If |x| + |y| = -x - y and xy does not equal 0, which of the following must be true?
a) x + y > 0
b) x + y < 0
c) x - y > 0
d) x - y < 0
e) x2 - y2 > 0
I chose D but it's wrong, OA after some discussion. I'm not convinced with explanation given in the source.
Thank,
Satish.
PS inequality
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Last edited by satish.nagdev on Thu Oct 08, 2009 6:37 pm, edited 1 time in total.
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I think D is right.
|X|+|Y| must be positive
-x*y can be positive in two ways:
1. x<0 and y>0;
2. x>0 and y<0
Both of these fit into D.
It's D.
I am wondering what could be another way to solve this Q?
|X|+|Y| must be positive
-x*y can be positive in two ways:
1. x<0 and y>0;
2. x>0 and y<0
Both of these fit into D.
It's D.
I am wondering what could be another way to solve this Q?
- riteshbindal
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Manhattan testriteshbindal wrote:What is the source of this question?
OA is B
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- riteshbindal
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Dude... You edited the question. It means whatever question we were solving was incorrect. Please check the question before posting.satish.nagdev wrote:If |x| + |y| = -x - y and xy does not equal 0, which of the following must be true?
a) x + y > 0
b) x + y < 0
c) x - y > 0
d) x - y < 0
e) x2 - y2 > 0
I chose D but it's wrong, OA after some discussion. I'm not convinced with explanation given in the source.
Thank,
Satish.
Now, this question is as simple as 1 + 2 = 3.
|x| + |y| = -x - y
=> |x| + |y| = -(x+y)
Now |x| and |y| will always be positive so their sum will also be positive.
Hence, -(x+y) > 0
=> x+y<0
That is option B Ans.
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yea sorry actually "-" character didn't appear properly earlier so i just re-entered it.riteshbindal wrote:Dude... You edited the question. It means whatever question we were solving was incorrect. Please check the question before posting.satish.nagdev wrote:If |x| + |y| = -x - y and xy does not equal 0, which of the following must be true?
a) x + y > 0
b) x + y < 0
c) x - y > 0
d) x - y < 0
e) x2 - y2 > 0
I chose D but it's wrong, OA after some discussion. I'm not convinced with explanation given in the source.
Thank,
Satish.
Now, this question is as simple as 1 + 2 = 3.
|x| + |y| = -x - y
=> |x| + |y| = -(x+y)
Now |x| and |y| will always be positive so their sum will also be positive.
Hence, -(x+y) > 0
=> x+y<0
That is option B Ans.
Wounded by GMAT but not dead
- riteshbindal
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