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# Is 3x − y + z greater than 2x − y + 2z?

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Joined: 25 Nov 2011
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Is 3x − y + z greater than 2x − y + 2z?  [#permalink]

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Updated on: 12 Jun 2013, 03:27
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25% (medium)

Question Stats:

75% (01:26) correct 25% (01:48) wrong based on 66 sessions

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Is 3x − y + z greater than 2x − y + 2z?

(1) x is positive.

(2) (x^2)*z is negative.

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-Aravind Chembeti

Originally posted by Chembeti on 25 Feb 2012, 23:50.
Last edited by Bunuel on 12 Jun 2013, 03:27, edited 2 times in total.
Edited the question
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Joined: 02 Sep 2009
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Re: Is 3x − y + z greater than 2x − y + 2z?  [#permalink]

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25 Feb 2012, 23:58
Is 3x − y + z greater than 2x − y + 2z?

Is $$3x-y+z>2x-y+2z$$? --> is $$x>z$$?

(1) x is positive. No info about $$z$$. Not sufficient.

(2) (x^2)*z is negative --> $$(x^2)*z<0$$ --> $$z$$ is negative, but limited info about $$x$$ (we only know that $$x\neq{0}$$). Not sufficient.

(1)+(2) From (1) $$x$$ is positive and from (2) $$z$$ is negative --> $$x>z$$. Sufficient.

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Re: Is 3x − y + z greater than 2x − y + 2z?  [#permalink]

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26 Feb 2012, 13:59
Bunuel wrote:
Is 3x − y + z greater than 2x − y + 2z?

Is $$3x-y+z>2x-y+2z$$? --> is $$x>z$$?

(1) x is positive. No info about $$z$$. Not sufficient.

(2) x^2*z is negative --> $$x^2*z<0$$ --> $$z$$ is negative, but limited info about $$x$$ (we only know that $$x\neq{0}$$). Not sufficient.

(1)+(2) From (1) $$x$$ is positive and from (2) $$z$$ is negative --> $$x>z$$. Sufficient.

How did you deduce z must be negative?

From exponent rule, $$b^(-n)$$ is $$1/b^n$$. This will always be positive if n is even, regardless of the base. Since $$n = 2z$$, $$x^(2*z)$$ will always be positive, which contradicts the original statement #2. Am I missing something here?
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Re: Is 3x − y + z greater than 2x − y + 2z?  [#permalink]

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26 Feb 2012, 14:04
PlayStation64 wrote:
Bunuel wrote:
Is 3x − y + z greater than 2x − y + 2z?

Is $$3x-y+z>2x-y+2z$$? --> is $$x>z$$?

(1) x is positive. No info about $$z$$. Not sufficient.

(2) x^2*z is negative --> $$x^2*z<0$$ --> $$z$$ is negative, but limited info about $$x$$ (we only know that $$x\neq{0}$$). Not sufficient.

(1)+(2) From (1) $$x$$ is positive and from (2) $$z$$ is negative --> $$x>z$$. Sufficient.

How did you deduce z must be negative?

From exponent rule, $$b^(-n)$$ is $$1/b^n$$. This will always be positive if n is even, regardless of the base. Since $$n = 2z$$, $$x^(2*z)$$ will always be positive, which contradicts the original statement #2. Am I missing something here?

Welcome to GMAT Club.

I think you misunderstood the statement: it's $$(x^2)*z<0$$ not $$x^{(2z)}<0$$: 2 is an exponent and z is a multiple and not that 2z is an exponent.

Hope it's clear.
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Re: Is 3x − y + z greater than 2x − y + 2z?  [#permalink]

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26 Feb 2012, 14:07
Ahhh! Thanks for the clarification! This changes everything!
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Re: Is 3x − y + z greater than 2x − y + 2z?  [#permalink]

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24 Nov 2017, 00:35
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Re: Is 3x − y + z greater than 2x − y + 2z? &nbs [#permalink] 24 Nov 2017, 00:35
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# Is 3x − y + z greater than 2x − y + 2z?

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