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# M11 #26

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24 Jan 2009, 13:18
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Is $$A^7 B^2 C^3 > 0$$ ?

1. $$BC < 0$$
2. $$AC > 0$$

Source: GMAT Club Tests - hardest GMAT questions
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24 Jan 2009, 13:38
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i will go with C

1) tells us that either C or B is <0

insuff we dont know about A..

2) AC>0 good...we dont know if B=0

together sufficient..we know B is not 0 and AC>0
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24 Jan 2009, 16:16
topmbaseeker wrote:
Is $$A^7 B^2 C^3 \gt 0$$ ?

1. $$BC \lt 0$$
2. $$AC \gt 0$$

(C) 2008 GMAT Club - m11#26

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

both will be reqd. just need to determine if both are adequate.
B^2*C^2>0
AC>0
So AB^2C^3>0
Now, A is also nonzero from 2.
So, A^6>0
Thus,
A^6*A*B^2*C^3>0
Hence, sufficient. C.
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14 Apr 2010, 06:57
A^7B^2C^3 >0

1.BC<0
2.AC>0

sol 1
assume B>0 C<0 then BC<0

if C<0, A also <0 bcs AC>0 from 1&2 A^7B^2C^3 >0

Assume B<0 C>0 then BC<0

if C>0, A also >0 bcs AC>0 from 1&2 A^7B^2C^3 >0
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14 Apr 2010, 08:59
My pick is C

Must have both statements to ensure that (1.) at least 1 variable is > 0 (2.) none of the other variables = 0
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14 Apr 2010, 22:39
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statement 1 :
==========
We can get combinations of B & C bt no clue abt A.So insuff

Statement 2:
==========
We can get combinations about A & C bt no clue abt B.So insuff

Combining both we get
Case 1:
======
A -ve
C -ve
B +ve

For the above combination,we get the answer as YES

Case 2:
======
A +ve
C +ve
B -ve

For case 2 combination also we get the answer as YES.

So I will go with option C

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15 Apr 2010, 01:02
IMO C.

A^7B^2C^3 can be reduced to AC as whatever be the value of B, B^2 will always be >0 while A and C have odd powers.

1) BC<0 (Insufficient) as we dont know the value of A.
2) AC>0 (InSufficient) as discussed above.

AC can be >0 only if both A and C are +ve or are -ve, but wait what if B=0.

I hope this helps.
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18 Apr 2011, 04:32
Considering only 1. since BC < 0 we dont know about A
for 2. AC>0 (both negative or both positive) but B could be zero.

If we take both 1. and 2. -
if C < 0, B > 0 and A < 0 therefore $$A^7B^2C^3 > 0$$
if C > 0, B < 0 and A > 0 and still $$A^7B^2C^3 > 0$$

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18 Apr 2011, 17:36
I simplified this equation to A*B^2*C > 0 to make it slightly less daunting. B will always be positive unless it is zero.

1) BC < 0 means either B or C is Negative, but they can't both be negative. Also, neither can be zero. No other clues are given about A. Insufficient.
2) AC > 0 means A and C are either both negative or both positive. Can't be zero. Insufficient.

(2) states that the product of AC is always positive, since B^2 is also always positive and B isn't zero, then the equation must be positive. (C) is the answer.
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18 Apr 2011, 18:24
I too had missed that B could be equal to 0 in (2).
(1) says :

BC < 0, So B is +ve and C is -ve, or B is -ve and C is +ve, we don't know about A, so not sufficient

(2)

AC > 0, so odd powers of A * odd powers of C are > 0

Because A^7 * C^3 = (AC)^3 * A^4 (AC is +ve and A^4 is always +ve)

B^2 is always non-negative, but B can be equal to 0, so not sufficient.

(1) and (2)

(1) indicates B is non-zero

So A^7 * B^2 * C^3 > 0

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20 Apr 2011, 17:06
(A^7 B^2 C^3)>0?

