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Re: If x + y > 0, is x > |y|? [#permalink]
08 Apr 2011, 09:19

1

This post received KUDOS

First strategy ------------------ You can throw in the values of x and y to be sure of the behavior of the inequality x > |y| 1. x > y and x + y > 0 x = 4 y = 3 the answer is yes x = 4 y = -3 the answer is yes sufficient

2. y < 0 and x + y > 0 y = -4 x = 4.5 the answer is yes sufficient

Answer D.

Second strategy : Doing algebra will lead to same inference. ----------------- Is x > |y| ? The question can be rephrased as - Is x > y and x > -y ? Adding the two we have x + x > y - y or 2x > 0 or x > 0 So the question becomes Is x > 0?

1. x > y x + y >0 Adding the above two. 2x +y > y or x > 0 sufficient

2. 0>y x + y > 0 Adding the above two. x + y > y or x > 0 sufficient

Re: If x + y > 0, is x > |y|? [#permalink]
08 Apr 2011, 17:42

1

This post received KUDOS

Good solution there buddy.

gmat1220 wrote:

First strategy ------------------ You can throw in the values of x and y to be sure of the behavior of the inequality x > |y| 1. x > y and x + y > 0 x = 4 y = 3 the answer is yes x = 4 y = -3 the answer is yes sufficient

2. y < 0 and x + y > 0 y = -4 x = 4.5 the answer is yes sufficient

Answer D.

Second strategy : Doing algebra will lead to same inference. ----------------- Is x > |y| ? The question can be rephrased as - Is x > y and x > -y ? Adding the two we have x + x > y - y or 2x > 0 or x > 0 So the question becomes Is x > 0?

1. x > y x + y >0 Adding the above two. 2x +y > y or x > 0 sufficient

2. 0>y x + y > 0 Adding the above two. x + y > y or x > 0 sufficient

Re: If x + y > 0, is x > |y|? [#permalink]
08 Apr 2011, 04:58

1

This post was BOOKMARKED

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

Please help me with approach. I am having tough time with inequality. What is the better way to solve such questions? Picking number or doing algebraically?

Here how i attempted this: Given: x+y > 0 , x > -y To prove: x > |y|

Stmt1: x > y. Now |y| = +y when y > 0 and |y| = -y when y < 0. As can be seen, x > +y and x > -y. Hence x > |y|. Sufficient.

Stmt2: y < 0. |y| = -y . So is x > -y? From what is given, x > -y. Sufficient.

Ans: D. _________________

My dad once said to me: Son, nothing succeeds like success.

Re: If x + y > 0, is x > |y|? [#permalink]
08 Apr 2011, 20:26

x+y> 0

this statement implies that either both are +ve or one is +ve and other negative , but with a condition that the absolute value of +ve number > than that of -ve. --------------------------------------1

1. x>y => x must be +ve. y can be either -ve or +ve.

when x,y are +ve is x > |y|? is true. when x +ve and y -ve then also is x > |y|? is true using ----------------------1

2. y<0 => x must be +ve and its absolute value > than that of Y.

Re: If x + y > 0, is x > |y|? [#permalink]
09 Apr 2011, 02:32

Thanks everyone. Different approaches gave me better idea on how to approach such problems. All i need now is some more inequality practice. _________________

My dad once said to me: Son, nothing succeeds like success.

Re: If x + y > 0, is x > |y|? [#permalink]
20 May 2011, 05:41

jamifahad wrote:

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

Please help me with approach. I am having tough time with inequality. What is the better way to solve such questions? Picking number or doing algebraically?

Here how i attempted this: Given: x+y > 0 , x > -y To prove: x > |y|

Stmt1: x > y. Now |y| = +y when y > 0 and |y| = -y when y < 0. As can be seen, x > +y and x > -y. Hence x > |y|. Sufficient.

Stmt2: y < 0. |y| = -y . So is x > -y? From what is given, x > -y. Sufficient.

Ans: D.

X+Y > 0 tells (Both are positive or X>Y or vice versa) For X >|Y|? True when X is positive and > Y, So first two cases

Stmt 1 tells X> Y so x has to be positive to satisfy X+Y > 0 - true Stmt2 : Y<0 , so x has to be positive to satisfy X+Y > 0 - true

Re: If x + y > 0, is x > |y|? [#permalink]
28 Nov 2013, 01:18

Bunuel wrote:

bibha wrote:

If x+y > 0, is x> |y| i. x>y ii.y<0

Thanks

If x + y >0, is x > |y|?

x>|y| means: A. x>-y, if y\leq{0}; B. x>y, if y>{0}. So we should check whether above two inequalities are true.

First inequality is given to be true in the stem (x>-y), so we should check whether x>y is true.

(1) x > y. Sufficient. (2) y < 0 --> |y|=-y. Question becomes is x>-y. This given to be true in the stem. Sufficient.

Answer: D.

Hi Bunuel, Can you please explain this step.I could not follow from your answer explanation. (2) y < 0 --> |y|=-y. Question becomes is x>-y. This given to be true in the stem. Sufficient.

Thanks in advance _________________

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Re: If x + y > 0, is x > |y|? [#permalink]
28 Nov 2013, 05:42

Expert's post

sunita123 wrote:

Bunuel wrote:

bibha wrote:

If x+y > 0, is x> |y| i. x>y ii.y<0

Thanks

If x + y >0, is x > |y|?

x>|y| means: A. x>-y, if y\leq{0}; B. x>y, if y>{0}. So we should check whether above two inequalities are true.

First inequality is given to be true in the stem (x>-y), so we should check whether x>y is true.

(1) x > y. Sufficient. (2) y < 0 --> |y|=-y. Question becomes is x>-y. This given to be true in the stem. Sufficient.

Answer: D.

Hi Bunuel, Can you please explain this step.I could not follow from your answer explanation. (2) y < 0 --> |y|=-y. Question becomes is x>-y. This given to be true in the stem. Sufficient.

Thanks in advance

From (2) since y<0, then |y|=-y. Thus the question becomes: is x>-y? or is x+y>0? The stem says that this is true. Therefore the second statement is sufficient.

Re: If x + y > 0, is x > |y|? [#permalink]
24 Jan 2014, 14:34

x + y > 0 => x > -y. Given x > -y, is x > |y|?

1. x >y. solving x > y and x > -y (adding), x > 0 => |y| has to be less than x (for the sum to be greater than zero) 2. y < 0. Since x + y > 0, x has to be larger than y (which is negative) and |y|.

Re: If x + y > 0, is x > |y|? [#permalink]
04 Jul 2014, 03:27

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