#15 - Local, Must Be True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4635)

The correct answer choice is (A)

If no employee attends H, then the 2-2-2-2-0 distribution is in effect, and each of the other talks are attended by exactly two people.

In this scenario, F must have two attendees, and the only two employees that can possibly attend F are R and S. Accordingly, answer choice (A) is correct.

[cojokeefe]

Hi,

I am really confused on Logic Games Game 3 on the December 2010 LSAT practice test. Can you explain on Question 15 why A is correct? Wouldn't D have to be correct as well?

Thanks!
Courtney

[Jon Denning]

Hi Courtney - thanks for the question. For 15, if no one attends H, then that is the 2-2-2-2 distribution. Now think about F: since Q cannot go there, then you also know T cannot go there (that would be T's first attendance and Q goes with T at T's first attendance). So the 2 people at F must be R and S (answer choice A).

Answer choice D does not have to be true. Consider this arrangement for F, G, I, and L: RS, QS, QT, and RT. That is a perfectly acceptable situation with T NOT attending G.

Hope that helps!

[cojokeefe]

Hi Jon,

Thanks so much! That makes way more sense now!

Courtney