Yes, if most A are B AND most A are C, then some B are C would be correct. That's different from the original question, at least the way I understood it.
I'm not sure I understand the original premise correctly, but if it's "most A's are B's" and "most B's are C's", then we can't say absolutely anything about the relationship between A and C.
You can have, say, 5 of the A's. 3 of those (most) are B's. But what if there are a total of 100 B's, of which 51 (most) are C's, there's no requirement for those 51 to include the 3 A's.
Example: most young women in China (A) are bike riders (B). Most bike riders (B) in China are men (C). These can both be correct, as we know there are more men than young women in China. Yet the conclusion "some young women in China are men" (some A are C) is clearly not correct.
Subscribe to unlock everything that 7Sage has to offer.
Hold on there, stranger! You need a free account for that.
We love that you want to get going. Just create a free account below—it only takes a minute—and then you can continue!
Hold on there, stranger! You need a free account for that.
We love that you came here to read all the amazing posts from our 300,000+ members. They all have accounts too! Just create a free account below—it only takes a minute—and then you’re free to discuss anything!
Hold on there, stranger! You need a free account for that.
We love that you want to give us feedback! Just create a free account below—it only takes a minute—and then you’re free to vote on this!
Subscribers can learn all the LSAT secrets.
Happens all the time: now that you've had a taste of the lessons, you just can't stop -- and you don't have to! Click the button.
14 comments
@abhere294 youre right. When I wrote that, I was referring to OP example. I should've been more clear. My fault.
oh yes thanks just saw :)
True....different intersection than OP
Yes, if most A are B AND most A are C, then some B are C would be correct. That's different from the original question, at least the way I understood it.
Well, actually if there was A most B and A most C, then B some C could be concluded.
not correct...please check video above
JY says..." A--most-> B-most->C ....... No valid conclusions!"
http://classic.7sage.com/lesson/invalid-argument-form-6-of-7/
I'm not sure I understand the original premise correctly, but if it's "most A's are B's" and "most B's are C's", then we can't say absolutely anything about the relationship between A and C.
You can have, say, 5 of the A's. 3 of those (most) are B's. But what if there are a total of 100 B's, of which 51 (most) are C's, there's no requirement for those 51 to include the 3 A's.
Example: most young women in China (A) are bike riders (B). Most bike riders (B) in China are men (C). These can both be correct, as we know there are more men than young women in China. Yet the conclusion "some young women in China are men" (some A are C) is clearly not correct.
Apologies if I misread the original premise.
2 most=some (some A are C)
Edit: Thanks, @nye887085 Check this out! http://classic.7sage.com/lesson/invalid-argument-form-6-of-7/
@kennedybj959 thanks for pointing me to it....
Nope. Two mosts don't form an inference.
7sage.com/lesson/7-common-invalid-argument-forms/