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

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Walkabout
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If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post  17 Dec 2012, 07:43
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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) \(60\sqrt{15}\)
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Re: If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post  17 Dec 2012, 07:50
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I'm writing the answer but Bunuel is too fast to reply, a machine :)
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If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post  17 Jan 2022, 10:17
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Given that \(x@y=\sqrt{xy}\) and we need to find the value of \((5@45)@60\)

Let's start by finding the value of \(5@45\)

To find \(5@45\) we need to compare \(5@45\) with x@y and see what is before the \(@\) what is after
=> x = 5 and y = 45

=> to find the value of \(5@45\) we need to replace x with 5 and y with 45 in \(x@y=\sqrt{xy}\)
=> \(5@45\)=\(\sqrt{5*45}\) = \(\sqrt{5*5*9}\) = \(\sqrt{5^2*3^2}\) = 5*3 = 15

=> \((5@45)@60\) = \(15@60\)

Similarly, \(15@60=\sqrt{15*60}\) = \(\sqrt{15*15*4}\) = \(\sqrt{15^2*2^2}\) = 15*2 = 30

So, Answer will be 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, 07:46
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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  24 Oct 2013, 20:11
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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 xy=(xy)^1/2 for all [#permalink] New post  21 Jun 2016, 10:29
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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) \(60\sqrt{15}\)


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] New post  10 Jul 2017, 09:45
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What is the point of these questions? I don't get what is being tested here
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If the operation @ is defined by x@y=(xy)^1/2 for all [#permalink] New post  Updated on: 25 May 2020, 13:28
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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: A

Originally posted by BrentGMATPrepNow on 28 Aug 2017, 15:39.
Last edited by BrentGMATPrepNow on 25 May 2020, 13:28, edited 1 time in total.
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