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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] New post 18 Dec 2010, 08:46
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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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Last edited by Bunuel on 20 May 2012, 11:29, edited 1 time in total.
Edited the question and added the OA
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Re: Symbols problem [#permalink] New post 18 Dec 2010, 09:02
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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.

Answer: E (II and III).

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Hope it's clear.
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Re: Symbols problem   [#permalink] 18 Dec 2010, 09:02
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