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:

Re: If mv < pv< 0, is v > 0? [#permalink]
21 May 2012, 00:17

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

If mv < pv< 0, is v > 0?

Given: \(mv<pv<0\) --> two cases:

If \(v>0\) then when dividing by \(v\) we would have: \(m<p<0\); If \(v<0\) then when dividing by \(v\) we would have: \(m>p>0\) (flip the sign when dividing by negative value).

(1) m < p --> we have the first case, so \(v>0\). Sufficient. (2) m < 0 --> we have the first case, so \(v>0\). Sufficient.

Re: If mv < pv< 0, is v > 0? [#permalink]
23 May 2012, 01:49

If mv < pv< 0, is v > 0? from here, we know that: mv < 0 -> if m is +, V must be -, if m is - then v is + (they must be opposite to be negative) pv < 0 -> if p is +, V must be - ,if p is -, then v is + (they must be opposite to be negative) mv < pv -> if v is +, then m < p,if v is - then m > p (divide both side by v,dont forget to flip inequality sign when v is -)

(1) m < p --> so v>0. Sufficient. (2) m < 0 --> so v>0. Sufficient.

Re: If mv < pv< 0, is v > 0? [#permalink]
18 Oct 2014, 01:06

Bunuel wrote:

If mv < pv< 0, is v > 0?

Given: \(mv<pv<0\) --> two cases:

If \(v>0\) then when dividing by \(v\) we would have: \(m<p<0\); If \(v<0\) then when dividing by \(v\) we would have: \(m>p>0\) (flip the sign when dividing by negative value).

(1) m < p --> we have the first case, so \(v>0\). Sufficient. (2) m < 0 --> we have the first case, so \(v>0\). Sufficient.

Answer: D.

Hope it's clear.

I understood Bunuel's explanation for statement-1 but following values makes statement-1 insufficient. Please help me understand this:

(1) m<p

lets take v=1, m=-3, p=-2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0, so v is +ve here lets take v=-1, m=3, p=2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0 but v is -ve here

Re: If mv < pv< 0, is v > 0? [#permalink]
18 Oct 2014, 01:13

Expert's post

1

This post was BOOKMARKED

HKD1710 wrote:

Bunuel wrote:

If mv < pv< 0, is v > 0?

Given: \(mv<pv<0\) --> two cases:

If \(v>0\) then when dividing by \(v\) we would have: \(m<p<0\); If \(v<0\) then when dividing by \(v\) we would have: \(m>p>0\) (flip the sign when dividing by negative value).

(1) m < p --> we have the first case, so \(v>0\). Sufficient. (2) m < 0 --> we have the first case, so \(v>0\). Sufficient.

Answer: D.

Hope it's clear.

I understood Bunuel's explanation for statement-1 but following values makes statement-1 insufficient. Please help me understand this:

(1) m<p

lets take v=1, m=-3, p=-2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0, so v is +ve here lets take v=-1, m=3, p=2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0 but v is -ve here

Thanks

m = 3 and p = 2 violate the first statement, which says that m < p. _________________

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...