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# If x + y > 0, is x > |y|? (1) x > y (2) y < 0

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Manager
Joined: 26 Aug 2003
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If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink]  05 Nov 2003, 07:33
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If x + y > 0, is x > |y|?
(1) x > y
(2) y < 0
Senior Manager
Joined: 23 Sep 2003
Posts: 294
Location: US
Followers: 1

Kudos [?]: 2 [0], given: 0

B?

This is what I have:

x + y > 0; which means x> -y

Is x > |y|?

1) is insufficient. if x = 2, and y = -2, X = |y|; if x = 3 and y = -2, x>|y|
2) is sufficient. If y <0 then x> -(y) because y <0 so x >|y|. An example. y = -13. This means that x > -(-13); x > 13, so x >|y|.

Last edited by Makky07 on 05 Nov 2003, 10:19, edited 1 time in total.
Director
Joined: 28 Oct 2003
Posts: 503
Location: 55405
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Kudos [?]: 11 [0], given: 0

I'm thinking 1 is sufficient, though it's counter to my intuition and I'm probably wrong:

Given that x+y>0 we can subtract y from both sides, and determine that x>-y

From 1, we know that x>y.

So if x>y, and x>-y, doesn't x have to be greater than |y|?

and ndidi, you write that "if x = 2, and y = -2, X = |y|".
But, if x is 2 and y is negative 2, x plus y IS NOT greater than zero!

Like I said, I could be wrong...
Director
Joined: 28 Oct 2003
Posts: 503
Location: 55405
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Kudos [?]: 11 [0], given: 0

I misstated my reply to ndidi--
In ndidi's example, s/he uses the plug method, and he s/he states:
"if x = 2, and y = -2"
but that violates the parameters of the problem, which states that "x + y > 0". If you plug those numbers, you get 2+(-2)>0, which is untrue.
Senior Manager
Joined: 23 Sep 2003
Posts: 294
Location: US
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Kudos [?]: 2 [0], given: 0

Yep, you're right. I messed up with the x = 2 and y = -2 example.

The answer should be D. Both statements are sufficient to answer the question.
Manager
Joined: 26 Aug 2003
Posts: 233
Location: United States
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Kudos [?]: 2 [0], given: 0

Answer is D. I thougth it would be B as well but apparently it's not the answer. Good one guys!
Director
Joined: 28 Oct 2003
Posts: 503
Location: 55405
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Kudos [?]: 11 [0], given: 0

stolyar: You wrote "consider x=10 and y=-20"

But you can't consider those values. The problem states that x+y>0

10+ (-20) is not greater than zero.
SVP
Joined: 03 Feb 2003
Posts: 1607
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Kudos [?]: 84 [0], given: 0

agree with D--I was wrong and eliminated my previous post
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