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![If [x, y, z]=(z+x)/(2z-y), then what is [3, -1, 2]?](https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If [x, y, z]=(z+x)/(2z-y), then what is [3, -1, 2]?"){kind=icon}
23 Sep 2019, 00:59
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[GMAT math practice question]
If \([x, y, z]=\frac{(z+x)}{(2z-y)}\), then what is \([3, -1, 2]\)?
\(A. -3\)
\(B. -1\)
\(C. 1\)
\(D. 3\)
\(E. 5\)
If \([x, y, z]=\frac{(z+x)}{(2z-y)}\), then what is \([3, -1, 2]\)?
\(A. -3\)
\(B. -1\)
\(C. 1\)
\(D. 3\)
\(E. 5\)
23 Sep 2019, 01:07
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IMO the whole trick lies in y=-1
In the denominator, you have: 2z-y
z=2 y=-1
2*2-(-1)=5
Even if you misinterpret and just deduct 4-1=3, the whole expression becomes 5/3 and you wont find this in answer choices, automatically throwing you back and checking where you went wrong
IMO
Ans: C
In the denominator, you have: 2z-y
z=2 y=-1
2*2-(-1)=5
Even if you misinterpret and just deduct 4-1=3, the whole expression becomes 5/3 and you wont find this in answer choices, automatically throwing you back and checking where you went wrong
IMO
Ans: C
sashiim20
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Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
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23 Sep 2019, 10:36
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MathRevolution
[GMAT math practice question]
If \([x, y, z]=\frac{(z+x)}{(2z-y)}\), then what is \([3, -1, 2]\)?
\(A. -3\)
\(B. -1\)
\(C. 1\)
\(D. 3\)
\(E. 5\)
If \([x, y, z]=\frac{(z+x)}{(2z-y)}\), then what is \([3, -1, 2]\)?
\(A. -3\)
\(B. -1\)
\(C. 1\)
\(D. 3\)
\(E. 5\)
\([x, y, z]=\frac{(z+x)}{(2z-y)}\)
\([3, -1, 2] = \frac{(2+3)}{(2*2-(-1))} = \frac{5}{4+1} = \frac{5}{5} = 1\)
Answer C...
