PT71.S3.Q11 - Secondary school students achieve broad mastery

youbbyun

So for this tricky SA question ...

I was looking for a statement that would say “D”

A and D ——> M
Not M ——-> Not A or Not D
Conclusion: Not M --> Not A

Would a way to conclude Not A is if we have D?

Would this be logically sound?

Can another right AC be just D? Why did they decide to go with A -> D?

Any explanation would be greatly appreciated.

Thanks!

Admin note: edited title and added link
https://7sage.com/lsat_explanations/lsat-71-section-3-question-11/

Cant Get Right

Yeah, you're right that establishing D would result in /A. We can't have both of them, so if we establish that we do have D, we must kick A out. That would be sufficient, but it's not the only way. What they chose to do instead is to connect the two in a way that no matter which one you don't have, you don't have A. If you kick A out, your conclusion is good. What's the alternative? Kicking out D. So the correct AC gives us if not D then not A. Now, no matter which one we choose to boot, we're covered. If A is out then A is out. And if D is out, A is out.

It's a clever question, not quite as cookie cutter as we're used to. That's becoming something of a hallmark of the contemporary test though, so definitely worth studying.