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11 Sep 2019, 01:02
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C
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Difficulty:
35%
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Question Stats:
79% (02:45) correct
21%
(03:11)
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based on 135
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[GMAT math practice question]
\(<x, y>\) is defined as \(\frac{(2x+y)}{(x-2y)}.\) What is the value of \(<\frac{1}{2}, \frac{2}{3} >- <\frac{1}{3} , \frac{1}{4} >\)?
\(A. \frac{5}{2}\)
\(B. \frac{5}{3}\)
\(C. \frac{7}{2}\)
\(D. \frac{3}{2}\)
\(E. \frac{8}{3}\)
\(<x, y>\) is defined as \(\frac{(2x+y)}{(x-2y)}.\) What is the value of \(<\frac{1}{2}, \frac{2}{3} >- <\frac{1}{3} , \frac{1}{4} >\)?
\(A. \frac{5}{2}\)
\(B. \frac{5}{3}\)
\(C. \frac{7}{2}\)
\(D. \frac{3}{2}\)
\(E. \frac{8}{3}\)
sashiim20
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11 Sep 2019, 06:43
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MathRevolution
\(<x, y>\) is defined as \(\frac{(2x+y)}{(x-2y)}\)
\(<\frac{1}{2}, \frac{2}{3} > = \frac{2(\frac{1}{2})+\frac{2}{3}}{\frac{1}{2}-2(\frac{2}{3})}\) \(=\frac{(1)+(\frac{2}{3})}{(\frac{1}{2})-(\frac{4}{3})} = \frac{\frac{(3+2)}{3}}{\frac{(3-8)}{6}} = \frac{5}{1}*\frac{2}{(-5)} = -2\)
\(<\frac{1}{3}, \frac{1}{4} > = \frac{2(\frac{1}{3})+\frac{1}{4}}{\frac{1}{3}-2(\frac{1}{4})}\) \(=\frac{(\frac{2}{3})+(\frac{1}{4})}{(\frac{1}{3})-(\frac{1}{2})} = \frac{\frac{(8+3)}{12}}{\frac{(2-3)}{6}} = \frac{11}{2}*\frac{1}{(-1)} = -\frac{11}{2}\)
Hence,
\(<\frac{1}{2}, \frac{2}{3} >- <\frac{1}{3} , \frac{1}{4} > =\) \(-2-(\frac{11}{2})\) \(= -2+\frac{11}{2}= \frac{-4+11}{2} = \frac{7}{2}\)
Answer C...
13 Sep 2019, 08:17
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=>
<1/2,2/3> = (2*(1/2) + (2/3))/((1/2)-2(2/3)) = (1 + (2/3))/((1/2) – (4/3)) = (5/3) / (-(5/6)) = -2.
<1/3, 1/4> = (2*(1/3) + (1/4))/((1/3)-2(1/4)) = (2/3+1/4)/(1/3-1/2) = (11/12)/(-(1/6)) = -(11/2)
Thus, <1/2,2/3>-<1/3,1/4> = (-2)-(-(11/2)) = 7/2
Therefore, C is the answer.
Answer: C
<1/2,2/3> = (2*(1/2) + (2/3))/((1/2)-2(2/3)) = (1 + (2/3))/((1/2) – (4/3)) = (5/3) / (-(5/6)) = -2.
<1/3, 1/4> = (2*(1/3) + (1/4))/((1/3)-2(1/4)) = (2/3+1/4)/(1/3-1/2) = (11/12)/(-(1/6)) = -(11/2)
Thus, <1/2,2/3>-<1/3,1/4> = (-2)-(-(11/2)) = 7/2
Therefore, C is the answer.
Answer: C
