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:

I originally came up with B - but I'm obviously wrong

I think I've violated a rule of inequalities here, so, at the risk of looking stupid, I'll write out my train of thought:

I rephrased the given into:

1- is x - y > x + y ? (I think that is where I am fundamentally wrong - are you allowed to cross-multiply inequalities?)

then

2- is x > x + 2y ?

then

3- is 0 > 2y or is Y < 0 ?

I immediately spotted "b" as answering the given, so I selected it...

I'm gonna go have a look in the MGMAT inequalities guide, but I thought I'd post the question up here anyway for other peoples' benefit
_________________

Its E. My method was to multiply top and bottom with (x+y)/(x+y) = 1 You get (x^2 - y^2) / (x+y)^2

In this case it doesn't matter if you know x is positive and y is negative because it all squares out to positive. So you can't tell is top or bottom is bigger so maybe less than 1 or bigger than 1.

I originally came up with B - but I'm obviously wrong

I think I've violated a rule of inequalities here, so, at the risk of looking stupid, I'll write out my train of thought:

I rephrased the given into:

1- is x - y > x + y ? (I think that is where I am fundamentally wrong - are you allowed to cross-multiply inequalities?)

then

2- is x > x + 2y ?

then

3- is 0 > 2y or is Y < 0 ?

I immediately spotted "b" as answering the given, so I selected it...

I'm gonna go have a look in the MGMAT inequalities guide, but I thought I'd post the question up here anyway for other peoples' benefit

The most important thing to remember about inequalities is that you have to reverse the inequality if you multiply or divide by a negative. In this case you can't tell what the sign of x + y will be, so cross multiplying is dangerous - you have to consider two scenarios.

Q139) if "x" is not equal to "-y", is x-y / x+y > 1?

(1) X > 0 (2) Y < 0

We know that we cannot multiply an inequality by an unknown if we don't know whether the unknown is a positive value or a negative value. The inequality behaves differently in the two cases. If the unknown is positive, the inequality stays as it is. If the unknown is negative, the inequality flips sign. Therefore, x-y > x+y is not an option. Next, '>1' is much more complicated that '>0' where we just need to consider whether the variables are positive or negative. Therefore, bring whatever is on the right hand side to the left hand side.

Question: Is \(\frac{(x - y)}{(x + y)} > 1\) Is \(\frac{(x - y)}{(x + y)} - 1 > 0\) Is \(\frac{-2y}{(x + y)} > 0\)

For this expression to be positive, y/(x+y) should be negative. Therefore, either 'y should be negative and (x+y) should be positive' or 'y should be positive and (x+y) should be negative'.

Statement 1: X > 0 No info about y so not sufficient. Statement 2: Y < 0 No info about x so not sufficient.

Both together, we know that y is negative. We now need to know the sign of (x+y). Just knowing that x is positive doesn't tell us the sign of (x+y) because we don't know which one out of x and y has greater absolute value. If absolute value of x is greater than that of y, x+y is positive. If absolute value of x is less than that of y, x+y is negative. We do not know the sign of x+y so we still cannot say whether \(\frac{-2y}{(x + y)}\) is positive. Not sufficient. Answer E.
_________________

hey karishma , can u tel the exact solution for this problem??? Is x > 0 ? (1) /x - 3/ < 5 (2) /x + 2/ > 5

Put it in a new post.
_________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

Statement 2: |x + 2| > 5 This translates to x < -7 or x > 3. x could be positive or negative. Not sufficient.

Both together, we get that 3 < x < 8. These will be the only values satisfying both conditions. If this is not intuitive, draw the two inequalities on the number line. x should lie between -2 and 8 and it should be either less than -7 or greater than 3. The only overlapping region is from 3 to 8. x will be positive so sufficient.

Statement 1: X > 0 X positive but Y negative gives 1 solution and Y positive gives different solution so NS

Statement 2: Y < 0

Y big negative and X small positive gives NO. Y little negative and X big positive gives yes.

combining: from S2 you can conclude that the answer varies on the values of X and Y. again X big positive and Y little negative the answer is 0>y(y+x) i.e 0> -1(-1+8)

X little positive and Y big negative answer is 0<y(y+x) i.e 0< -8(-8+3).

in the second Q.(the modulus) i recieved 2 solutions for S1:

3<x<8 and -2<x

1. do i have to combine the solution of S1 to -2<x<8.

How did you get this? Even if you want to use the two step algebra: Step 1: Assume x >= 3 (x - 3) < 5 x < 8 So 3 <= x < 8

Step 2: Assume x < 3 and -(x - 3) < 5 gives x > -2 -2 < x < 3

So basically x can lie from -2 to 3 and from 3 to 8 i.e. it can lie from -2 to 8. OR Just say that if |x - 3| < 5, then -5 < x - 3 < 5. Adding 3 to the inequality, -2 < x < 8.

dimri10 wrote:

for S2 I got: x>3 [highlight]and[/highlight] -7>x

How do i combine both statements to 3<x<8?

The highlighted part has to be 'OR'. When you have |x + 2| > 5, then x+2 > 5 OR x+2 < -5 i.e. x > 3 OR x < -7 To satisfy the following two conditions: 1. -2 < x < 8 2. x>3 OR -7>x

Attachment:

Ques3.jpg [ 4.45 KiB | Viewed 1565 times ]

From the figure, which are the only values of x satisfying both the conditions? Between 3 and 8, right? So 3 < x < 8
_________________

i am sorry karishma. just got confused. the number system approach is way better to solve such questions. i have seen some other posts in that issue (yours as well as Bunuel) and now it's a cake. 11 out of 10.

Re: If x is not equal to -y, is (x-y)/(x+y) > 1? [#permalink]

Show Tags

26 Dec 2015, 18:21

Hello from the GMAT Club BumpBot!

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.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...