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If the operation @ is defined by x@y=(xy)^1/2 for all

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If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 17 Dec 2012, 06:43
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90% (01:49) correct 10% (00:59) wrong based on 436 sessions
If the operation @ is defined by \(x@y=\sqrt{xy}\) for all positive numbers x and y, then (5@45)@60 =

(A) 30
(B) 60
(C) 90
(D) \(30\sqrt{15}\)
(E) \(60\sqrt{15}\)
[Reveal] Spoiler: OA
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 17 Dec 2012, 06:46
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Walkabout wrote:
If the operation @ is defined by \(x@y=\sqrt{xy}\) for all positive numbers x and y, then (5@45)@60 =

(A) 30
(B) 60
(C) 90
(D) \(30\sqrt{15}\)
(E) \(60\sqrt{15}\)


\((5@45)@60 =(\sqrt{5*45})@60=(\sqrt{225})@60=15@60=\sqrt{15*60}=30\).

Answer: A.
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 17 Dec 2012, 06:50
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 24 Oct 2013, 19:11
I tried to keep the numbers as manageable as possible and use numbers that I knew the perfect squares of.

\((5@45)@60\)= \(\sqrt{5*(5*9)}\) \(=5*3=15\)

\(15@60=\) \(\sqrt{(3*5)*(3*4*5)}\) \(= 3*5*2= 30\)
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 01 Nov 2014, 04:13
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all   [#permalink] 01 Nov 2014, 04:13
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