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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]

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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) $$60\sqrt{15}$$
[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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17 Dec 2012, 07:46
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Expert's post
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) $$60\sqrt{15}$$

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

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

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

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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 x@y=(xy)^1/2 for all [#permalink]

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01 Nov 2014, 05:13
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Re: If the operation is defined by xy=(xy)^1/2 for all [#permalink]

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21 Jun 2016, 10:29
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

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Re: If the operation is defined by xy=(xy)^1/2 for all   [#permalink] 21 Jun 2016, 10:29
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