Hi,
The correct answer is D) but why 2) only can prove?
From the text, x>y and asking if X is larger than the absolute value of Y.
So, for example, 4>  2  OK
4 >  6 Not OK
I know 1) can prove it by telling us both are positive and still X is larger than Y, but what dose 2) tell you to prove this?? hmm. Can anyone please explain?
If x + y >0, is x > y?
(1) x > y
(2) y < 0
DS 500 TET7#22
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 jayhawk2001
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In your example, 4 > 6 does not satisfy the original constraint x+y>0dunkin77 wrote:Hi,
The correct answer is D) but why 2) only can prove?
From the text, x>y and asking if X is larger than the absolute value of Y.
So, for example, 4>  2  OK
4 >  6 Not OK
I know 1) can prove it by telling us both are positive and still X is larger than Y, but what dose 2) tell you to prove this?? hmm. Can anyone please explain?
If x + y >0, is x > y?
(1) x > y
(2) y < 0
1  sufficient. When y is positive, x > y yields x > y. When y is negative,
we know that y < x since x + y > 0.
2  sufficient. y is negative. x has to be positive since x+y > 0.
Also y < x else x+y cannot be > 0. For example if y = 2, x has
to be greater than 3 else x+y will be negative.
Hence D

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Suppose y = 14, x has to be greater than or equal to 15 for the sum to be more than 1.
14+15=1
8+9=1
15 is greater than absolute value of 14.
9 is greater than absolute value of 8
If the absolute value of y>= x then we can never satisfy the given equation i.e. x+y>0
Try it
If y = 9 and x = 5; in this case absolute value of y is >=value of x; but it negates the equation i.e. x+y>0
Hence D
14+15=1
8+9=1
15 is greater than absolute value of 14.
9 is greater than absolute value of 8
If the absolute value of y>= x then we can never satisfy the given equation i.e. x+y>0
Try it
If y = 9 and x = 5; in this case absolute value of y is >=value of x; but it negates the equation i.e. x+y>0
Hence D