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Senior Manager
Joined: 10 Jan 2013
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If x^3y<0, and y/z>0, then which of the following must be less than 1?
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11 Aug 2019, 09:20
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61% (02:32) correct 39% (02:15) wrong based on 87 sessions
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If \(x^3y<0\), and \(y/z>0\), then which of the following must be less than 1? A. \(\sqrt[3]{x}\) B. \(y/(x)^2\) C. \(x^3z^4\) D. \(x^2yz\) E. \(xy^2z^3\)
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Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?
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11 Aug 2019, 09:32
as z and y will have same sign , x and y with opposite sign so only e option will have negative soln in all condition and is less than 1 in all cases.



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Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?
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12 Aug 2019, 13:02
\(x^3y<0\)
\(x^3\) and \(x\) have the same sign so \(xy<0\), which means \(x\) and \(y\) have opposite signs
\(\frac{y}{z}>0\), which means \(y\) and \(z\) have the same sign
Therefore \(x\) and \(z\) have opposite signs and so \(xz<0\) and \(\frac{x}{z}<0\)
Note that at this point we do not know the exact sign of \(x\), \(y\) or \(z\) but we do know how the sign of one affects the other
A. \(\sqrt[3]{x}\)
\(\sqrt[3]{x}\) and \(x\) have the same sign but we don't know what that sign is
B. \(\frac{y}{x^2}\)
It is possible that \(x\) is negative and \(y\) is positive in which case \(\frac{y}{x^2}\) is positive
C. \(x^3z^4\)
It is possible that \(x\) is positive and \(z\) is negative in which case \(x^3z^4\) is positive
D. \(x^2yz\)
Since \(y\) and \(z\) have same sign \(yz>0\) and \(x^2>0\), so \(x^2yz>0\)
E. \(xy^2z^3\)
\(xy^2z^3\) = \((xz)(yz)^2\)
\((yz)^2\) is always positive and as we established above \(xz\) is negative
This means \((xz)(yz)^2\) = \(xy^2z^3\) is negative and therefore less than 1
Answer is (E)
Hit Kudos if this helped!



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Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?
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15 Aug 2019, 16:25
Picking numbers here is a bit tedious. Just list out the 2 possible cases. Case 1: X=pos Y=neg Z=neg Case 2: X=neg Y=pos Z=pos
Just putting the signs in choice E shows that if X is positive, Z will be negative and hence the answer will be negative (<1) Likewise, if X is negative, the entire expression will be negative and hence <1.



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Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?
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12 Aug 2019, 10:02
saurabh9gupta wrote: If \(x^3y<0\), and \(y/z>0\), then which of the following must be less than 1?
A. \(\sqrt[3]{x}\)
B. \(y/(x)^2\)
C. \(x^3z^4\)
D. \(x^2yz\)
E. \(xy^2z^3\) Bunuel chetan2u  your approach please



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Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?
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12 Aug 2019, 11:52
Alternate approach:
Given: \(x^{3}\)y < 0 and \(\frac{y}{z}\)>0
Let x = 1, y = 1, and z = 2 ==> test these numbers in the answer choices and you can quickly eliminate B (1) and D (2). A (1), C (16) & E (8) are < 1 so keep them and pick new set of numbers.
Let x =1, y = 1, and z = 2 ==> test these numbers in the remaining answer choices and you will get A(1), C (16), and E (8 which is < 1), so E is the answer.




Re: If x^3y<0, and y/z>0, then which of the following must be less than 1?
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12 Aug 2019, 11:52




