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If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0

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If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0 [#permalink] New post   17 Jul 2019, 04:42
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If \(x^3 + yz > 0\), is \(xy^3z^2 > 0\)?

(1) y > 0 > z
(2) x > 0
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A
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Re: If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0 [#permalink] New post   18 Jul 2019, 08:07
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Since y^2 z^2 is positive, we can divide the question on both sides by y^2 z^2 without needing to worry about whether we should reverse the inequality. So the question is just asking "is xy > 0?"

Statement 1 tells us y is positive. It also tells us yz is negative. Since x^3 + yz > 0, then x^3 > -yz, and if yz itself is negative, -yz is positive, so this tells us "x^3 is greater than some positive number", and x^3 must be positive. That means x is positive, so x and y are both positive, and xy > 0, and Statement 1 is sufficient.

Statement 2 is not sufficient since y can be positive or negative, so the answer is A.
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Re: If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0 [#permalink] New post   13 Nov 2020, 07:22
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Dillesh4096 wrote:
If \(x^3 + yz > 0\), is \(xy^3z^2 > 0\)?

(1) y > 0 > z
(2) x > 0




(1) y is positive but z is negative. So from \(x^3 + yz > 0\) we can deduce x is positive. \(xy^3z^2 > 0\) holds true. Sufficient.

(2) No information about y. INSUFFICIENT.

The Answer is A
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Re: If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0 [#permalink] New post   13 Nov 2020, 07:59
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IanStewart wrote:
Since y^2 z^2 is positive, we can divide the question on both sides by y^2 z^2 without needing to worry about whether we should reverse the inequality. So the question is just asking "is xy > 0?"

Statement 1 tells us y is positive. It also tells us yz is negative. Since x^3 + yz > 0, then x^3 > -yz, and if yz itself is negative, -yz is positive, so this tells us "x^3 is greater than some positive number", and x^3 must be positive. That means x is positive, so x and y are both positive, and xy > 0, and Statement 1 is sufficient.

Statement 2 is not sufficient since y can be positive or negative, so the answer is A.



Your reasoning is Y and Z are non-zero. Right? Because if x is negative and either y or z is zero then\( x^3\)+yz will not be greater than "0"
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Re: If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0 [#permalink] New post   13 Nov 2020, 09:41
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MHIKER wrote:
Your reasoning is Y and Z are non-zero. Right?


Statement 1 tells us y and z are nonzero.
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Re: If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0 [#permalink] New post   13 Jul 2023, 04:12
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Re: If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0 [#permalink]
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