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x > 0 and y >0
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05 Jan 2020, 01:31
05 Jan 2020, 01:31
Expert Reply
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Question Stats:
84% (00:38) correct
15% (01:13) wrong based on 65 sessions
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\(x > 0\) and \(y >0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation[/m]
Quantity A |
Quantity B |
\((\sqrt{\frac{x}{y}}\) \(+ \sqrt{\frac{y}{x}})\)\(( \sqrt{\frac{x}{y}} -\) \(\sqrt{\frac{y}{x}})\) |
\(\frac{(x+y)(x-y)}{xy}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation[/m]
Re: x > 0 and y >0
[#permalink]
05 Jan 2020, 05:01
05 Jan 2020, 05:01
2
Carcass wrote:
\(x > 0\) and \(y >0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation[/m]
Quantity A |
Quantity B |
\((\sqrt{\frac{x}{y}}\) \(+ \sqrt{\frac{y}{x}})\)\(( \sqrt{\frac{x}{y}} -\) \(\sqrt{\frac{y}{x}})\) |
\(\frac{(x+y)(x-y)}{xy}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation[/m]
Here,
QTY A:: \((\sqrt{\frac{x}{y}}\) \(+ \sqrt{\frac{y}{x}})\)\(( \sqrt{\frac{x}{y}} -\) \(\sqrt{\frac{y}{x}})\)
= \(\frac {x}{y} - \frac {y}{x} \) (since, \((a+b)(a-b) = a^2-b^2\))
= \(\frac {x^2 - y^2}{xy}
\)
= \(\frac {(x+y)(x-y)}{xy}\)
Therefore, QTY A = QTY B
Option C
gmatclubot
Re: x > 0 and y >0 [#permalink]
05 Jan 2020, 05:01
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