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655-705 (Hard)|   Algebra|      
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MathRevolution
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b Updated on: 12 Dec 2025, 07:24
Originally posted by MathRevolution on 28 Nov 2017, 00:58.
Last edited by Bunuel on 12 Dec 2025, 07:24, edited 2 times in total.
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65% (hard)

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61% (02:14) correct 39% (02:58) wrong based on 348 sessions

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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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A
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 28 Nov 2017, 01:00
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MathRevolution
[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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MathRevolution, please edit your question to make it clearer to read...
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 28 Nov 2017, 01:05
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MathRevolution
[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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Similar question: https://gmatclub.com/forum/when-x-y-are ... 19963.html
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 30 Nov 2017, 01:28
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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 ).

Therefore, the answer is A.
Answer: A
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 06 Dec 2017, 12:42
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MathRevolution
=>
=>


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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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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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 17 Dec 2017, 00:19
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arrugev93
MathRevolution
=>
=>


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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Sum of reciprocals is much easier than other methods.
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 19 Dec 2017, 11:12
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I do not understand the solution.
I am getting 1+a/ (b+c)
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 07 Jun 2021, 11:56
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I don't understand why it is not 1+a/ (b+c)
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For positive integers x and y, xy is defined by xy=(x+y)/xy. If a, b 09 Dec 2025, 15:38
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