DS 500 TET7#22

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DS 500 TET7#22

by dunkin77 » Sun Apr 08, 2007 2:30 pm
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

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Re: DS 500 TET7#22

by jayhawk2001 » Sun Apr 08, 2007 9:09 pm
dunkin77 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
In your example, 4 > -6 does not satisfy the original constraint x+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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by Cybermusings » Thu Apr 12, 2007 8:41 am
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

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by maihuna » Wed Apr 15, 2009 7:53 am
To prove: x>|y|

if y<0 x>-y or x+y >0 which is the stem so the ans is yes.