September 2016 LSAT
Section 3
Question 23

September 2016 LSAT
Section 3
Question 23

Replies

BenMingov on December 18, 2019

Hi Ryan, thanks for the question!The idea is that if P is the only computer that transmits to two others, then it must transmit to both R and S. This is because R and S must get the virus from the same computer.

Then we have two options that can give the virus to P. This is computers T and U. We can create two diagrams. One in which T - P. One in which U - P.

So now we have the following:

T - P - R and S, (in this scenario, R must give the virus to Q because T cannot anymore. So R gives it to Q and S gives it to U)

U - P - R and S (in this scenario, S must give the virus to T because it cannot give it to the leftover Q. The remaining Q can receive the virus from R or T)

This gives us 3 diagrams to answer our question.

T - P - R - Q

- S - U

U - P - R - Q

- S - T

U - P - R

- S - T - Q

In all scenario, Q is last (answer choice C)

Answer choice A is incorrect because it is possible that S transmits the virus to U. (First diagram).

I hope this helps. Please let me know if you have any other questions!

saskipper on June 24, 2020

How do you know that a computer must give it to P? Couldn't there be more diagrams if P was the first infected? or am I overlooking a rule here?saskipper on June 24, 2020

Nevermind.GET $100