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Quick Question PT 41 S3 Q10
Posted: Thu Nov 03, 2016 8:42 pm
by New_Spice180
I was able to get the answer choice on this question, but the explanation of why answer C is wrong is interesting. On the Manhattan forums it states that it "must be true". The poster later states that if you are to take the statement " those who have the ability to fully concentrate are always of above average intelligence and contrapose it you get AAI (some)---> ~FC. I'm not sure as to where the "some came from is it from negating all to "not all"? Further clarification would be appreciated!
Re: Quick Question PT 41 S3 Q10
Posted: Thu Nov 03, 2016 10:27 pm
by nimbus cloud
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Re: Quick Question PT 41 S3 Q10
Posted: Fri Nov 04, 2016 11:40 am
by New_Spice180
nimbus cloud wrote:Let me try. So we know that: FC -> AAI.
From this we can infer that AAI some FC. But it is possible that AAI many or even most ~FC.
Analogy: Those who have the ability to solve differential equations are of above-average intelligence. However, it could be (and is) true that many people of above-average intelligence are unable to solve them (and I am sure you know them).
So I disagree that it MBT, but the question is asking what CBT, and yes, "C" could be true.
The correct answer on the other hand directly contradicts FC --> AAI. It says FC some ~AAI.
That was my reasoning initially. I knew that it certainly COULD BE TRUE not that it must be true given the premises. Anyway, thanks for reinforcing my reasoning!