Formal Logic: Some and Most Combinations

[lsat_novice]

Page 452-453 of the Logical Reasoning Bible says that A ("most arrow" pointing toward A) B C leads to "some A's are C's"...in other words, A C.

This makes sense to me because I assumed that A ("most arrow" pointing toward A) B C
= A B C
= A C.

BUT page 452 also says that A B C does not yield an inference.

May someone please explain? Thank you!

P.S. Sorry if this is confusing because of the way it's written but I didn't see an icon for the Most Arrow pointing to the left.

[Dave Killoran]

Hi Novice,

I understood you just fine

At the bottom of page 456, there's a note that says:

• 5. The only relationship with two Mosts that will produce an inference is A <--m-- B --m--> C. The inference is A C.

On the LSAT, this is the only way you will see a double-most configuration that yields an inference. Any other combination of Some or Most produces no inference. If you keep that quick rule in mind, you won't run into any issues on test day.

The reason the above works is because the two Mosts both originate from the exact same group (B, in the case above). Imagine the following two statements:

• Most doctors are smart. ( D S )
  Most doctors are wealthy. ( D W )

If there are 100 doctors, then at least 51 are smart and at least 51 are wealthy. So, there has to be an overlap there, leading to the inference that Some smart doctors are wealthy.

Now, let's change it to:

• Some doctors are smart. ( D S )
  Most doctors are wealthy. ( D W )

If there are 100 doctors, then at least 1 is smart and at least 51 are wealthy. Does that scenario force the two groups to overlap? No. They could, but they don't have to, so no inference.

Does that help a bit? Please let me know. Thanks!

[lsat_novice]

Yes, thank you so much!