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TAGS: Retired Moderator Joined: 07 Jun 2014
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x > y xy ≠ 0 [#permalink]
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Expert's post 00:00

Question Stats: 45% (01:35) correct 55% (01:47) wrong based on 60 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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Sandy
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Re: x > y xy ≠ 0 [#permalink]
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]
Expert's post
Great explanation. Thank you.

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