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March 29, 2024, 04:57:09 pm

Author Topic: Verbal Reasoning  (Read 20644 times)  Share 

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che0106

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Re: Verbal Reasoning
« Reply #15 on: February 28, 2011, 09:44:39 pm »
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Its from the practice tests for selective entry high school :|
which high school?

nacho

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Re: Verbal Reasoning
« Reply #16 on: March 08, 2011, 07:52:10 pm »
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The closest answer is B:
Some Hiffs are Ziffs, but the actual answer should be that ALL Hiffs are Ziffs.
Defs not A

Edit:
Nvm, I read the post below, think i overlooked E D:
« Last Edit: March 08, 2011, 08:41:59 pm by nacho »
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funkyducky

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Re: Verbal Reasoning
« Reply #17 on: March 08, 2011, 08:24:28 pm »
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The answer is E -

If all PIFS are ZIFS, then you can consider PIF as being a subset of ZIF, as in, if you had a PIF, it would definitely be a ZIF, but the ZIF would not necessarily be a PIF, because the statements given to us are not commutative, ie. "All PIFS are ZIFS" does not imply that all ZIFS are PIFS. Hence, we can state with certainty that some ZIFS are PIFS ie. E is correct.
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simplyme

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Re: Verbal Reasoning
« Reply #18 on: April 22, 2011, 06:17:57 pm »
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I am pretty sure the answer, A, is incorrect, and A B C D E are all incorrect.

simplyme

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Re: Verbal Reasoning
« Reply #19 on: April 22, 2011, 06:21:36 pm »
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My German cousin was trying to help (FYI he's a mathematician :buck2: :buck2:)....There is something confusing to me, if you can make sense out of it..ENJOY LOL


About this Question 20:  

All HIFS are PIFS: becomes in short form    H>>a>>P
Some SIFS are RIFS: becomes in short form  S>>e>>R
All ZIFS are SIFS: becomes in short form Z>>a>>S
All PIFS are ZIFS: becomes in short form P>>e>>Z.

Now i start with  H>>a>>P and search for other connections that either star with P or ends with H and connect them with each other:  H>>a>>P>>a>>Z>>a>>S>>e>>R
in this case they are all connected to a straight line.

But it could also be possible, that we have for example P>>a>>Z , P>>e>>S , P>>a>>R, P>>e>>H
in this case you would draw four arrows from P to the other sets.

Which of the following do we know is correct.

All ZIFS are RIFS? Z>>a>>R?  We know: Z>>a>>S>>e>>R, so there could be one ZIF that is also a SIF but this ZIF might not be a RIF as we only know that there exist some (but not all) SIFS that are also RIFS.
The "rules of connections" state that from Z>>a>>S>>e>>R follows Z>>e>>R. We can not follow that Z>>a>>R.

Some HIFS are ZIFS?  H>>e>>Z? From H>>a>>P>>a>>Z follows H>>a>>Z and H>>e>>Z.
So this is true.

In some sense >>e>> is weaker than >>a>>:
If we have a very long chain of for example 200 elements
M1>>a>>M2>>a>>M3....>>a>>M200
gives  always a connection M1>>a>>M200 and also M1>>e>>M200.

But if only one connection of the chain is  >>e>> then the whole >>a>>..>>a>>.. chain is destroyed by this >>e>>  and we only get M1>>e>>M2.
All RIFS are SIFS?: R>>a>>S? We see that R is the last element of this chain. So this is not true.
All HIFS are RIFS? H>>a>>R? H>>a>>P>>a>>Z>>a>>S>>e>>R we see the >>a>> chain is broken by S>>e>>R so this is not true.
Some ZIFS are PIFS? Z>>a>>P?  We see P>>a>>Z so we can change the orientation by exchanging a by e:
P<<e<<Z so this is true.
« Last Edit: April 22, 2011, 06:28:43 pm by simplyme »

simplyme

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Re: Verbal Reasoning
« Reply #20 on: April 24, 2011, 04:02:47 pm »
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this is my take...

E - Some ZIFS are PIFS is valid conditionally
All HIFS are PIFS
All PIFS are ZIFS

however we can't say some ZIFS are PIFS - it will not be universally true- as there are, also, the other members or a member of ZIFS which does not belongs to PIFS.

