bp shinners wrote:nmop_apisdn wrote:While doing a bundle of MBT questions, I missed PT11-S2-Q12.
This isn't the best example of these types of questions, as this one relies on more than your ability to combine quantified statements.
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.
Thank you! I will definitely review your response more thoroughly once I am home.