Re: bp shinners’ semi-weekly office hours
Posted: Tue Jun 26, 2012 10:35 pm
Thank you! I will definitely review your response more thoroughly once I am home.bp shinners wrote:This isn't the best example of these types of questions, as this one relies on more than your ability to combine quantified statements.nmop_apisdn wrote:While doing a bundle of MBT questions, I missed PT11-S2-Q12.
However, in general, my rules for combining quantifiers:
the SUFFICIENT condition of your STRONGEST statement must be SHARED. (Lot of S's for mnemonic effect). You usually end up with a SOME statement (one exception, noted below). Again, the S repeats: easy to remember - Sufficient of Strongest gives you a Some. If both statements are of equal strength, you need both to have the same sufficient condition.
You can't combine a Most and a Some or two Somes. Ever.
The one exception is if the sufficient condition of an All statement overlaps with the Necessary condition of a Most.
Most Michael Bay movies have Ben Affleck.
All Ben Affleck films are terrible.
I have a most and an all; the sufficient of the stronger is shared; however, I end up with a most statement:
Most Michael Bay movies are terrible.
I can get this stronger because of the way it's set up; let's make up some numbers.
Michael Bay has made exactly 100 movies. Based on my first statement, at least 51 feature Ben Affleck. Of those 51+ Michael Bay movies featuring Ben Affleck, 100% (i.e. all 51+) are terrible. That means that of Michael Bay's 100 movies, at least 51+ are terrible; or Most Michael Bay movies are terrible.
Learning these can suck, and it is going to require a bit of memorization. If you remember that the sufficient of the strongest has to be shared to give you a sum (with an asterisk for the situation that gives you a most), you're more than halfway there.
Now, on to that question.
Here, I'm told that all T->Older than M
S-m->Older than M
S-s->Younger than M (because of the not all - don't forget this one, as it's important)
M -> Older than D
I don't share terms here in a way that would allow me to combine them, but my Older than M is comparative to M, which does let me combine them, to a certain extent. This is actually closer to ordering rules in a Logic Game than it is a LR question. If I'm arranging them from oldest to youngest, I'd have:
T and some S are older than M is older than D and some S
A) MBF - All T older than all M which is older than all D; some D can't be as old as any T
B) Could be true - I know that some sycamores and all dogwoods are younger than some maples; however, I don't get a comparison ever between those young S and the D, so I can't say B for sure.
C) Again, CBT for the same reason as B. This and B are essentially the same answer, which should have been a tipoff that neither are correct.
D) CBT again. And for the same reasons as B and C. I never compare these groups (oldest S and T), so I can't say this for sure.
E) Now, I've got something. Look back at my chain. I have T older than M and some S younger than M. That means that some S have to be younger than T. I have compared these groups because I've compared them both to M, and they fell on different sides. All the other comparisons that could be true (B C and D) talked about things on the same side of M on my chain.