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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 \(x@y=\sqrt{xy}\) 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]

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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]

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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]

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Re: If the operation is defined by xy=(xy)^1/2 for all [#permalink]

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New post 21 Jun 2016, 09:29
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) \(60\sqrt{15}\)


We are given that the operation @ is defined by x@y = √(xy) for all positive numbers.

We are next given that (5@45)@60. Following PEMDAS rules, we want to begin with the operation inside the parentheses. According to the operation, x will be replaced with 5, and y will be replaced with 45. Thus, we have:

5@45 = √(5 *45) = √225 = 15

We have determined that (5@45) = 15, so we substitute 15 for (5@45) to obtain 15@60.

According to the operation, x will now be replaced with 15 and y will now be replaced with 60. Thus we have:

15@60 = √(15*60) = √900 = 30

The answer is A.
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Re: If the operation is defined by xy=(xy)^1/2 for all [#permalink]

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New post 10 Jul 2017, 08:45
What is the point of these questions? I don't get what is being tested here

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Re: If the operation is defined by xy=(xy)^1/2 for all [#permalink]

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New post 28 Aug 2017, 14:39
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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) \(60\sqrt{15}\)


I call these "Strange Operator" questions.
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:
[Reveal] Spoiler:
A


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Re: If the operation is defined by xy=(xy)^1/2 for all   [#permalink] 28 Aug 2017, 14:39
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