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Is x + y > 0 ?

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Source: — Data Sufficiency |
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by papgust » Sun Nov 22, 2009 3:36 am
Abdulla wrote:Is x + y > 0?


(1) x-y > 0

(2) x2-y2 > 0

OA is C
Ans is C only.

Qn can be phrased as "is x > -y?"

1) x-y > 0
x > y.
Assume, x = 3, y = 1, 3 > 1.
3 > -1
x = -1, y = -2, -1 > -2 (Yes)
-1 > -(-2). -1 < 2 (No)
Insufficient.

2) x2-y2 > 0
(x+y)(x-y) > 0
Insufficient.

Combining, (x+y)(x-y) > 0
We know that x-y > 0. If x-y>0, then x+y should also be > 0 in order to satisfy (x+y)(x-y) > 0
Hence C
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by Abdulla » Sun Nov 22, 2009 11:55 am
papgust wrote:
Abdulla wrote:Is x + y > 0?


(1) x-y > 0

(2) x2-y2 > 0

OA is C
Ans is C only.

Qn can be phrased as "is x > -y?"

1) x-y > 0
x > y.
Assume, x = 3, y = 1, 3 > 1.
3 > -1
x = -1, y = -2, -1 > -2 (Yes)
-1 > -(-2). -1 < 2 (No)
Insufficient.

2) x2-y2 > 0
(x+y)(x-y) > 0
Insufficient.

Combining, (x+y)(x-y) > 0
We know that x-y > 0. If x-y>0, then x+y should also be > 0 in order to satisfy (x+y)(x-y) > 0
Hence C
Good work Papgust !!
Abdulla
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by mehravikas » Tue Nov 24, 2009 6:41 pm
sorry but I don't get it. I think statement 2 should be sufficient.

(x + y) (x - y) > 0 can be broken down into -

x + y > 0
x - y > 0

papgust wrote: 2) x2-y2 > 0
(x+y)(x-y) > 0
Insufficient.

Combining, (x+y)(x-y) > 0
We know that x-y > 0. If x-y>0, then x+y should also be > 0 in order to satisfy (x+y)(x-y) > 0
Hence C
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