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M11 #26

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Re: M11 #26 [#permalink] New post 24 Apr 2012, 21:50
From the given question I found that
IF A>0 than C>0
IF A<0 than C<0

(I answered this question in less than a minute and choose B, because I
didn't consider B to be equal to zero)

Good question for GMAT starter, easy to get into trap... :(
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Re: M11 #26 [#permalink] New post 25 Apr 2012, 17:17
ilhom1986 wrote:
From the given question I found that
IF A>0 than C>0
IF A<0 than C<0

(I answered this question in less than a minute and choose B, because I
didn't consider B to be equal to zero)

Good question for GMAT starter, easy to get into trap... :(


I did this sometime back...bt I dont seem to have considered that B may be equal to 0, and still chose C. :lol:

CAn anyone show how B=0 case make a difference?
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Re: M11 #26 [#permalink] New post 26 Apr 2012, 17:33
Bunuel wrote:
charlemagne wrote:
ilhom1986 wrote:
From the given question I found that
IF A>0 than C>0
IF A<0 than C<0

(I answered this question in less than a minute and choose B, because I
didn't consider B to be equal to zero)

Good question for GMAT starter, easy to get into trap... :(


I did this sometime back...bt I dont seem to have considered that B may be equal to 0, and still chose C. :lol:

CAn anyone show how B=0 case make a difference?


Sure. For example if we put b\neq{0} in the stem then the answer would become B, instead of C.

IF THE QUESTION WERE:

If b\neq{0} is a^7*b^2*c^3>0 ?

Since b\neq{0} then b^2>0, so we can reduce by it (b^2 does not affect given inequality at all) and the question becomes: is a^7*c^3>0? So, the question basically asks whether a and c have the same sign. (Notice that if b\neq{0} were not given then b^2\geq{0}, and we could not reduce by it)

(1) bc<0. Not sufficient.

(2) ac>0 --> a and c have the same sign. Sufficient.

Answer: B.

Hope it's clear.

P.S. Solution to he original question is given here: m11-75085.html#p1073868



Yes....much clearer....Thank you!!!

Another kudos to you.... :-D
BTW, with so many of these already on your name, you must be going crazy with the amount of auto-notification emails coming your way??
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Re: M11 #26 [#permalink] New post 03 May 2012, 18:36
Bunuel wrote:
charlemagne wrote:

Yes....much clearer....Thank you!!!

Another kudos to you.... :-D
BTW, with so many of these already on your name, you must be going crazy with the amount of auto-notification emails coming your way??


That's not a problem.

Also one should consider kudos not only as a "thank you" to user whose post was helpful, but also as a tool to distinguish a valuable post. Notice that posts with more than 2 kudos have different color, also notice that total # of kudos is shown just beside the topic name, so by giving a kudos to a post you are drawing an attention of other users to the helpful material and thus are contributing to the community.


Gee...didnt know that...thanks for the info bunuel!

Quote:
lso notice that total # of kudos is shown just beside the topic name,



Didn't get this part though. :P
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Re: M11 #26 [#permalink] New post 03 May 2012, 20:19
Expert's post
charlemagne wrote:
Quote:
also notice that total # of kudos is shown just beside the topic name,


Didn't get this part though. :P


Total # of kudos accumulated by topics:
Attachments

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Kudos.png [ 29.9 KiB | Viewed 3946 times ]


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Re: M11 #26 [#permalink] New post 23 Apr 2013, 05:48
Here is how: Question: Is (A^7) * (B^2) * (C^3) > 0?

Statement 1: BC < 0
Evaluate the Y/N question:
Is (A^7) * (B*C)^2 * (C) > 0?
(B*C)^2 > 0
But, if A > 0 and C > 0 then expression > 0 => Yes
However, if A > 0 and C < 0 then expression < 0 => No
So, S1 is not sufficient. Eliminate A and D.

Statement 2: AC > 0
Evaluate the Y/N question:
Is expression (A^7) * (B^2) * (C^3) > 0?

Combine A and C.

Is (A*C)^3 * A^4 * B^2 > 0

S2 says A*C > 0 ; so (A*C)^3 > 0
A^4 is always positive even if A were negative; so A^4 > 0
B^2 is always positive even if B were negative; so B^2 > 0
So, the expression is always positive even if A and/or B were negative.
expression > 0

Only exception to this is: if B = 0, then expression = 0
S2 is NOT sufficient.

Take both S1 and S2.
This eliminate the possibility that B may be 0.
So, C is correct.

