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:

If \(v*m*t\) not \(= 0\), is \(v^2*m^3*t^{-4} > 0\)?

1) \(m>v^2\) 2) \(m>t^{-4}\)

Shalva's solution is correct.

First of all: knowing that \(v*m*t\neq{0}\) implies that none of the variables equals to zero.

Second of all: \(v^2*m^3*t^{-4}>0\)? --> \(\frac{v^2*m^3}{t^4}> 0\)? Now when this inequality holds true? Obviously as \(v^2\) and \(t^4\) are positive this inequality will hold true if and only \(m^3>0\), or, which is the same, when \(m>0\).

(1) \(m>v^2\) --> \(m\) is more than some positive number (\(v^2\)), hence \(m\) is positive. Sufficient.

(2) \(m>t^{-4}\) --> \(m>\frac{1}{t^4}\) --> Again \(m\) is more than some positive number (\(\frac{1}{t^4}\) ), hence \(m\) is positive. Sufficient.

in both conditions the only variable that can change the sign of the expression is M. Since in both conditions we are told that M most be greater than some always positive number than in both condition we can answer the question if M is positive and thus the expression is positive.

Re: If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ? [#permalink]

Show Tags

18 Dec 2013, 05:30

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. _________________

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...