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 get why 1 is NSF, no info on x. But with stmt 2, isnt it also NSF b/c we cant confirm if x is neg or positive or not?

If \(x^2y^3=200\), what is \(xy\)?

(1) \(y\) is an integer --> \(y\) can be any positive integer and there will be two values of \(x\) for each \(y\) to satisfy \(x^2y^3=200\). For example \(y=10\) --> \(x^2y^3=x^2*1000=200\) --> \(x=\frac{1}{\sqrt{5}}\) or \(x=-\frac{1}{\sqrt{5}}\). So there are infinite values of \(xy\). Not sufficient.

(2) \(\frac{x}{y}=2.5\) --> \(x=2.5y\). We have two equations and two variables --> \(x^2y^3=6.25y^5=200\) --> \(y^5=32\) --> \(y=2\), \(x=5\) --> \(xy=10\). Sufficient.

First of all \(x^2y^3=2^3*5^2\) has infinitely many solutions for \(x\) and \(y\). For ANY (nonzero) value of \(x\) there exist some \(y\) to satisfy \(x^2y^3=200\) and vise-versa. For example \(x=1\) and \(y=\sqrt[3]{200}\), or \(x=10\) and \(y=\sqrt[3]{2}\), ... As you can see it's not necessary for \(x\) and \(y\) to be integers (3 and 2) to satisfy the given equation.

As for (1): \(y\) is an integer. \(y\) can be any positive integer and there will be two values of \(x\) for each \(y\) to satisfy \(x^2y^3=200\). For example \(y=10\) --> \(x^2y^3=x^2*1000=200\) --> \(x=\frac{1}{\sqrt{5}}\) or \(x=-\frac{1}{\sqrt{5}}\). So there are infinite values of \(xy\). Not sufficient.

I got this problem wrong. Bunuel, can you explain if I approached it the wrong way.

I factored out 200, which = \(2^3 * 5^2\).

After this its obvious \(xy = 10\)

What am i missing here?

Be careful when 1.the number type is not stated-that could be any type,integer,real etc 2.you see factorization type question-they are not always integer or prime number,unless explicitly stated.

common mistakes for me..even after months of studying

No, that's totally wrong. You cannot solve x/y = 2.5 for x and y. Why do you say x = 5 and y = 2? Why not x = 2.5 and y = 1? Or x = 25 and y = 10? x/y = 2.5 has infinitely many solutions for x and y - we have one equation and two unknowns!
_________________

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.
_________________