- PowerScore Staff
- Posts: 8950
- Joined: Feb 02, 2011
- Tue Jul 03, 2018 1:32 pm
#47391
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4635)
The correct answer choice is (E)
This question is similar to question #17, with the exception that R is now the only employee to attend I.
Again the 2-2-2-1-1 distribution applies, and the inference pattern follows a somewhat similar route to question #17.
H can only have 0 or 1 attendees, and so H is the other talk with just one attendee (which must be S or T), and F, G, and I each have two attendees:
F, G, and L each have two attendees, and thus R and S must attend F (they are the only two possible candidates):
If G has two attendees, then it must be either Q and S or Q and T. If it were Q and S, then S would attend both F and G, leaving T to attend H. But, this causes a violation of the fourth rule because there is no room for Q to attend H as well. Thus, Q and S cannot attend G in this question, and instead Q and T must be the attendees at G:
At this point, the attendees of F, G, and I are fixed, leaving only H and L still unresolved. H must be attended by S or T, but there is no way to determine which one attends. The remainder of S/T attends L, along with Q, who has not yet been assigned to a second talk. This leads to the final diagram for this question:
The only uncertainty involves S and T, and so, as in the last question you should immediately examine the answer choices that contain S or T (again, as in the prior question, these answers are (C), (D), and (E). This is not surprising because the test makers want you to have to work through more answer choices as you approach the end of a game). Because T can attend L instead of H, answer choice (E) is not necessarily true and is therefore correct.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4635)
The correct answer choice is (E)
This question is similar to question #17, with the exception that R is now the only employee to attend I.
Again the 2-2-2-1-1 distribution applies, and the inference pattern follows a somewhat similar route to question #17.
H can only have 0 or 1 attendees, and so H is the other talk with just one attendee (which must be S or T), and F, G, and I each have two attendees:
F, G, and L each have two attendees, and thus R and S must attend F (they are the only two possible candidates):
If G has two attendees, then it must be either Q and S or Q and T. If it were Q and S, then S would attend both F and G, leaving T to attend H. But, this causes a violation of the fourth rule because there is no room for Q to attend H as well. Thus, Q and S cannot attend G in this question, and instead Q and T must be the attendees at G:
At this point, the attendees of F, G, and I are fixed, leaving only H and L still unresolved. H must be attended by S or T, but there is no way to determine which one attends. The remainder of S/T attends L, along with Q, who has not yet been assigned to a second talk. This leads to the final diagram for this question:
The only uncertainty involves S and T, and so, as in the last question you should immediately examine the answer choices that contain S or T (again, as in the prior question, these answers are (C), (D), and (E). This is not surprising because the test makers want you to have to work through more answer choices as you approach the end of a game). Because T can attend L instead of H, answer choice (E) is not necessarily true and is therefore correct.
You do not have the required permissions to view the files attached to this post.