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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8167
GMAT 1: 760 Q51 V42 GPA: 3.82
For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 56% (02:17) correct 44% (03:09) wrong based on 103 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. $$\frac{3}{abc}$$

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Originally posted by MathRevolution on 28 Nov 2017, 00:58.
Last edited by MathRevolution on 28 Nov 2017, 01:02, 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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1
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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Math Expert V
Joined: 02 Sep 2009
Posts: 59265
Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b  [#permalink]

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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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Formatting issue led to confusion
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8167
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b  [#permalink]

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=>
=>

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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Joined: 13 Nov 2017
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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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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.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8167
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b  [#permalink]

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arrugev93 wrote:
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.

Sum of reciprocals is much easier than other methods.
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GMAT 1: 710 Q49 V49 Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b  [#permalink]

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I do not understand the solution.
I am getting 1+a/ (b+c) Re: For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b   [#permalink] 19 Dec 2017, 11:12
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