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Given that x y<0 and |x|>|y|
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12 Sep 2026, 02:26
12 Sep 2026, 02:26
Expert Reply
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Question Stats:
50% (01:09) correct
50% (01:25) wrong based on 4 sessions
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Given that \(x y<0\) and \(|x|>|y| \)
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\( \frac{x}{y}+\frac{y}{x}\) |
\(-2\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
Given that x y<0 and |x|>|y|
[#permalink]
30 Sep 2026, 23:56
30 Sep 2026, 23:56
Expert Reply
Since \(x y<0, x \) and $y$ have opposite signs, so both \(\frac{x}{y}\) and \(\frac{y}{x}\) are negative.
Let
\(t=\frac{|x|}{|y|}>1\)
Then
\(\frac{x}{y}=-\frac{|x|}{|y|}=-t, \quad \frac{y}{x}=-\frac{|y|}{|x|}=-\frac{1}{t}\)
So
\(\frac{x}{y}+\frac{y}{x}=-\left(t+\frac{1}{t}\right)\)
For \(t>1\) ,
\(t+\frac{1}{t}>2\)
hence
\(-\left(t+\frac{1}{t}\right)<-2\)
So Quantity A is less than Quantity B.
Quantity B is greater.
Answer: (B)
Let
\(t=\frac{|x|}{|y|}>1\)
Then
\(\frac{x}{y}=-\frac{|x|}{|y|}=-t, \quad \frac{y}{x}=-\frac{|y|}{|x|}=-\frac{1}{t}\)
So
\(\frac{x}{y}+\frac{y}{x}=-\left(t+\frac{1}{t}\right)\)
For \(t>1\) ,
\(t+\frac{1}{t}>2\)
hence
\(-\left(t+\frac{1}{t}\right)<-2\)
So Quantity A is less than Quantity B.
Quantity B is greater.
Answer: (B)
gmatclubot
Given that x y<0 and |x|>|y| [#permalink]
30 Sep 2026, 23:56
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