Two scenerios: (1) A > B and (2) A < B.

(1) Suppose if 1/x > 1/y:

(1/x $ 1/y) = 1/x + 1/y = (x+y)/(xy).

(1/y $ 1/x) = 1/x - 1/y = (y-x)/(xy).

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{1}{x} + \frac{1}{y}) $ (\frac{1}{x} - \frac{1}{y})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y}{xy}) $ (\frac{y-x}{xy})\)

Here \((\frac{x+y}{xy})\) must be > \((\frac{y-x}{xy})\). If so, then \((\frac{x+y}{xy}) $ (\frac{y-x}{xy})\) = \((\frac{x+y}{xy}) + (\frac{y-x}{xy})\). Therefore,

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y}{xy}) + (\frac{y-x}{xy})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y+y-x}{xy})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2y}{xy})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2}{x})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = 2A if we suppose 1/x = A.

(1) Suppose if 1/x < 1/y:

(1/x $ 1/y) = 1/y - 1/x = (x-y)/(xy).

(1/y $ 1/x) = 1/y + 1/x = (x+y)/(xy).

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{1}{y} - \frac{1}{x}) $ (\frac{1}{y} + \frac{1}{y})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x-y}{xy}) $ (\frac{x+y}{xy})\)

Here \((\frac{x+y}{xy})\) must be > \((\frac{x-y}{xy})\). If so, then \((\frac{x-y}{xy}) $ (\frac{x+y}{xy})\) = \((\frac{x+y}{xy}) - (\frac{x-y}{xy})\). Therefore,

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y}{xy}) - (\frac{x-y}{xy})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y-x+y}{xy})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2y}{xy})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2}{x})\)

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = 2A if we suppose 1/x = A.

Therefore, in short, the value of the operation is equal to 2A. A = 1/X in each case and highest A is 1/(1/5) = 5 in D. So 2A in D is 10.

Hope it is clear.

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Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html

Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

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