________________________________________________________________________________________

B - SOME HIFS are ZIFS could be most possible correct answer.

ALL HIFS are PIFS
ALL PIFS are ZIFS

Of course, ALL HIFS are ZIFS but I am not sure if it's acceptable to say SOME HIFS are ZIFS

in my concluded view, B should be the closet answer... :laugh: :D :P

« Last Edit: April 24, 2011, 09:35:22 pm by simplyme »

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Re: Verbal Reasoning
« Reply #21 on: April 24, 2011, 06:32:40 pm »
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Let me know when someone finds out the answer for certain.

dilks

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Re: Verbal Reasoning
« Reply #22 on: February 26, 2012, 01:00:02 pm »
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Q) Assume that:
All HIFS are PIFS
Some SIFS are RIFS
All ZIFS are SIFS
All PIFS are ZIFS

The Long Method

First determine the following thereoms:

All HIFS are PIFS
All PIFS are ZIFS
All ZIFS are SIFS
Therefore:
All HIFS are ZIFS
All HIFS are SIFS

Some SIFS are RIFS
Therefore:
Some HIFS are RIFS
Some PIFS are RIFS

That leaves us with these thereoms:

All HIFS are PIFS
All PIFS are ZIFS
All ZIFS are SIFS
All HIFS are ZIFS
All HIFS are SIFS
Some SIFS are RIFS
Some HIFS are RIFS
Some PIFS are RIFS

Now:
Some SIFS are RIFS
Some HIFS are RIFS
Some PIFS are RIFS
Therefore the reverse is true:
Some RIFS are SIFS
Some RIFS are HIFS
Some RIFS are PIFS

A: All ZIFS are RIFS
B: Some HIFS are ZIFS
C: All RIFS are SIFS
D: All HIFS are RIFS
E: Some ZIFS are PIFS

We can now eliminate answers B through D, because these all contradict our thereoms which leaves us with two potential answers, A and E.

All RIFS that are HIFS are also PIFS
All RIFS that are HIFS are also ZIFS
All RIFS that are HIFS are also SIFS
All RIFS that are PIFS are also ZIFS
All RIFS that are PIFS are also SIFS
All RIFS that are ZIFS are also SIFS

Therefore:
Some but not all PIFS are HIFS
Some but not all ZIFS are HIFS
Some but not all SIFS are HIFS
Some but not all ZIFS are PIFS
Some but not all SIFS are PIFS
Some but not all SIFS are ZIFS

Therefore  A is wrong and E is right.

"Some but not all ZIFS are PIFS"
E: Some ZIFS are PIFS

Part 2: Doing this kind of problem quickly.
As a matter of fact it is actually really intuitive that the answer would be E, because of the answers provided, and the nature of the relationships between each element. First of all let's rearrange the starting functions in chronological order.

All HIFS are PIFS
All PIFS are ZIFS
All ZIFS are SIFS
Some SIFS are RIFS

First of all, we know that it is impossible for All HIFS to also be PIFS, ZIFS and SIFS, because only some SIFS are RIFS and since all elements are contained in the set called RIFS, that would constitute a contradiction.

Second of all we know that the system is hierarchical: HIFS come before PIFS which come before SIFS which come before RIFS.

So we know from the start that it would be impossible for All PIFS to be HIFS or something like that, because it comes below it in the hierachy. We don't even need to prove that E is right because it is the only answer which conforms to the hierarchial modelling of the system. Because it says that only some of the lower tier elements will be higher tier elements.

Also re-reading the post the correct reasoning is given pretty cogently here:

The answer is E -

If all PIFS are ZIFS, then you can consider PIF as being a subset of ZIF, as in, if you had a PIF, it would definitely be a ZIF, but the ZIF would not necessarily be a PIF, because the statements given to us are not commutative, ie. "All PIFS are ZIFS" does not imply that all ZIFS are PIFS. Hence, we can state with certainty that some ZIFS are PIFS ie. E is correct.
« Last Edit: February 26, 2012, 08:13:38 pm by dilks »
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Shenz0r

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Re: Verbal Reasoning
« Reply #23 on: February 26, 2012, 08:33:12 pm »
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I think it would be much easier to solve the problem if you drew out a venn-diagram for yourself.