Last edited by vshrivastava on 23 Apr 2013, 05:59, edited 1 time in total.
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Re: M11 #26 [#permalink] New post 23 Apr 2013, 08:08
Answer C.

Rule out A, B and D immediately as both will be required.

So there are two outcomes based on (1) and (2):

Outcome 1: A = +ve, B = -ve, C = +ve

Gives: (+ve)^7 x (-ve)^2 x (+ve)^3 = (+ve) x (+ve) x (+ve) = +ve

OR

Outcome 2: A = -ve, B = +ve, C = -ve (i.e. opposite of outcome 1)

Gives: (-ve)^7 x (+ve)^2 x (-ve)^3 = (-ve) x (+ve) x (-ve) = +ve

So answer is C, as both possible options give the answer i.e. A^7 x B^2 x C^3 > 0

BUT TOOK ME 4:00mins TO GET THERE!!!
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Re: M11 #26 [#permalink] New post 23 Apr 2013, 17:50
topmbaseeker wrote:
Is A^7 B^2 C^3 > 0 ?

1. BC < 0
2. AC > 0


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I did the question too quick to fall in the trap. Very good question!.
C is correct.

A^7*B^2*C^3 = (A^6*B^2*C^2)*(C*A)

If B = 0, (A^6*B^2*C^2)*(C*A) = 0.
If B #0, the sign of (A^6*B^2*C^2)*(C*A) depends on (C*A)

We must combine statement 1 and 2 to conclude: B #0 and (C*A) > 0
==> (A^6*B^2*C^2)*(C*A) > 0.
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Re: M11 #26 [#permalink] New post 23 Apr 2013, 22:44
[quote="topmbaseeker"]Is A^7 B^2 C^3 > 0 ?

1. BC < 0
2. AC > 0

for AC>0
either both are negative or positive
we dont have depend on value of B,bcause square of anything is positive
in this case if A and c ARE NEGATIVE THEN 2 NEGATIVE SIGNS LEFT CANCEL OUT EACH OTHER AND SQUARE OF B IS POSITIVE,SO ANSWER IS POSITIVE
AND IF A AND C ARE POSITIVE AGAIN SQUARE OF B IS POSITIVE
so answer is B
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Re: M11 #26 [#permalink] New post 24 Apr 2013, 00:17
Expert's post
haulem wrote:
topmbaseeker wrote:
Is A^7 B^2 C^3 > 0 ?

1. BC < 0
2. AC > 0

for AC>0
either both are negative or positive
we dont have depend on value of B,bcause square of anything is positive
in this case if A and c ARE NEGATIVE THEN 2 NEGATIVE SIGNS LEFT CANCEL OUT EACH OTHER AND SQUARE OF B IS POSITIVE,SO ANSWER IS POSITIVE
AND IF A AND C ARE POSITIVE AGAIN SQUARE OF B IS POSITIVE
so answer is B


The correct answer is C, not B.

Check here: m11-75085.html#p1073868 and here: m11-75085-20.html#p1215974
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Re: M11 #26 [#permalink] New post 07 May 2013, 06:54
I think the answer is B, S2 alone is sufficient.

A^7, B^2, C^3 can be considered individually.

S1 cannot account for A at all, and A^7 could be either positive or negative so S1 is out.

Looking at S2, B^2 is automatically positive, so we have only A^7 and C^3 to worry about. Since AC>0, then (AC)^3>0 so we have only A^4 to worry about, but A^4 is positive by definition.

So S2 alone is sufficient. S1 doesn't work at all.
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Re: M11 #26 [#permalink] New post 08 May 2013, 05:24
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ClubMember wrote:
I think the answer is B, S2 alone is sufficient.

A^7, B^2, C^3 can be considered individually.

S1 cannot account for A at all, and A^7 could be either positive or negative so S1 is out.

Looking at S2, B^2 is automatically positive, so we have only A^7 and C^3 to worry about. Since AC>0, then (AC)^3>0 so we have only A^4 to worry about, but A^4 is positive by definition.

So S2 alone is sufficient. S1 doesn't work at all.


The correct answer is C, not B.

Check here: m11-75085.html#p1073868 and here: m11-75085-20.html#p1215974
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Re: M11 #26 [#permalink] New post 13 May 2014, 22:35
From 1. BC < 0 => (BC)^2 > 0

From 2. AC > 0
from this we can get when A < 0 then C < 0 AND when A > 0 then C > 0

for A> 0 and C>0 we have A^7C > 0

for A < 0 and C < 0 => A^6 * (AC) > 0

SO A^7B^2C^3 > 0
Re: M11 #26   [#permalink] 13 May 2014, 22:35
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