Linking Quantifiers Help

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CoGar

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Linking Quantifiers Help

Postby CoGar » Wed Sep 21, 2016 6:50 pm

Hey all,

I am still struggling to link/bridge different quantifiers such as "some", "most" "not all" etc. etc. and was wondering if anyone could give a brief and simple ways of when you can make deductions and when you cant. It seems like a ton of MBT questions stem from understanding these and being able to properly infer when you can connect these when they're presented in the stim.

Thanks alot

speedwagon

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Re: Linking Quantifiers Help

Postby speedwagon » Wed Sep 21, 2016 7:11 pm

Ok I think I am quoting the PS bible here which I found really did it for me.

The lineup/ladder (which is right from their list) goes like this:

All (all)
|
Most (half plus one, including all) - careful! things like "almost all" "usually" etc also rank in here
|
Some (at least 1, including all)

You can only infer down the list, ie, if it says "all" you can assume "some" and if it says "most" you can assume some but if it says some you don't know anything aside from "more than one."

Helpful?

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kindofcanuck

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Re: Linking Quantifiers Help

Postby kindofcanuck » Wed Sep 21, 2016 7:50 pm

CoGar wrote:Hey all,

I am still struggling to link/bridge different quantifiers such as "some", "most" "not all" etc. etc. and was wondering if anyone could give a brief and simple ways of when you can make deductions and when you cant. It seems like a ton of MBT questions stem from understanding these and being able to properly infer when you can connect these when they're presented in the stim.

Thanks alot


An example of a question you're struggling with would be helpful, but broadly those words will always have their common meanings. "most" is a majority. "some" is any quantity. "not all" is - well, not all, but anything up to that point. If "most" of a group have a quality and you have the entire group, then most of the ones you have have that quality. If you have some, or not all of the group, then you cannot make that inference, as your group may be drawn mostly or entirely from the subsection that didn't have that quality.

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34iplaw

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Re: Linking Quantifiers Help

Postby 34iplaw » Wed Sep 21, 2016 7:56 pm

A most B
B most C
therefore, A some C

A some B
B some C
no conclusion

A most B
B some C
no conclusion

A->B
BsomeC
no conclusion

A->B
AsomeC
BsomeC

A->B
AmostC
BsomeC

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Deardevil

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Re: Linking Quantifiers Help

Postby Deardevil » Wed Sep 21, 2016 9:09 pm

Not sure what else I can add, so here is an example from the first PrepTest.

[+] Spoiler
Some people are Montagues and some people are Capulets.
No Montague can be crossed in love.
All Capulets can be crossed in love.
Therefore, Capulets are not Montagues.
Anyone who is not a Montague is intemperate.

Assume that all of the statements in the passage are true.
If it is also true that no Montague is intemperate,
then which one of the following must be true?

(A) The only people who can be crossed in love are intemperate Capulets.
(B) Anyone who is not a Capulet is a Montague.
(C) All intemperate people can be crossed in love.
(D) All intemperate people are Capulets.
(E) All Capulets are intemperate.

Breakdown

P some M
P some C
M -> ~CL
C -> CL
~M -> IT
M -> ~IT

(A) claims that the ONLY ones capable of being crossed in love are intemperate Capulets,
but the world does not revolve around Romeo and Juliet; there are OTHER regular people who can be crossed in love, too!
(B) states that if one is not a Juliet groupie, one must be from the side of Romeo; WE, who are from neither family, still exist!
(C) says if you're intemperate, you could be crossed in love, which is straight-up WRONG;
the only thing we can conclude from the given information is that you must be a Montague (mutual exclusivity),
and all Montagues are incapable of being crossed in love.
(D) now mistakes all intemperate people to be Capulets when there is no data whatsoever to compare the two;
this may or may not be the case, and we just have no clue.
(E) has to be it, then; just connect the dots: C -> CL -> ~M (contrapositive of M -> ~CL) -> IT, which boils down to C -> IT.


This probably doesn't help, but if you can post a specific problem, you're welcome to.

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Deardevil

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Re: Linking Quantifiers Help

Postby Deardevil » Wed Sep 21, 2016 9:46 pm

I think this is a good example from PT 11.

[+] Spoiler
In the Centerville Botanical Gardens, all tulip trees are older than any maples.
A majority, but not all, of the garden's sycamores are older than any of its maples.
All the garden's maples are older than any of its dogwoods.

If the statements above are true, which one of the following must also be true of trees in the Centerville Botanical Gardens?

(A) Some dogwoods are as old as the youngest tulip trees.
(B) Some dogwoods are as old as the youngest sycamores.
(C) Some sycamores are not as old as the oldest dogwoods.
(D) Some tulip trees are not as old as the oldest sycamores.
(E) Some sycamores are not as old as the youngest tulip trees.

Breakdown

This is an interesting inference question;
almost like it was pulled from the LG section, so I'll diagram the ages of the trees as such:

    T --- M --- D
    S --- M --- S
(if a majority, but NOT ALL, sycamores are older than maples,
SOME are older while the other portion is younger, though I suppose they can also be of the same age, too)

(A) is wrong because all dogwoods are younger than maples, whereas all maples are younger than tulip trees.
(B) is wrong simply because we don't know if dogwoods are related to sycamores in terms of age.
(C) is wrong because of the reason above; no inferences can be drawn regarding these trees.
(D) is wrong for the same reason as the previous two choices.
(E) is TCR; doesn't matter if some sycamores are younger or as old as maples since all maples are younger than tulip trees,
making these sycamores "not as old" as even the youngest tulips.



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