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# For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b

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Math Revolution GMAT Instructor
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b [#permalink]

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27 Nov 2017, 23:58
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Difficulty:

85% (hard)

Question Stats:

49% (01:40) correct 51% (02:21) wrong based on 51 sessions

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[GMAT math practice question]

For positive integers $$x$$ and $$y$$, $$x@y$$ is defined by $$x@y=\frac{(x+y)}{xy}$$. If $$a, b$$, and $$c$$ are positive integers, what is the value of $$\frac{1}{a}$$@1/$$(\frac{1}{b}@\frac{1}{c})$$?

A. $$a+b+c$$
B. $$\frac{1}{abc}$$
C. $$\frac{1}{(a+b+c)}$$
D. $$\frac{1}{(ab+bc+ca)}$$
E. $$\frac{3}{abc}$$
[Reveal] Spoiler: OA

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Last edited by MathRevolution on 28 Nov 2017, 00:02, edited 1 time in total.

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Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b [#permalink]

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28 Nov 2017, 00:00
MathRevolution wrote:
[GMAT math practice question]

For positive integers $$x$$ and $$y$$, $$x@y$$ is defined by $$x@y=\frac{(x+y)}{xy}$$. If $$a, b$$, and $$c$$ are positive integers, what is the value of $$\frac{1}{a}@\frac{1}{([fraction]1/b}@\frac{1}{c})[/fraction]$$?

A. $$a+b+c$$
B. $$\frac{1}{abc}$$
C. $$\frac{1}{(a+b+c)}$$
D. $$\frac{1}{(ab+bc+ca)}$$
E. $$\frac{3}{abc}$$

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Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b [#permalink]

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28 Nov 2017, 00:05
Expert's post
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MathRevolution wrote:
[GMAT math practice question]

For positive integers $$x$$ and $$y$$, $$x@y$$ is defined by $$x@y=\frac{(x+y)}{xy}$$. If $$a, b$$, and $$c$$ are positive integers, what is the value of $$\frac{1}{a}$$@1/$$(\frac{1}{b}@\frac{1}{c})$$?

A. $$a+b+c$$
B. $$\frac{1}{abc}$$
C. $$\frac{1}{(a+b+c)}$$
D. $$\frac{1}{(ab+bc+ca)}$$
E. $$\frac{3}{abc}$$

Similar question: https://gmatclub.com/forum/when-x-y-are ... 19963.html
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Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b [#permalink]

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28 Nov 2017, 10:59
Formatting issue led to confusion

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Math Revolution GMAT Instructor
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Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b [#permalink]

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30 Nov 2017, 00:28
=>
=>

We can simplify the definition of the operation in the following way:
x@y=(x+y)/xy = x/(xy) + y/(xy) = 1/x+1/y.
So,
1/a@1/(1/b@1/c) = a + (1/b @1/ c) = a + ( b + c ).

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Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b [#permalink]

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06 Dec 2017, 11:42
MathRevolution wrote:
=>
=>

x@y=(x+y)/xy = x/(xy) + y/(xy) = 1/x+1/y.
So,
1/a@1/(1/b@1/c) = a + (1/b @1/ c) = a + ( b + c ).

Hello,

I don't get why you are performing a sum instead of the division you asked in the question.

You are saying: 1/a@1/(1/b@1/c) = a + (1/b @1/ c) = a + ( b + c ).

But as I see, it would be= 1/a@1/(1/b@1/c) = 1+a / b+c

Please tell me what I am doing wrong.

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Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b   [#permalink] 06 Dec 2017, 11:42
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