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x > 0 and y >0

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Carcass
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x > 0 and y >0 [#permalink] New post   05 Jan 2020, 01:31
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\(x > 0\) and \(y >0\)


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.

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C
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pranab223
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Re: x > 0 and y >0 [#permalink] New post   05 Jan 2020, 05:01
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Carcass wrote:
\(x > 0\) and \(y >0\)


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
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gmatclubot
Re: x > 0 and y >0 [#permalink]
05 Jan 2020, 05:01
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