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If \(x < y < z\) but \(x^2 > y^2 > z^2 > 0\), which of the following must be positive?

A.\(x^3\) \(y^4 z^5\)

B. \(x^3 y^5 z^4\)

C. \(x^4 y^3 z^5\)

D. \(x^4 y^5 z^3\)

E. \(x^5 y^4 z^3\)

Think of the cases in which '\(x < y < z\) but \(x^2 > y^2 > z^2 > 0\)' happens.

A simple case I can think of is all negative numbers: -5 < -4 < -3 but 25 > 16 > 9 > 0 Another thing that comes to mind is that z can be positive as long as its absolute value remains low: -5 < -4 < 3 but 25 > 16 > 9 > 0

We need to find the option that must stay positive:

A.\(x^3\) \(y^4 z^5\) Will be negative in this case: -5 < -4 < 3 i.e. x negative, y negative, z positive

B. \(x^3 y^5 z^4\) Will be positive in both the cases.

C. \(x^4 y^3 z^5\) Will be negative in this case: -5 < -4 < 3 i.e. x negative, y negative, z positive

D. \(x^4 y^5 z^3\) Will be negative in this case: -5 < -4 < 3 i.e. x negative, y negative, z positive

E. \(x^5 y^4 z^3\) Will be negative in this case: -5 < -4 < 3 i.e. x negative, y negative, z positive

Notice that for an expression to stay positive, we need the power of both x and y to be either even or both to be odd since x and y are both negative. Also, we need the power of z to be even so that it doesn't affect the sign of the expression. Only (B) satisfies these conditions. We don't need to consider any other numbers since we have already rejected 4 options using these numbers. The fifth must be positive in all cases.
_________________

If \(x < y < z\) but \(x^2 > y^2 > z^2 > 0\), which of the following must be positive?

A.\(x^3\) \(y^4 z^5\)

B. \(x^3 y^5 z^4\)

C. \(x^4 y^3 z^5\)

D. \(x^4 y^5 z^3\)

E. \(x^5 y^4 z^3\)

First of all: \(x^2 > y^2 > z^2 > 0\) means that \(|x|>|y|>|z|>0\) (we can take even roots from all parts of an inequality, if all parts are non-negative).

Thus we have that \(x < y < z\) and \(|x|>|y|>|z|>0\). This implies that both \(x\) and \(y\) must be negative numbers: \(x\) to be less than \(y\) and at the same time to be further from zero than \(y\) is, it must be negative. The same way \(y\) to be less than \(z\) and at the same time to be further from zero than \(z\) is, it must be negative. Notice here, that \(z\) may be positive as well as negative. For example if \(x=-3\), \(y=-2\), then \(z\) can be -1 as well as 1. Since we don't know the sign of \(z\), then in order to ensure (to guarantee) that the product will be positive its power in the expression must be even. Only answer choice B fits.

Re: If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo [#permalink]

Show Tags

21 Aug 2013, 05:37

Bunuel wrote:

carcass wrote:

If \(x < y < z\) but \(x^2 > y^2 > z^2 > 0\), which of the following must be positive?

A.\(x^3\) \(y^4 z^5\)

B. \(x^3 y^5 z^4\)

C. \(x^4 y^3 z^5\)

D. \(x^4 y^5 z^3\)

E. \(x^5 y^4 z^3\)

First of all: \(x^2 > y^2 > z^2 > 0\) means that \(|x|>|y|>|z|>0\) (we can take even roots from all part of an inequality, if all parts are non-negative).

Thus we have that \(x < y < z\) and \(|x|>|y|>|z|>0\). This implies that both \(x\) and \(y\) must be negative numbers: \(x\) to be less than \(y\) and at the same time to be further from zero than \(y\) is, it must be negative. The same way \(y\) to be less than \(z\) and at the same time to be further from zero than \(z\) is, it must be negative. Notice here, that \(z\) may be positive as well as negative. For example if \(x=-3\), \(y=-2\), then \(z\) can be -1 as well as 1. Since we don't know the sign of \(z\), then in order to ensure (top guarantee) that the product will be positive its power in the expression must be even. Only answer choice B fits.

Answer: B.

Hope it's clear.

Thanx Bunuel for the explanation..it's now crystal clear..

Re: If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo [#permalink]

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28 Mar 2014, 10:25

Play Smart, try to look for similarities across answer choices

Z doesn't necessarily have to be negative. See we have that x and y have their directions reversed when we square them, hence they are negative strictly speaking/ But x could be positive because we are not told that z<0 in the first equation. Therefore only B works

Re: If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo [#permalink]

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16 Sep 2015, 04:20

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Re: If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo [#permalink]

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13 Oct 2016, 20:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo [#permalink]

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21 Apr 2017, 10:00

1

This post received KUDOS

A trick to solve this question in 20 secondes. Because you are looking for answer choice that must be positive (assuming the question stem), let's remove positve variables in each of the answer choices.

For example, in Option A - The sign of (x^3)*(y^4)*(z^5) is the same as sign of x*z Using similar reasoning, you can rework the answer choices are follows: we are looking for

A- x*z B- x*y C- y*z D- y*z E- x*y

Notice that: A and E are the same, and C and D are the same.

Therefore only B can be correct.
_________________

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