1. Not sufficient as we dont know anything about A. (B<0,C>0 or B>0 , C<0)

2. Not sufficient as we dont know anything about B. ( A>0,C>0 or A<0, C<0)

Together we have B<0 , C>0 , A>0 (case 1) or B> 0 , C<0, A<0(Case 2)

In both the cases we have (A^7 B^2 C^3)>0. So thats sufficient.

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14 Apr 2012, 23:39
Here 3 unknown in equations i.e. A, B and C.
so to solve this equations require details of all three.
(1) statement 1 provides information about B and C
(2) Statement 2 provides information about A and C

Over all by multiplication using statement 1 and 2 we can get the answers.

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14 Apr 2012, 23:51
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Expert's post
topmbaseeker wrote:
Is $$A^7 B^2 C^3 > 0$$ ?

1. $$BC < 0$$
2. $$AC > 0$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Inequality $$a^7*b^2*c^3>0$$ to be true $$a$$ and $$c$$ must be either both positive or both negative AND $$b$$ must not be zero (in order $$a^7*b^2*c^3$$ not to equal to zero).

(1) $$bc<0$$ --> $$b\neq{0}$$. Don't know about $$a$$ and $$c$$. Not sufficient.

(2) $$ac>0$$ --> $$a$$ and $$c$$ are either both positive or both negative. Don't know about $$b$$. Not sufficient.

(1)+(2) Sufficient.

Hope it helps.
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21 Apr 2012, 05:42
topmbaseeker wrote:
Is $$A^7 B^2 C^3 > 0$$ ?

1. $$BC < 0$$
2. $$AC > 0$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

i followed graph method( it takes time, i dont know any easier method to do this), i got the answer C
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21 Apr 2012, 09:41
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harshavmrg wrote:
topmbaseeker wrote:
Is $$A^7 B^2 C^3 > 0$$ ?

1. $$BC < 0$$
2. $$AC > 0$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

i followed graph method( it takes time, i dont know any easier method to do this), i got the answer C

Easier method is given here: m11-75085.html#p1073868

Similar questions to practice:
is-x-2-y-5-z-0-1-xz-y-0-2-y-z-98341.html
is-x-7-y-2-z-3-0-1-yz-0-2-xz-127692.html
m21-q30-96613.html

Hope it helps.
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22 Apr 2012, 19:15
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topmbaseeker wrote:
Is $$A^7 B^2 C^3 > 0$$ ?

1. $$BC < 0$$
2. $$AC > 0$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

1. BC < 0 implies either B or C is negative, but not both
2. AC > 0 implies neither A nor C are negative, but both A and C may be negative

Using both 1 and 2, we have two options:

I - A is negative; B is positive; C is negative => $$A^7 B^2 C^3 > 0$$ is positive
II - A is positive; B is negative; C is positive => $$A^7 B^2 C^3 > 0$$ is positive

Hence Option C.! :D
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23 Apr 2012, 06:48
I will split the question as
[(A)^6]*[(BC)^2]*(AC)

So (BC)^2>0 or =0
A^6>0 or =0
AC can be either >0 <0 or 0

so {[(A)^6]*[(BC)^2]*(AC)} will look like {(+/0) * (+/0) *(+/-/0)}

1:)BC<0 =>[(BC)^2]>0 but we are unsure about AC wheter its >0 or <0 so reject

strike off options A,D

Remaining options B,C,E

2.)AC>0 now this implies
[(A)^6]>0 (as A,C!=0)
[(BC)^2]>=0(we are unsure of B, whether its 0 or not)
(AC)>0
so result will be >=0, but its insufficient, as its not >0 but we have dual case of >=0

Strike off option B

options E,C are left, lets go for C ,combining
BC<0 and AC>0
implies that
[(BC)^2]>0
[(A)^6]>0
[AC]>0 so end result will be >0

So C is the ANS...
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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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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.

CAn anyone show how B=0 case make a difference?
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25 Apr 2012, 20:53
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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.

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.

Hope it's clear.

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