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x > y xy ≠ 0

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x > y xy ≠ 0 [#permalink] New post 23 Jun 2018, 12:33
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Question Stats:

43% (01:40) correct 56% (01:47) wrong based on 58 sessions
\(x > y\)
\(xy \neq 0\)

Quantity A
Quantity B
\(x^2 \div (y+\frac{1}{y})\)
\(y^2 \div (x+\frac{1}{x})\)


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: x > y xy ≠ 0 [#permalink] New post 26 Jun 2018, 07:45
sandy wrote:
\(x > y\)
\(xy \neq 0\)

Quantity A
Quantity B
\(x^2 \div (y+\frac{1}{y})\)
\(y^2 \div (x+\frac{1}{x})\)


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Here since no value are give let us take

both x and y be positive integer such that x= 3 and y =2

From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{18}{5}\) = 3.6 (putting the value of x and y in the equation)

From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{12}{10}\)= 1.2 (putting the value of x and y in the equation)

From above we get statement 1 > Statement 2

Now if we consider negative integer such that x= -2 and y = -3 (since x>y)

From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{- 12}{10}\)= -1.2 (putting the value of x and y in the equation)

From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{- 18}{5}\) = -3.6(putting the value of x and y in the equation)

From above we have Statement 1 > Statement 2

But we have to look into every possibilities

Now if we consider negative integer such that x= 2 and y = -2 (since x>y)

From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{- 4}{5}\)= -0.8 (putting the value of x and y in the equation)

From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{4}{5}\) = 0.8(putting the value of x and y in the equation)

From above we have Statement 2 > Statement 1.

Hence the option is D
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Re: x > y xy ≠ 0 [#permalink] New post 08 Jan 2020, 09:51
Expert's post
Great explanation. Thank you.

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Re: x > y xy ≠ 0   [#permalink] 08 Jan 2020, 09:51
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