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PS inequality

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PS inequality

by satish.nagdev » Tue Oct 06, 2009 10:54 pm
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.
Last edited by satish.nagdev on Thu Oct 08, 2009 6:37 pm, edited 1 time in total.
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by CrackGMAC » Wed Oct 07, 2009 3:38 am
No Clue m8 !!! :!:
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by Katrusya » Wed Oct 07, 2009 11:21 am
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?
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by riteshbindal » Wed Oct 07, 2009 1:18 pm
What is the source of this question?
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by satish.nagdev » Thu Oct 08, 2009 6:42 pm
riteshbindal wrote:What is the source of this question?
Manhattan test

OA is B
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Re: PS inequality

by riteshbindal » Thu Oct 08, 2009 6:52 pm
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.
Dude... You edited the question. It means whatever question we were solving was incorrect. Please check the question before posting.
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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Re: PS inequality

by satish.nagdev » Thu Oct 08, 2009 6:56 pm
riteshbindal wrote:
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.
Dude... You edited the question. It means whatever question we were solving was incorrect. Please check the question before posting.
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.
yea sorry actually "-" character didn't appear properly earlier so i just re-entered it.
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by riteshbindal » Thu Oct 08, 2009 6:58 pm
Thats fine!
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