PS inequality

No Clue m8 !!! :!:

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?

What is the source of this question?

riteshbindal wrote:What is the source of this question?

Manhattan test

OA is B

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.

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.

Thats fine!