Its from the practice tests for selective entry high school :|

MHS/Mac.Rob verbal reasoning was much easier. Its just definitions of random words I failed at...

Actually, they changed the format of the entrance exam the year I took the test (back in 2008) because Edutest started writing it, these types of questions certainly did pop up
« Last Edit: February 26, 2012, 08:35:24 pm by Shenz0r »
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dilks

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Re: Verbal Reasoning
« Reply #24 on: February 26, 2012, 09:11:23 pm »
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Good point, a Venn diagram does sound like the simplest solution.
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VivaTequila

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Re: Verbal Reasoning
« Reply #25 on: March 31, 2012, 10:13:56 pm »
+1
It's E. I don't get what is so hard... and I don't think I've ever done any kind of logical reasoning. Hold on, going to make a paint picture showing why...

(and as I go to do this, I'm thinking "famous last words" to myself)



Now;


A: All ZIFS are RIFS
All ZIFS are part of the domain which encloses SIFS, and some SIFS can also be RIFS (but not all).

Here, there is an undisclosed probability that all of the ZIFS will occupy an arbitrary segment of the SIFS which HAPPEN to have the entire arbitrary segment of this temporal SIFS body within the area of RIFS.

However, it may very well be that the part of SIFS which the ZIFS is occupying (arbitrarily defined) is NOT part of the RIFS segment.

This is a MAYBE, because the question hasn't been structured properly.

B: Some HIFS are ZIFS
ALL HIFS are PIFS. ALL PIFS are ZIFS.
The question is inherently true; some HIFS are indeed ZIFS, but in fact it is significant because ALL OF THEM are ZIFS - not some of them.

This is incorrect on a tautology.

C: All RIFS are SIFS

No... only some SIFS are RIFS. This is given in the assumed statements.

D: All HIFS are RIFS

HIFS only take up a small segment of the SIFS area, and only PART of the SIFS area is shared with RIFS. If it so happens again that the small segment (arbitrarily defined) happens to be part of the SIFS area which is common to RIFS, then it could be true. However, this is only a probability. It is highly unlikely that the HIFS which are 100% part of SIFS would be in the segment of SIFS which is shared with RIFS.

This is likely to be incorrect, but only because of poor question structure.

E: Some ZIFS are PIFS

Indeed this is the most logical answer, as all PIFS are ZIFS but not nessecarily all ZIFS are PIFS.

Unless the set containing the elements PIFS=ZIFS then this statement (which would still technically be true if they were equal although tautologically incorrect) is true.

It's kinda like saying all apples are fruits, but not all fruits are apples because there are other types of fruits like mangos and grapes.

Can someone confirm? I would select E, although TECHNICALLY!!!! all of them except D are correct
« Last Edit: March 31, 2012, 10:52:47 pm by VivaTequila »

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Re: Verbal Reasoning
« Reply #26 on: March 31, 2012, 10:43:58 pm »
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I still don't know why B also can't be considered known, unless you are defining some as not all (~all). 
But if this is verbal reasoning, and not reasoning limited to any particular formal logic,
then some often means "at least 1 or more". 

So, using this sense, if "all humans are animals", it follows that "some humans are animals".

so using assumptions 1 and 4:
all hifs are pifs. all pifs are zifs :. all hifs->zifs :. some hifs are zifs

Kanon

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Re: Verbal Reasoning
« Reply #27 on: March 31, 2012, 10:58:31 pm »
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Declaring
         Dim a = HIFS
          Dim b = RIFS
          Dim c = PIFS
          Dim d = SIFS
          Dim e = ZIFS
         
Conditions:
All a are c
Some d are b
All c are d
All c are e

Result:

A is all c, d, e
B is some of a, c, e
C is all a, d, e
D is some b AND is all c, d, e
E is all c, d, a

Multiple Choice:
Which of the following do we know is correct?
A: All e are b
B: Some a are e
C: All b are d
D: All a are b
E: Some e are c

Elimination:
A, B, C, D, C

Conclusion:
I have failed you all.
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Re: Verbal Reasoning
« Reply #28 on: March 31, 2012, 11:28:57 pm »
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Yeah, I still don't think there is an answer :P

VivaTequila

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Re: Verbal Reasoning
« Reply #29 on: March 31, 2012, 11:39:17 pm »
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Look at my post - all of them but C are correct and E is the most correct.