Is (x^7)(y^2)(z^3) > 0? (m25#34) : GMAT Data Sufficiency (DS)
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# Is (x^7)(y^2)(z^3) > 0? (m25#34)

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Is (x^7)(y^2)(z^3) > 0? (m25#34) [#permalink]

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06 Aug 2009, 19:55
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Is (x^7)(y^2)(z^3) > 0?

(1) yz < 0
(2) xz > 0

OPEN DISCUSSION OF THE QUESTION IS HERE: is-x-7-y-2-z-3-0-1-yz-0-2-xz-127692.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Mar 2012, 00:46, edited 1 time in total.
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Re: is it > 0 [#permalink]

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06 Aug 2009, 20:00
B for me.

We need to know the signs of x & z to determine if the given product will be +ve. (y^2 will be always positive, no matter what the sign of y is)

1. No info about x. Insuff.
2. xz> 0 means either both x & z are -ve or both are positive, i.e. their signs are the same.

If they have the same sign, the (x^7)(y^2)(z^3) will always be > 0. SUFF.
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Re: is it > 0 [#permalink]

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06 Aug 2009, 22:11
OA B

Given : as y^2 will always be +ve ....we have to determine signs of X & Z....possibilities are (both +ve, or Both -ve)

Statement 1: talks about Y & Z hence we deduce....(Y,Z) => (+,-) or (-,+).....and still leaving X undefined.....hence insufficient

Statement 2 : aptly relates X & Y as ( +,+) or (-,-)....hence sufficient

if like my post...consider it for Kudos
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Re: is it > 0 [#permalink]

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06 Aug 2009, 23:17
[quote="sprtng"]Is (x^7)(y^2)(z^3) > 0?
1. yz < 0
2. xz > 0

is x,z have the same sign

from 1

y,x have different signs...insuff

from 2
x,z have the same sign ....suff...B
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Re: is it > 0 [#permalink]

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07 Aug 2009, 11:52
sprtng wrote:
OA: C note the situation in statement 2 does not cover when y = 0

haha!

Nice! tricky tricky!!
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Re: is it > 0 [#permalink]

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07 Aug 2009, 11:58
Jivana wrote:
sprtng wrote:
OA: C note the situation in statement 2 does not cover when y = 0

haha!

Nice! tricky tricky!!

i know, i got it wrong too...got too distracted by the squares..didn't consider y could be 0 at all. haha
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Re: is it > 0 [#permalink]

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07 Aug 2009, 12:47
very nice question....what is the source, Sprtng?
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Re: is it > 0 [#permalink]

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07 Aug 2009, 12:49
So ..here my 700 vanish in the air and welcome to the 600s .....
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Re: is it > 0 [#permalink]

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07 Aug 2009, 14:07
sdrandom1 wrote:
very nice question....what is the source, Sprtng?

haha one of the questions from gmatclub in the math tests.
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Re: is it > 0 [#permalink]

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07 Aug 2009, 14:07
yezz wrote:
So ..here my 700 vanish in the air and welcome to the 600s .....

I wouldn't say that, but it's good to be brought down to earth once in a while. Even though I missed this problem, I didn't get pissed off, unlike how I would have reacted a month ago...I just smiled and said to myself, 'Don't do these silly s##t next time.'

Bring on the 700s I say!
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Re: is it > 0 [#permalink]

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07 Aug 2009, 20:43
Jivana wrote:
yezz wrote:
So ..here my 700 vanish in the air and welcome to the 600s .....

I wouldn't say that, but it's good to be brought down to earth once in a while. Even though I missed this problem, I didn't get pissed off, unlike how I would have reacted a month ago...I just smiled and said to myself, 'Don't do these silly s##t next time.'

Bring on the 700s I say!

Ride on on Bro......thats wut am talking about
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Re: is it > 0 [#permalink]

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11 Jul 2011, 09:43
+1 B, y^2 is always +ve.
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Re: is it > 0 [#permalink]

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11 Jul 2011, 09:54
pkmme wrote:
+1 B, y^2 is always +ve.

y^2 can be "0". Answer should be "C".
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Re: is it > 0 [#permalink]

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13 Jul 2011, 04:54
totally agree with C. I was distracted by all of those "B is correcT"
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Re: is it > 0 [#permalink]

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14 Jul 2011, 01:02
B is the ans.
From option B we know eithe both X and Z = -Ve
or Both of them are positive
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Re: is it > 0 [#permalink]

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14 Jul 2011, 04:18
Damn you GMATClub question makers!!

Good one!
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Re: Is (x^7)(y^2)(z^3) > 0? (m25#34) [#permalink]

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23 Mar 2012, 00:38
This is question is so easy that i dont even feel like givin explanation.
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Re: Is (x^7)(y^2)(z^3) > 0? (m25#34) [#permalink]

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23 Mar 2012, 00:44
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karankabu wrote:
This is question is so easy that i dont even feel like givin explanation.

Welcome to GMAT Club.

It seems that the question is not that easy since the correct answer is C, not B. Refer to the solution below:

Is x^7*y^2*z^3 > 0 ?

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

(1) $$yz<0$$ --> $$y\neq{0}$$. Don't know about $$x$$ and $$z$$. Not sufficient.

(2) $$xz>0$$ --> $$x$$ and $$z$$ are either both positive or both negative. Don't know about $$y$$. Not sufficient.

(1)+(2) Sufficient.

OPEN DISCUSSION OF THE QUESTION IS HERE: is-x-7-y-2-z-3-0-1-yz-0-2-xz-127692.html (In case of any additional question).
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Re: Is (x^7)(y^2)(z^3) > 0? (m25#34)   [#permalink] 23 Mar 2012, 00:44
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# Is (x^7)(y^2)(z^3) > 0? (m25#34)

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