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18 Nov 2016, 00:07
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E
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Difficulty:
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Question Stats:
77% (01:35) correct
23%
(01:37)
wrong
based on 817
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If \(x^2y=z^3\), is \(z^3 > 0\)?
(1) \(xy^2 > 0\)
(2) yz > 0
(1) \(xy^2 > 0\)
(2) yz > 0
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29 Nov 2016, 16:31
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Mbawarrior01
We are given that (x^2)y = z^3, and we need to determine whether z^3 > 0. Recall that the values of z^3 and z have the same sign; thus, we need to determine whether z > 0.
Statement One Alone:
x(y^2) > 0
Since y^2 is always nonnegative, x(y^2) > 0 means x > 0. However, since we cannot determine the sign of y, we cannot determine whether z^3 = (x^2)y is greater than 0.
For example, if y > 0, then z^3 > 0; however, if y < 0, then z^3 < 0. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.
Statement Two Alone:
yz > 0
Since yz > 0, we know that either both y and z are positive or both y and z are negative. However, since we cannot determine the sign of y, we cannot determine whether z^3 = (x^2)y is greater than 0.
For example, if y > 0, then z > 0, and hence z^3 > 0; however, if y < 0, then z < 0, and hence z^3 < 0. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.
Statements One and Two Together:
Using the two statements together, we still cannot determine the sign of y and thus cannot determine whether z^3 > 0.
Answer: E
18 Nov 2016, 00:42
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If \(x^2y=z^3\), is \(z^3 > 0\)?
The question asks whether \(x^2y > 0\). Notice that this inequality will be true if \(x \neq 0\) AND \(y>0\).
(1) \(xy^2 > 0\) --> \(x>0\) and \(y \neq 0\). Not sufficient.
(2) yz > 0 --> \(y^3z^3 > 0\) --> \(y^3*x^2y > 0\) --> \(x^2y^4 > 0\) --> \(xy \neq 0\). Not sufficient.
(1)+(2) Nothing new: \(x>0\) and \(y \neq 0\). Not sufficient.
Answer: E.
The question asks whether \(x^2y > 0\). Notice that this inequality will be true if \(x \neq 0\) AND \(y>0\).
(1) \(xy^2 > 0\) --> \(x>0\) and \(y \neq 0\). Not sufficient.
(2) yz > 0 --> \(y^3z^3 > 0\) --> \(y^3*x^2y > 0\) --> \(x^2y^4 > 0\) --> \(xy \neq 0\). Not sufficient.
(1)+(2) Nothing new: \(x>0\) and \(y \neq 0\). Not sufficient.
Answer: E.
