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10 Jun 2015, 04:29
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x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?
A. \(\sqrt{x}-\sqrt{y}\)
B. \(\sqrt{x}+\sqrt{y}\)
C. \(\frac{\sqrt{x}}{\sqrt{y}}\)
D. \(2\sqrt{xy}\)
E. \(2\sqrt{x-y}\)
Kudos for a correct solution.
A. \(\sqrt{x}-\sqrt{y}\)
B. \(\sqrt{x}+\sqrt{y}\)
C. \(\frac{\sqrt{x}}{\sqrt{y}}\)
D. \(2\sqrt{xy}\)
E. \(2\sqrt{x-y}\)
Kudos for a correct solution.
10 Jun 2015, 08:09
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Bunuel
x and y are positive integers such that x > y. If \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\), which of the following is equivalent to 2b?
A. \(\sqrt{x}-\sqrt{y}\)
B. \(\sqrt{x}+\sqrt{y}\)
C. \(\frac{\sqrt{x}}{\sqrt{y}}\)
D. \(2\sqrt{xy}\)
E. \(2\sqrt{x-y}\)
Kudos for a correct solution.
A. \(\sqrt{x}-\sqrt{y}\)
B. \(\sqrt{x}+\sqrt{y}\)
C. \(\frac{\sqrt{x}}{\sqrt{y}}\)
D. \(2\sqrt{xy}\)
E. \(2\sqrt{x-y}\)
Kudos for a correct solution.
Given: \(2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}\)
i.e. \(2(\sqrt{x}-\sqrt{y})=\frac{\sqrt{x}^2-\sqrt{y}^2}{b}\)
i.e. \(2(\sqrt{x}-\sqrt{y})=\frac{(\sqrt{x}-\sqrt{y})*(\sqrt{x}+\sqrt{y})}{b}\)
i.e. \(2=\frac{(\sqrt{x}+\sqrt{y})}{b}\)
i.e. \(2b=(\sqrt{x}+\sqrt{y})\)
Answer: Option
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GMAT 1: 750 Q49 V42
Posts: 170
11 Jun 2015, 03:37
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\(2\sqrt{x}-2\sqrt{y}=\frac{(x-y)}{b}\)
1. Factor 2 out:
\(2(\sqrt{x}-\sqrt{y})=\frac{(x-y)}{b}\)
2. Multiply both sides by b:
\(2b(\sqrt{x}-\sqrt{y})=x-y\)
3. Isolate 2b, by dividing both sides by \((\sqrt{x}-\sqrt{y})\):
\(2b=\frac{(x-y)}{(\sqrt{x}-\sqrt{y})}\)
4. Multiply right side by conjugate to eliminate square roots from denominator:
\(2b=\frac{(x-y)*}{(\sqrt{x}-\sqrt{y})} * (\sqrt{x}+\sqrt{y})/(\sqrt{x}+\sqrt{y})\)
\((\sqrt{x}-\sqrt{y})\) and \((\sqrt{x}+\sqrt{y})\) simplify to "x-y" and that cross cancels with numerator, hence leaving us with \((\sqrt{x}+\sqrt{y})\)
Answer:
1. Factor 2 out:
\(2(\sqrt{x}-\sqrt{y})=\frac{(x-y)}{b}\)
2. Multiply both sides by b:
\(2b(\sqrt{x}-\sqrt{y})=x-y\)
3. Isolate 2b, by dividing both sides by \((\sqrt{x}-\sqrt{y})\):
\(2b=\frac{(x-y)}{(\sqrt{x}-\sqrt{y})}\)
4. Multiply right side by conjugate to eliminate square roots from denominator:
\(2b=\frac{(x-y)*}{(\sqrt{x}-\sqrt{y})} * (\sqrt{x}+\sqrt{y})/(\sqrt{x}+\sqrt{y})\)
\((\sqrt{x}-\sqrt{y})\) and \((\sqrt{x}+\sqrt{y})\) simplify to "x-y" and that cross cancels with numerator, hence leaving us with \((\sqrt{x}+\sqrt{y})\)
Answer:
