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If wxyz ≠ 0, does wxyz = 1? (1) w = 1/x, y = 1/z (2) wx^2 = 1/xyz

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If wxyz ≠ 0, does wxyz = 1? (1) w = 1/x, y = 1/z (2) wx^2 = 1/xyz  [#permalink]

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New post 01 Aug 2018, 05:00
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Re: If wxyz ≠ 0, does wxyz = 1? (1) w = 1/x, y = 1/z (2) wx^2 = 1/xyz  [#permalink]

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New post 01 Aug 2018, 20:24
Bunuel wrote:
If \(wxyz ≠ 0\), does \(wxyz = 1\)?


(1) \(w = \frac{1}{x}\), \(y = \frac{1}{z}\)

(2) \(wx^2 = \frac{1}{xyz}\)



Can someone please explain this one? I understand why A is sufficient, but for some reason I'm getting stuck on proving out (2) as insufficient

Thanks.
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If wxyz ≠ 0, does wxyz = 1? (1) w = 1/x, y = 1/z (2) wx^2 = 1/xyz  [#permalink]

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New post 01 Aug 2018, 20:42
Bunuel wrote:
If \(wxyz ≠ 0\), does \(wxyz = 1\)?


(1) \(w = \frac{1}{x}\), \(y = \frac{1}{z}\)

(2) \(wx^2 = \frac{1}{xyz}\)


St1:- \(w = \frac{1}{x}\), \(y = \frac{1}{z}\)
Substituting x and y in question stem, we have
wxyz=\(\frac{1}{x}*x*\frac{1}{z}*z\)=\((\frac{1}{x}*x)*(\frac{1}{z}*z)\)=1*1=1
Sufficient.
St2:- \(wx^2 = \frac{1}{xyz}\)
Cross-multiplying,
\(wx^2*xyz =1\)
When x=w=y=z=1, then LHS=RHS=1. So, \(wxyz=1\)
When \(w=\frac{1}{8}\),x=2,y=z=1, then LHS=RHS=1. So, \(wxyz=\frac{1}{8}*2*1*1={\frac{1}{4}}\neq1\)
Insufficient.

Ans. (A)
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If wxyz ≠ 0, does wxyz = 1? (1) w = 1/x, y = 1/z (2) wx^2 = 1/xyz &nbs [#permalink] 01 Aug 2018, 20:42
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