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27 Jul 2018, 00:14
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (02:12) correct
60%
(01:44)
wrong
based on 78
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Not Attempted Yet
Is \(\frac{|x|}{y}>\frac{x}{y}\), where \(xy\neq{0}\)?
(1) \(y<0\)
(2) \(x=y\)
New question!!!..
self made
(1) \(y<0\)
(2) \(x=y\)
New question!!!..
self made
27 Jul 2018, 00:32
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Given xy \(\neq\) 0
=> x \(\neq\) 0 and y \(\neq\) 0
Statement 1
y < 0
=> \(\frac{1}{y}\) < 0
We know for any value of x, |x| \(\geq\) x
=> If we multiply above equation with \(\frac{1}{y}\) which is < 0, the inequality sign reverses
=> \(\frac{|x|}{y} \leq \frac{x}{y}\)
Hence statement 1 is sufficient to say \(\frac{|x|}{y}\) is not greater than \(\frac{x}{y}\)
Statement 2
x = y
=> \(\frac{x}{y}\) = 1
=> \(\frac{x}{y}\) is always positive
=> \(\frac{|x|}{y}\) is either positive with a maximum value of \(\frac{x}{y}\) or negative based on sign of y
In both the cases \(\frac{|x|}{y} \leq \frac{x}{y}\)
Statement 2 is sufficient
Hence statement s is sufficient to say \(\frac{|x|}{y}\) is not greater than \(\frac{x}{y}\)
Hence option D
=> x \(\neq\) 0 and y \(\neq\) 0
Statement 1
y < 0
=> \(\frac{1}{y}\) < 0
We know for any value of x, |x| \(\geq\) x
=> If we multiply above equation with \(\frac{1}{y}\) which is < 0, the inequality sign reverses
=> \(\frac{|x|}{y} \leq \frac{x}{y}\)
Hence statement 1 is sufficient to say \(\frac{|x|}{y}\) is not greater than \(\frac{x}{y}\)
Statement 2
x = y
=> \(\frac{x}{y}\) = 1
=> \(\frac{x}{y}\) is always positive
=> \(\frac{|x|}{y}\) is either positive with a maximum value of \(\frac{x}{y}\) or negative based on sign of y
In both the cases \(\frac{|x|}{y} \leq \frac{x}{y}\)
Statement 2 is sufficient
Hence statement s is sufficient to say \(\frac{|x|}{y}\) is not greater than \(\frac{x}{y}\)
Hence option D
vaibhav1221
Joined: 19 Nov 2017
Last visit: 24 Jul 2025
Posts: 292
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Given Kudos: 50
Location: India
Schools: IIMA '25 (A) Columbia CBS'26 (D) Fuqua '26 (D) ISB '25 (WL) Darden '26 (D) Kenan-Flagler '26 (WL)
GMAT 1: 710 Q49 V38
GPA: 3.25
WE:Account Management (Advertising and PR)
Schools: IIMA '25 (A) Columbia CBS'26 (D) Fuqua '26 (D) ISB '25 (WL) Darden '26 (D) Kenan-Flagler '26 (WL)
GMAT 1: 710 Q49 V38
Posts: 292
Kudos:
410
27 Jul 2018, 00:41
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I think it should be D.
Statement 1: y<0
x>0 or x<0
If x>0
\(\frac{|+x|}{-y} = \frac{+x}{-y}\)
If x<0
\(\frac{|-x|}{-y} < \frac{-x}{-y}\)
Sufficient.
\(\frac{|x|}{y} ≤ \frac{x}{y}\)
Statement 2: x=y
If x and y are less than 0
\(\frac{|-x|}{-y} < \frac{-x}{-y}\)
If x and y are greater than 0
\(\frac{|+x|}{+y} = \frac{+x}{+y}\)
Sufficient.
\(\frac{|x|}{y} ≤ \frac{x}{y}\)
Statement 1: y<0
x>0 or x<0
If x>0
\(\frac{|+x|}{-y} = \frac{+x}{-y}\)
If x<0
\(\frac{|-x|}{-y} < \frac{-x}{-y}\)
Sufficient.
\(\frac{|x|}{y} ≤ \frac{x}{y}\)
Statement 2: x=y
If x and y are less than 0
\(\frac{|-x|}{-y} < \frac{-x}{-y}\)
If x and y are greater than 0
\(\frac{|+x|}{+y} = \frac{+x}{+y}\)
Sufficient.
\(\frac{|x|}{y} ≤ \frac{x}{y}\)
