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 lichenfarmer
  • Posts: 7
  • Joined: May 01, 2020
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#75163
For this question, the diagram is as shown:

I :most: J :dbl: K :arrow: L

One of the inferences I made that wasn't listed was:

L :some: J

I'm not sure why this wouldn't be included as it seems logically correct to me. If all K's are L's, then some L's are K's. If all K's are J's (bound by a double arrow), then wouldn't some L's be J's?

What is the key thing I'm missing in this case?
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 Dave Killoran
PowerScore Staff
  • PowerScore Staff
  • Posts: 5972
  • Joined: Mar 25, 2011
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#75195
Hi lichenfarmer,

Thanks for the question! That's not listed because it's an inherent inference of J :arrow: L, and the drill notes that inherent inferences won't be charted out in the answer key :-D This has come up before, though, and so in the answer key I added the following note on the inference list:

  • J :arrow: L (This inherently includes the inference L some J)
So, you can actually see the reference there to L :some: J, which confirms you were correct in that relationship.

Please let me know if that helps. Thanks!
 lichenfarmer
  • Posts: 7
  • Joined: May 01, 2020
|
#75208
Yes, I must've missed that! Thanks so much! :)

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