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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 27 Jul 2020, 09:00
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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 27 Jul 2020, 09:02
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 27 Jul 2020, 09:51
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x@y = \sqrt{xy}
5@45 = \sqrt{5*45} = \sqrt{5*5*9} = 5*3 = 15
15@60 = \sqrt{15*50} = \sqrt{15*15*4} = 15*2 = 30

Ans : A) 30
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 27 Jul 2020, 12:29
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Carcass 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) \(60\sqrt{15}\)


We want to evaluate: (5 @ 45) @ 60

Start with (5 @ 45)
(5 @ 45) = √[(5)(45)]
= √225
= 15


So, (5 @ 45) @ 60 = 15 @ 60
From here, 15 @ 60 = √[(15)(60)]
= √900
= 30

Answer: A
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post 28 Jul 2020, 01:15
(5@45)@60

\(\sqrt{\sqrt{5 \times 45} \times 60}\)
\(\sqrt{\sqrt{225} \times 60}\)
\(\sqrt{900}\)
= 30
Re: If the operation @ is defined by x@y=(xy)^1/2 for all   [#permalink] 28 Jul 2020, 01:15
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If the operation @ is defined by x@y=(xy)^1/2 for all

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