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27 Apr 2020, 02:10
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[GMAT math practice question]
What is the equivalent expression of \((\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+zx})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)?
A. \(\frac{1}{x+y+z } \)
B. \(\frac{1}{xy+yz+zx} \)
C. \(\frac{1}{xyz}\)
D. \((\frac{1}{xyz})^2 \)
E. \(\frac{xy+yz+zx}{x+y+z}\)
What is the equivalent expression of \((\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+zx})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)?
A. \(\frac{1}{x+y+z } \)
B. \(\frac{1}{xy+yz+zx} \)
C. \(\frac{1}{xyz}\)
D. \((\frac{1}{xyz})^2 \)
E. \(\frac{xy+yz+zx}{x+y+z}\)
ArunSharma12
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27 Apr 2020, 02:26
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[GMAT math practice question]
What is the equivalent expression of \((\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+zx})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)?
A. \(\frac{1}{x+y+z } \)
B. \(\frac{1}{xy+yz+zx} \)
C. \(\frac{1}{xyz}\)
D. \((\frac{1}{xyz})^2 \)
E. \(\frac{xy+yz+zx}{x+y+z}\)
What is the equivalent expression of \((\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+zx})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)?
A. \(\frac{1}{x+y+z } \)
B. \(\frac{1}{xy+yz+zx} \)
C. \(\frac{1}{xyz}\)
D. \((\frac{1}{xyz})^2 \)
E. \(\frac{xy+yz+zx}{x+y+z}\)
\((\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+zx})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)=
\((\frac{1}{x+y+z})(\frac{xy+yz+xz}{xyz})(\frac{1}{xy+yz+zx})(\frac{x+y+z}{xyz})\)= \((\frac{1}{xyz})^2\)
29 Apr 2020, 05:36
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=>
\((\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+zx})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)
\(=(\frac{1}{x+y+z})(\frac{xy+yz+zx}{xyz})(\frac{1}{xy+yz+zx})(\frac{x+y+z}{xyz})\)
\(=(\frac{1}{xyz})^2\)
Therefore, D is the answer.
Answer: D
\((\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+zx})(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)
\(=(\frac{1}{x+y+z})(\frac{xy+yz+zx}{xyz})(\frac{1}{xy+yz+zx})(\frac{x+y+z}{xyz})\)
\(=(\frac{1}{xyz})^2\)
Therefore, D is the answer.
Answer: D
