December 2010 LSAT Section 4 Question 15

# If none of the SoftCorp employees attends Handling People, then which one of the following must be true?

2 Replies

Irina on February 16 at 01:43AM

@SorooshKosha,If I am understanding your question correctly, you are asking why we cannot have the following assignment of attendees to talks:

I: TR

L: QS

The short answer is that this is also a valid scenario, my guess is that the video only demonstrates one of several possible hypotheticals. In fact, for this particular question, almost any combination of Q R S T is possible for I & L as long as Q and R do not attend the same talk.

We have 5 talks total, and each employee - Q R S T - attends exactly 2 talks

R Q X Q/R R/Q

S S/T X __ __

F G H I L

For this question, no one attends H, which means every employee attends exactly two talks.

Since Q cannot attend F, and S &T cannot attend the same talk together, R must be one of the attendees, and S must attend R's first talk.

Same for G, since R cannot attend G and S&T cannot attend the same talk, one of the attendees must definitely be Q and the other one S or T.

Q and R each still have one talk to attend and they cannot attend the same talk together because it would result in ST combination for the remaining talk, which is against the rules. It leaves us with two possible scenarios:

(1) If Q &S attend the second talk, then we can determine the complete assignment of all the attendees:

R Q X Q R

S S X T T

(2) If Q & T attend the second talk, then any combination of Q R S T are all free variables for the last two talks as long as Q and R attend separate talks:

R Q X Q/R R/Q

S T X S/T T/S

Let me know if this answers your question.

Soroosh on March 11 at 02:33PM

Thanks a lot!