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Is x^3*y^2*z<0?

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Is x^3*y^2*z<0? [#permalink]

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New post 10 Jul 2017, 00:00
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Is \(x^3y^2z<0?\)

1) \(x^2y<0\)
2) \(yz<0\)
[Reveal] Spoiler: OA

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Re: Is x^3*y^2*z<0? [#permalink]

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New post 10 Jul 2017, 00:06
Y will always be positive so can safely be ignored for now. All we need to do is prove that x and z are opposite signs.

1) (x^2)y < 0
X will always be positive this inequality => y < 0 but that doesn’t help.
Insufficient.

2) yz < 0
Both are opposite signs. Nothing about x. Insufficient.

1+2)
We know y < 0 and z >0 but nothing about x. Insufficient.

E.

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Re: Is x^3*y^2*z<0? [#permalink]

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New post 12 Jul 2017, 00:08
==> If you modify the original condition and the question, when xyz≠0, you get xz<0?. There are 3 variables (x,y,z), and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), you get x^2y<0, y<0 and yz<0, z>0. Since x is unknown, it is not sufficient.

Therefore, the answer is E.
Answer: E
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Re: Is x^3*y^2*z<0? [#permalink]

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New post 31 Aug 2017, 10:00
MathRevolution wrote:
Is \(x^3y^2z<0?\)

1) \(x^2y<0\)
2) \(yz<0\)


Statement 1 states that y<0. We don't know anything about x. --> insufficient
Statement 2 states that either of y or z is <0. --> insufficient

Statement 1 & 2: y<0 and z>0; but still no idea about the sign of x. --> insufficient

Answer: E
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Re: Is x^3*y^2*z<0?   [#permalink] 31 Aug 2017, 10:00
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