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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]

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New post 17 Jul 2019, 04:42
2
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

62% (01:21) correct 38% (01:13) wrong based on 26 sessions

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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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Posts: 1827
Re: If x3 + yz > 0, is xy3z2 > 0? (1) y > 0 > z (2) x > 0  [#permalink]

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New post 18 Jul 2019, 08:07
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] 18 Jul 2019, 08:07
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