LSAT Logic Games Answers

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1. C.

To answer this question, simply compare each rule to the answer choices, and cross off any choice that violates the rule. Rule #1 is violated by answer choice B, so B cannot be correct. Rule #3 is violated by answer choice A, so A cannot be correct. Rule #4 is violated by answer choice E, so E cannot be correct. Rule #5 is violated by answer choice D, so D cannot be correct. This leaves only answer choice C, which must be the correct answer.

2. A.

Begin by drawing seven dashes and placing M on the third dash. We know that O must be next to M, and cannot go into period 4, since period 4 must be either S or P.  Thus O must be placed in period 2 and S must be placed in period 1. This will leave P to be placed in period 4. Since U and T cannot be in consecutive periods, they will need to go into spots 5 and 7 in either order, leaving N for period 6.

3. B.

Choice A is incorrect because S must be scheduled before M. Choice C is incorrect because S must be scheduled before M, O and U. Choice D is incorrect because if T is placed in 6, there will be no valid space left for U. Choice E in incorrect because S, M and O must be scheduled before U.

4. E.

Begin by drawing seven dashes and placing N on the third dash. We know that M and O will need to go on the fifth and sixth dashes. (They cannot occupy the fourth period, since that period must be reserved for either S or P. Neither can they occupy the sixth and seventh periods, since U must be scheduled after them. They also cannot be scheduled for the first and second periods, since S must be scheduled before them.) Since M and O must be scheduled – in either order – for the fifth and sixth periods, U will have to be scheduled for the seventh period. Thus, U cannot be scheduled for the sixth period.

5. E.

Begin by drawing seven dashes and placing T on the seventh dash. If T is the seventh period, S cannot be the fourth period, because there would not be enough room to schedule M, O, and U after S but before T. Thus, P must be the fourth period. The only periods now available to M and O are second and third. (They cannot occupy the fifth and sixth periods, since there would be no room to place U, and they also cannot be scheduled for the first and second periods, since S must be scheduled before them.) This will force S into the first period. Since U and T cannot occupy consecutive periods, U must be in period 5, the only remaining open period that is not consecutive with T’s period.