Quick Question PT 41 S3 Q10

Prepare for the LSAT or discuss it with others in this forum.
User avatar
New_Spice180

Bronze
Posts: 127
Joined: Mon Feb 08, 2016 11:01 am

Quick Question PT 41 S3 Q10

Postby New_Spice180 » Thu Nov 03, 2016 8:42 pm

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!

User avatar
nimbus cloud

Silver
Posts: 534
Joined: Wed Jan 28, 2015 2:19 am

Re: Quick Question PT 41 S3 Q10

Postby nimbus cloud » Thu Nov 03, 2016 10:27 pm

.
Last edited by nimbus cloud on Sun Mar 19, 2017 1:34 pm, edited 2 times in total.

User avatar
New_Spice180

Bronze
Posts: 127
Joined: Mon Feb 08, 2016 11:01 am

Re: Quick Question PT 41 S3 Q10

Postby New_Spice180 » Fri Nov 04, 2016 11:40 am

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!



Return to “LSAT Prep and Discussion Forum�

Who is online

Users browsing this forum: No registered users and 13 guests