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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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

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.

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.

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.

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

\(x>|y|\) 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 _________________

--------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way

\(x>|y|\) 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.

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|.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...