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(The complete setup for this game can be found here: lsat/viewtopic.php?t=12912)
The correct answer choice is (A)
If exactly four variables are assigned, our solution must adhere to Template 2A (Template 1 uses at least five variables, whereas Template 2B uses all six). The stem asks us to identify which variable must be assigned to Silva. According to Template 2A, the correct answer choice should be either F or H.
This question can also be solved without the use of templates, although it would take a bit more work. If exactly four variables are assigned, the numerical distribution would be fixed at exactly two variables per group (2-2). We already know three of the variables that must always be assigned—F, H, and K—leaving us room for only one more variable.
It is clear that if K is not assigned to Thorne, the solution would violate the 2-2 distribution required by the stem. This is because if K is not assigned to Thorne, we would need to assign at least three variables to Thorne in compliance with the first and the fourth rules:
Clearly, satisfying the necessary condition would be impossible in a 2-2 distribution, forcing K to be assigned to Thorne. The FH-block, meanwhile, must be assigned to Silva:
Answer choice (A) is the correct answer choice, because F must be assigned to Silva. See Template 2A, as well as the local solution above.
Answer choice (B) must be false. If G is assigned to Silva, then L must be assigned to Thorne. Since F, H, and K must always be assigned (main inference), this answer choice would require us to assign at least five variables to the two ceremonies. Therefore, G cannot be assigned to Silva.
Answer choice (C) must be false. If K is assigned to Silva, then F, H, and M must all be assigned to Thorne (first and fourth rules). K cannot be alone in her group, forcing us to use at least one other variable, for a total of at least five. Therefore, K cannot be assigned to Silva.
Answer choice (D) must also be false, because in a 2-2 distribution L can only be assigned to Thorne, not Silva. See Template 2A. If L were assigned to Silva, then the 2-2 distribution would force K to be assigned to Silva as well, with F and H be assigned to Thorne. This would violate the provisions of the last rule, proving that L cannot be assigned to Silva.
Answer choice (E) is also false. If M is assigned to Silva, then K must be assigned to Thorne in accordance with the contrapositive of the fourth rule. Since F and H must be together, this assignment would violate the 2-2 distribution required by the question stem. Therefore, M cannot be assigned to Silva.