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

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New post Updated on: 28 Nov 2017, 01:02
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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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New post 28 Nov 2017, 01:00
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. \(\frac{3}{abc}\)


MathRevolution, please edit your question to make it clearer to read...
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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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New post 28 Nov 2017, 01:05
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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New post 28 Nov 2017, 11:59
Formatting issue led to confusion
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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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New post 30 Nov 2017, 01:28
=>
=>

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 ).

Therefore, the answer is A.
Answer: A
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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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New post 06 Dec 2017, 12:42
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 ).

Therefore, the answer is A.
Answer: A


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]

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New post 17 Dec 2017, 00:19
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 ).

Therefore, the answer is A.
Answer: A


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

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New post 19 Dec 2017, 11:12
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 &nbs [#permalink] 19 Dec 2017, 11:12
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