jan 11, 2017 6:18 pm
#32019
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15203)
The correct answer choice is (C)
If V and Z are assigned to the same group as each other, that group cannot be P. Why? Because Z and Y must both be assigned to P if either one of them is (last rule), but V cannot be assigned to the same group as Y (third rule). Therefore, V and Z cannot be assigned to P.
Given that the VZ block can be assigned to either L or M, it is best to quickly create two separate local diagrams in solving this question. Note that Y cannot be assigned to P in either solution in compliance with the last rule, nor can it be assigned to the same group as V. Therefore, Y must also be assigned to either L or M, alternating away from the VZ block.
Note that in Local Setup 1, the ST block (second rule) can only be assigned to P, with the remaining variable (R) assigned to M. The same limitation does not apply to Local Setup 2, where the ST block can be assigned to either L or P, as can R. All three solutions are represented below (you may not need to sketch each of these out on the test, but we show them here for the sake of clarity):
(Note that in Local Setup 2b, R can be in EITHER L or P - we have placed it in L just to show that it can go there, as Local Setup 2a already showed it could be in P)
Answer choice (A): This answer choice is incorrect, because R can be assigned to L (Local Setup 2b).
Answer choice (B): This answer choice is incorrect, because S can be assigned to L (Local Setup 2a).
Answer choice (C): This is the correct answer choice, because the ST block cannot be assigned to M in any of the three solutions. Filling M with S and T would force either Y or V and Z into P (as Y and P strive to avoid each other), meaning all three of Y, V, and Z would then be assigned to P (from the final rule). This inevitably places Y and V together, which cannot occur (rule 3).
Answer choice (D): This answer choice is incorrect, because R can be assigned to P (Local Setup 2a).
Answer choice (E): This answer choice is incorrect, because S can be assigned to P (Local Setups 1 and 2a).
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15203)
The correct answer choice is (C)
If V and Z are assigned to the same group as each other, that group cannot be P. Why? Because Z and Y must both be assigned to P if either one of them is (last rule), but V cannot be assigned to the same group as Y (third rule). Therefore, V and Z cannot be assigned to P.
Given that the VZ block can be assigned to either L or M, it is best to quickly create two separate local diagrams in solving this question. Note that Y cannot be assigned to P in either solution in compliance with the last rule, nor can it be assigned to the same group as V. Therefore, Y must also be assigned to either L or M, alternating away from the VZ block.
Note that in Local Setup 1, the ST block (second rule) can only be assigned to P, with the remaining variable (R) assigned to M. The same limitation does not apply to Local Setup 2, where the ST block can be assigned to either L or P, as can R. All three solutions are represented below (you may not need to sketch each of these out on the test, but we show them here for the sake of clarity):
(Note that in Local Setup 2b, R can be in EITHER L or P - we have placed it in L just to show that it can go there, as Local Setup 2a already showed it could be in P)
Answer choice (A): This answer choice is incorrect, because R can be assigned to L (Local Setup 2b).
Answer choice (B): This answer choice is incorrect, because S can be assigned to L (Local Setup 2a).
Answer choice (C): This is the correct answer choice, because the ST block cannot be assigned to M in any of the three solutions. Filling M with S and T would force either Y or V and Z into P (as Y and P strive to avoid each other), meaning all three of Y, V, and Z would then be assigned to P (from the final rule). This inevitably places Y and V together, which cannot occur (rule 3).
Answer choice (D): This answer choice is incorrect, because R can be assigned to P (Local Setup 2a).
Answer choice (E): This answer choice is incorrect, because S can be assigned to P (Local Setups 1 and 2a).
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