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 350,000 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:

C is the correct answer.
We can simplify the stem. x$y = (x + y) * (x - y ) / (x-y)
We can simplify it further and will end up with x$y = x + y (when x =/=y is not equal to). Therefore, insufficient if x = y.

Statement 2 does not help us much. Insufficient.

Combining the two statements, we get that x+y is 3 and x is not equal to y. It is sufficient as it is right now, but if you would like to solve for x and y, you can. x = 2.5 and y = 0.5;

I am not sure about this OA/OE , can some Math experts validate/invalidate this OE with a better example.

Normally, the question will already contain x != y. In addition, the equation exists and is given: so we cannot have x=y. It is not as we are searching the solution and we are transforming an equation.

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y).
That's it ...............
We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. _________________

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig u mean to say that when x=y the function is not defined.
Now 1 says x+y =3
I will take x=1.5 and y=1.5...............
Then where is the function to calculate with the above values , i mean when they are same the function is not defined.
So i say that 1 is not sufficient _________________

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient

Yes... To me, we have a system of equations :

(1) x$y = (x^2-y^2)/(x-y) (implying by existence of the equation x!=y)
(2) x+y =3

So, in my eyes, (1) implies an equation (3) x!=y that implies the couple (x;y) = (1,5;1,5) a not valid possibility

In the real GMAT, the problem will be clarified by the writings of x!=y.

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient

Yes... To me, we have a system of equations :

(1) x$y = (x^2-y^2)/(x-y) (implying by existence of the equation x!=y) (2) x+y =3

So, in my eyes, (1) implies an equation (3) x!=y that implies the couple (x;y) = (1,5;1,5) a not valid possibility

In the real GMAT, the problem will be clarified by the writings of x!=y.

That's what i am precisely saying........... we definitely need that information.......... _________________

The problem states an equation that needs to avoid certain values of x. Since an equation can exist only for values of x that makes this equation works, impossible values of x are, even if not stated, implicitely banned by the existence of the equation.

In real GMAT, the impossible values of x will be precised next to the equation and not in the statment 1 and 2

Last edited by Fig on 10 Oct 2006, 13:32, edited 2 times in total.

I agree with Fig and disagree with the OE. My thumb of rules for GMAT question is that if something appears in the denominator then it is not zero. However, if you don't know whether it is not zero, you can not divide both sides of an equation by it.

For example, if the stem says x/y=1. We can assume that y is not zero. However, if the stem says x=y. We can not say x/y=1 because y may be zero.

By the same token, if the stem says xy/y=1, we can say that x=1 and y<>0. But if the stem says xy=y, then all we can do is xy-y=0, or y(x-1)=0. In other words there are two solutions to this equation: x=1 and y=anything, or y=0 and x=anything.

What do you all think? _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Hey, Last week I started a few new things in my life. That includes shifting from daily targets to weekly targets, 45 minutes of exercise including 15 minutes of yoga, making...

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...