LG Game 4 Question 21 Explanation, by LSATHacks
This question asks who can’t transfer to two computers. Here’s the setup for quick reference:
This is the question I got wrong in timed conditions. Looking at it again, I noticed a shortcut: anything that can go before R/S can make two transfers.
And, prima facie, there’s no obvious reason any of T, U, Q and P can’t transfer to R/S. Which leaves R and S for our consideration. (Maybe T, U, Q and P couldn’t transfer twice, but they shouldn’t be our first suspects)
S obviously can only transfer to one computer. But, sadly it isn’t an available answer. This leaves us with R. Here’s what the diagram looks like if R makes two transfers:
So, this requires an addition to the diagram. Right away, that should make us think we’re on the right track. Any addition makes things more rigid. Notice also that when you count letters + unknown slots, 6 computers are listed. So, placing two after R requires a fixed structure.
Let’s see if we can fill it in though. Start by fulfilling rules. For example, Q needs R or T before it. So, let’s put Q after R:
The problem comes when we consider the next rule: P needs T or U before it. That would require two open slots back to back. We don’t have that, so there’s no way to fit in T – U or P – U.
So, C is CORRECT.
Since this is a hard question, here are diagrams to disprove the other answers.
This diagram disproves A:
This diagram disproves B and D. As drawn, it shows that T can transmit to R and S. Q and T are almost interchangeable, because rule 4 allows either of them to transfer to R. We just need one of T or R to transmit to Q. So move T one to the left, have T transmit to Q, and then Q transmits to R and S:
This diagram disproves E:
Hi Graeme, for B in this question, I think your reasoning might be slightly off, you said “Q and T are interchangeable, because rule 4 allows either of them to transfer to R. So, you can swap T/Q.”
If you swap T/Q you end up with Q as the originating computer which is not possible under rule 4 (you have to have R-Q or T-Q).
The diagram I ended up with looked almost the same as yours, you just have to bump T one position to the left, and have Q infect R/S. S then infects U which infects P, and R doesn’t infect anything.
You’re quite right, thank you for catching that. I’ll update the explanation.
Awesome! Thanks Graeme. I got a 163 on the January 2023 LSAT as a first time writer thanks to your website that I am quite happy with, thank you so much for all that you do!
Incredible, congratulations! Good luck with applications :)
Sorry, you can delete that comment – I see that T can pass to Q and then Q can pass to R and S. The wording was just confusing me.
I don’t see how your diagram disproves option B. How can Q and T be interchangeable in your diagram? The rules tell us that Q must first receive the virus from either R or T. So how can Q ever give the virus twice?
How is “B” not also a correct answer? I can’t produce I diagram where “Q” transmits to 2 computers without violating a rule. Please show me the diagram where “Q” transmits to 2 other computers without violating a rule. Thanks so much!