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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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91% (01:48) correct 9% (00:42) wrong based on 235 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}
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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] 24 Oct 2013, 19:11
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