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C
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Difficulty:
55%
(hard)
Question Stats:
65% (02:18) correct
35%
(02:27)
wrong
based on 3557
sessions
History
Date
Time
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Not Attempted Yet
Is xy > 0?
(1) x - y > -2
(2) x - 2y < -6
(1) x - y > -2
(2) x - 2y < -6
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
05 Jun 2011, 05:20
Kudos
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(1)
x = 1, y = 0
xy = 0 although x-y = 1 > -2
x = -3, y = -2
xy = 6 > 0 and x-y = -1 > -2
Insufficient
(2)
x - 2y < -6
x = 1, y = 4
x - 2y = 1 - 8 = -7 < -6 and xy > 0
x = 0, y = 4
x - 2y = 0 - 8 = -8 < -6 and xy = 0
Insufficient
(1) + (2)
x - y > -2
2y - x > 6
y > 4
Now x has to be such that x > y - 2
So x is positive, because y is 4.1, 5 etc.
and x is > 2.1, 3 etc.
So xy = positive
Answer - C
x = 1, y = 0
xy = 0 although x-y = 1 > -2
x = -3, y = -2
xy = 6 > 0 and x-y = -1 > -2
Insufficient
(2)
x - 2y < -6
x = 1, y = 4
x - 2y = 1 - 8 = -7 < -6 and xy > 0
x = 0, y = 4
x - 2y = 0 - 8 = -8 < -6 and xy = 0
Insufficient
(1) + (2)
x - y > -2
2y - x > 6
y > 4
Now x has to be such that x > y - 2
So x is positive, because y is 4.1, 5 etc.
and x is > 2.1, 3 etc.
So xy = positive
Answer - C
26 May 2012, 03:09
Kudos
Bookmarks
Is xy>0?
Note that question basically asks whether \(x\) and \(y\) have the same sign.
(1) x-y > -2 --> we can have an YES answer, if for example \(x\) and \(y\) are both positive (\(x=10\) and \(y=1\)) as well as a NO answer, if for example \(x\) is positive and \(y\) is negative (\(x=10\) and \(y=-10\)). Not sufficient.
(2) x-2y <-6 --> again it' easy to get an YES answer, if for example \(x\) and \(y\) are both positive (\(x=1\) and \(y=10\)) as well as a NO answer, if for example \(x\) is negative and \(y\) is positive (\(x=-1\) and \(y=10\)). Not sufficient.
You can get that the the two statement individually are not sufficient in another way too: we have (1) \(y<x+2\) and (2) \(y>\frac{x}{2}+3\). We are asked whether \(x\) and \(y\) have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line \(y=x+2\) (for 1) and all (x, y) points above the line \(y=\frac{x}{2}+3\) cannot lie only in I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.
(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): \(x-y-(x-2y)>-2-(-6)\) --> \(y>4\). As \(y>4\) and (from 1) \(x>y-2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y-2)\) --> \(x>2\)) --> we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient.
Answer: C.
Note that question basically asks whether \(x\) and \(y\) have the same sign.
(1) x-y > -2 --> we can have an YES answer, if for example \(x\) and \(y\) are both positive (\(x=10\) and \(y=1\)) as well as a NO answer, if for example \(x\) is positive and \(y\) is negative (\(x=10\) and \(y=-10\)). Not sufficient.
(2) x-2y <-6 --> again it' easy to get an YES answer, if for example \(x\) and \(y\) are both positive (\(x=1\) and \(y=10\)) as well as a NO answer, if for example \(x\) is negative and \(y\) is positive (\(x=-1\) and \(y=10\)). Not sufficient.
You can get that the the two statement individually are not sufficient in another way too: we have (1) \(y<x+2\) and (2) \(y>\frac{x}{2}+3\). We are asked whether \(x\) and \(y\) have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line \(y=x+2\) (for 1) and all (x, y) points above the line \(y=\frac{x}{2}+3\) cannot lie only in I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.
(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): \(x-y-(x-2y)>-2-(-6)\) --> \(y>4\). As \(y>4\) and (from 1) \(x>y-2\) then \(x>2\) (because we can add inequalities when their signs are in the same direction, so: \(y+x>4+(y-2)\) --> \(x>2\)) --> we have that \(y>4\) and \(x>2\): both \(x\) and \(y\) are positive. Sufficient.
Answer: C.
