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# The operation ⊗ is defined for all nonzero numbers a and b

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The operation ⊗ is defined for all nonzero numbers a and b [#permalink]  18 Dec 2010, 07:46
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The operation ⊗ is defined for all nonzero numbers a and b by a ⊗ b = a/b – b/a. If x and y are nonzero numbers, which of the following statements must be true?

I. x ⊗ xy = x(1 ⊗ y)
II. x ⊗ y = -(y ⊗ x)
III. 1/x ⊗ 1/y = y ⊗ x

A. I only
B. II only
C. III only
D. I and II
E. II and III
[Reveal] Spoiler: OA

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Ajit

Last edited by Bunuel on 20 May 2012, 10:29, edited 1 time in total.
Edited the question and added the OA
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Re: Symbols problem [#permalink]  18 Dec 2010, 08:02
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ajit257 wrote:
The operation ⊗ is defined for all nonzero numbers a and b by a ⊗ b = a/b – b/a. If x and y are nonzero numbers, which of the following statements must be true?
I. x ⊗ xy = x(1 ⊗ y)
II. x ⊗ y = -(y ⊗ x)
III. 1/x ⊗ 1/y = y ⊗ x

A. I only
B. II only
C. III only
D. I and II
E. II and III

Please can someone confirm the ans. I am not sure about the ans

Given: $$a@b=\frac{a}{b}-\frac{b}{a}$$, for all nonzero numbers $$a$$ and $$b$$.

I. $$x@(xy) = x(1@y)$$: $$LHS=x@(xy)=\frac{x}{xy}-\frac{xy}{x}=\frac{1}{y}-y$$ and $$RHSx(1@y)=x(\frac{1}{y}-y)$$ as you see LHS doen't equal to RHS;

II. $$x@y = -(y@x)$$: $$LHS=x@y=\frac{x}{y}-\frac{y}{x}$$ and $$RHS=-(y@x)=-(\frac{y}{x}-\frac{x}{y})=\frac{x}{y}-\frac{y}{x}$$ --> LHS=RHS;

III. $$(\frac{1}{x})@(\frac{1}{y}) = y@x$$: $$LHS=(\frac{1}{x})@(\frac{1}{y})=\frac{y}{x}-\frac{x}{y}$$ and $$RHS=y@x=\frac{y}{x}-\frac{x}{y}$$ --> LHS=RHS.

Hope it's clear.
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]  05 Nov 2013, 09:38
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]  05 Nov 2013, 09:39
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Re: The operation ⊗ is defined for all nonzero numbers a and b [#permalink]  06 Jun 2015, 13:23
Hello from the GMAT Club BumpBot!

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Re: The operation ⊗ is defined for all nonzero numbers a and b   [#permalink] 06 Jun 2015, 13:23
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