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September 2016 LSAT
If T did not transmit the virus to any other computer on the network, which one of the following must be true?
on December 17, 2019
Why is A correct? Why is D incorrect?
on December 21, 2019
@Ryan-Mahabir, happy to help! This is a difficult game.
First, let's look at what we can deduce from the setup of the game.
We know that S must transmit to exactly one other computer. It can't transmit the virus to itself, it can't transmit it to R since it receives the virus from the same computer that R does, it can't transmit it to Q because of Rule 4, and it can't transmit it to P because of Rule 4. This means that S can only transmit the virus to either T or U, as these are the only remaining options.
T and U are also the only options that can start the chain, as R, S, Q, and P all have rules that specify how they receive the virus from other computers. Since they're receiving the virus from other computers in the network, they cannot be the ones to start it.
Now, let's turn to this specific question.
If T cannot transit the virus to any other computer, we know that T cannot start the chain. Therefore, U is our only option for starting the chain.
This also means that T must receive the virus from S, since U can't.
Rule R states that either T or U must transmit the virus to P, and since T cannot transmit the virus to any computers in this scenario, we know that U must transmit to P.
This is our resulting set-up:
U - P - R - Q
- S - T
(A) is the only answer choice that is true in this scenario, so (B) through (E) can be eliminated.
Does that make sense? Please reach out with any other questions!
on March 3 at 01:54AM
can you explain what happened to the rule that says R and S must have the same computer in common. this rule is not mentioned in your diagram from above. since the set up is like this
the rule is not mentioned in your diagram so how can your answer be correct.
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