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If x/y>1 and x+y<0
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05 Mar 2021, 08:52
05 Mar 2021, 08:52
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49% (01:24) correct
49% (01:19) wrong based on 112 sessions
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If \(\frac{x}{y}>1\) and \(x+y<0\) , then which of the following must be true?
A. \(|x| >|y|\)
B. \(x<y\)
C. \(x>y\)
D. \(x^2>y^2\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A. \(|x| >|y|\)
B. \(x<y\)
C. \(x>y\)
D. \(x^2>y^2\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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If x/y>1 and x+y<0
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06 Mar 2021, 02:39
06 Mar 2021, 02:39
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Carcass wrote:
If \(\frac{x}{y}>1\) and \(x+y<0\) , then which of the following must be true?
A. \(|x| >|y|\)
B. \(x<y\)
C. \(x>y\)
D. \(x^2>y^2\)
A. \(|x| >|y|\)
B. \(x<y\)
C. \(x>y\)
D. \(x^2>y^2\)
If \(\frac{x}{y} > 1\) then, \(x\) and \(y\) must be both +ve or both -ve
And If, \(x+y<0\) then, \(x\) and \(y\) must be both -ve
Now, Since \(x\) and \(y\) are both -ve, and \(\frac{x}{y} > 1\), we can conclude that \(x < y\)
Lets take an example, \(x = -3\) and \(y = -2\)
A. \(|-3| >|-2|\) = YES
B. \(x<y\) = YES (that's what we concluded)
C. \(x>y\) = Can't be!
D. \((-3)^2>(-2)^2\) = YES
Hence, options A, B and D
Re: If x/y>1 and x+y<0
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07 Mar 2021, 17:37
07 Mar 2021, 17:37
1
Solution:
\(x/y>1\)
The above means two things:
1. x & y have same sign
2. the absolute value of x is greater than y.
x+y<0
This means that x and y are negative and x<y.
Thus, x<y, |x|>|y| and \(x^2>y^2\)
IMO A, B & D
Hope this helps
\(x/y>1\)
The above means two things:
1. x & y have same sign
2. the absolute value of x is greater than y.
x+y<0
This means that x and y are negative and x<y.
Thus, x<y, |x|>|y| and \(x^2>y^2\)
IMO A, B & D
Hope this helps
