It is currently 25 Sep 2017, 11:54

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 26 Aug 2003
Posts: 232

Kudos [?]: 13 [0], given: 0

Location: United States
If x + y > 0, is x > |y|? (1) x > y (2) y < 0 [#permalink]

### Show Tags

05 Nov 2003, 08:33
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 13 [0], given: 0

Senior Manager
Joined: 23 Sep 2003
Posts: 292

Kudos [?]: 4 [0], given: 0

Location: US

### Show Tags

05 Nov 2003, 09:04
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, 11:19, edited 1 time in total.

Kudos [?]: 4 [0], given: 0

Director
Joined: 28 Oct 2003
Posts: 501

Kudos [?]: 32 [0], given: 0

Location: 55405

### Show Tags

05 Nov 2003, 09:34
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...

Kudos [?]: 32 [0], given: 0

Director
Joined: 28 Oct 2003
Posts: 501

Kudos [?]: 32 [0], given: 0

Location: 55405

### Show Tags

05 Nov 2003, 09:40
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.

Kudos [?]: 32 [0], given: 0

Senior Manager
Joined: 23 Sep 2003
Posts: 292

Kudos [?]: 4 [0], given: 0

Location: US

### Show Tags

05 Nov 2003, 10:57
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.

Kudos [?]: 4 [0], given: 0

Manager
Joined: 26 Aug 2003
Posts: 232

Kudos [?]: 13 [0], given: 0

Location: United States

### Show Tags

06 Nov 2003, 06:54
Answer is D. I thougth it would be B as well but apparently it's not the answer. Good one guys!

Kudos [?]: 13 [0], given: 0

Director
Joined: 28 Oct 2003
Posts: 501

Kudos [?]: 32 [0], given: 0

Location: 55405

### Show Tags

06 Nov 2003, 07:01
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.

Kudos [?]: 32 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 299 [0], given: 0

### Show Tags

06 Nov 2003, 07:03
agree with D--I was wrong and eliminated my previous post

Kudos [?]: 299 [0], given: 0

06 Nov 2003, 07:03
Display posts from previous: Sort by