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If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo

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

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New post 10 Nov 2012, 14:40
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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. \(x^5 y^4 z^3\)

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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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New post 10 Nov 2012, 18:28
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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. \(x^5 y^4 z^3\)


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.
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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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New post 11 Nov 2012, 05:07
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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 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.

Answer: B.

Hope it's clear.
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If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo  [#permalink]

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New post 17 Jul 2014, 05:51
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\(x^ay^bz^c>0\)
There are two possible ways:
1) x<y<z<0
a+b+c must be even
2) x<y<0<z
a+b must be even, and B is the only answer
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If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo  [#permalink]

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New post Updated on: 22 Nov 2017, 13:37
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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*z

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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Originally posted by guialain on 21 Apr 2017, 10:00.
Last edited by guialain on 22 Nov 2017, 13:37, edited 1 time in total.
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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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New post 11 Nov 2012, 05:09
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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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New post 11 Nov 2012, 10:53
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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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New post 17 Jun 2013, 05:51
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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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New post 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..:)
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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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New post 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

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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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New post 21 Jun 2018, 06:09
Hi Bunuel. What does it happen if x and y are positive no intenger numbers?

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 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.

Answer: B.

Hope it's clear.
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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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New post 07 Jul 2019, 22:30
I plugged in values

first i considered all 3 as negative x,y,z = -3,-2,-1 and found all the options are positive

Next set of values -3,-2,1 and found only Option B is positive.
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Re: If x < y < z but x^2 > y^2 > z^2 > 0, which of the follo   [#permalink] 07 Jul 2019, 22:30
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