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 u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
25 Mar 2013, 20:24

11

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

PraPon wrote:

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

(1) \(u+v >0\) (2) \(v>0\)

Good Question. This is why GMAT is hard. The concepts are very simple but people get lost while trying to apply them. Let's try to think logically here.

\(u(u+v)\neq{0}\) implies that neither u nor (u + v) is 0. \(u >0\)

Take statement 2 first since it is simpler: (2) \(v>0\)

Consider \(\frac{1}{(u+v)} < \frac{1}{u} + v\) If v is positive, \(\frac{1}{(u+v)}\) is less than \(\frac{1}{u}\) whereas \(\frac{1}{u} + v\) is greater than \(\frac{1}{u}\). Hence right hand side is always greater. Sufficient.

(1) \(u+v >0\)

Since u is positive, v can be either 'positive' or 'negative but smaller absolute value than u'. We have already seen the 'v is positive' case. In that case, right hand side is greater than left hand side. Let's focus on 'v is negative but smaller absolute value than u'. Say u is 4 and v is -3. Right hand side becomes negative while left hand side is positive. Hence right hand side is smaller. Not sufficient.

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
26 Dec 2012, 01:42

7

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

Is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)? --> is \(\frac{-v}{u(u+v)} <v\)?

(1) \(u+v >0\). Since \(u+v >0\) and \(u >0\), then \(u(u+v)>0\). Now, if \(v>0\), then \(\frac{-v}{u(u+v)}<0 <v\) but if \(v\leq{0}\), then \(\frac{-v}{u(u+v)}\geq{0}\geq{v}\). Not sufficient.

(2) \(v>0\). Since \(u >0\) and \(v>0\), then \(\frac{-v}{u(u+v)}<0<v\). Sufficient.

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
26 Dec 2012, 05:40

PraPon wrote:

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

(1) \(u+v >0\) (2) \(v>0\)

is \(\frac{1}{(u+v)} < \frac{1}{u} +v\)or

is \((u+v) > \frac{u}{(1+uv)}\)

From (1), (u+v) is positive. So assume v is positive . If v is positive LHS is greater. Assume v=0. If v=0, then both are equal.So not sufficient.

We simply see that since \(v>0\) and\(u>0\), \(\frac{1}{u} > \frac{1}{(u+v)}\). So obviously the RHS is greater from (2) alone.

Note: with u positive, saying v is positive is more restrictive than saying (u+v) is positive because in the latter V can be positive or negative. So it is an indication that the answer is likely B. _________________

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
01 Jan 2013, 06:41

Hi Bunuel, if I further simplify your simplified question, i.e. -v/u(u+v) = v to.. by cancelling v on both sides... -1/u(u+v) < 1, then cross multiplying u, given that it is positive the sign does not change...and the question becomes is -1/u+v < u.. ? then I get answer from the condition A as well. given that LHS is negative.. it will be always that u, given that u>0.. what am I doing here wrong? can't I cancel v?

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
02 Jan 2013, 03:20

Expert's post

pavanpuneet wrote:

Hi Bunuel, if I further simplify your simplified question, i.e. -v/u(u+v) = v to.. by cancelling v on both sides... -1/u(u+v) < 1, then cross multiplying u, given that it is positive the sign does not change...and the question becomes is -1/u+v < u.. ? then I get answer from the condition A as well. given that LHS is negative.. it will be always that u, given that u>0.. what am I doing here wrong? can't I cancel v?

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot divide both parts of inequality -v/u(u+v)<v by v as you don't know the sign of this unknown: if v>0 you should write -1/u(u+v)>1 BUT if v<0 you should write -1/u(u+v) >1.

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
25 Mar 2013, 15:43

Bunuel wrote:

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

Is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)? --> is \(\frac{-v}{u(u+v)} <v\)?

(1) \(u+v >0\). Since \(u+v >0\) and \(u >0\), then \(u(u+v)>0\). Now, if \(v>0\), then \(\frac{-v}{u(u+v)}<0 <v\) but if \(v\leq{0}\), then \(\frac{-v}{u(u+v)}\geq{0}\geq{v}\). Not sufficient.

(2) \(v>0\). Since \(u >0\) and \(v>0\), then \(\frac{-v}{u(u+v)}<0<v\). Sufficient.

Answer: B.

Bunuel, how did you know to subtract 1/u first? For example, why didnt you add 1/u +v on the right hand side first? When I did this it became extremely messy and over two minutes so I had to guess, but I was hoping to be able to avoid this next time and easily see which form it needs to be in. Thanks for your help!

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
26 Mar 2013, 04:55

VeritasPrepKarishma wrote:

PraPon wrote:

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

(1) \(u+v >0\) (2) \(v>0\)

Good Question. This is why GMAT is hard. The concepts are very simple but people get lost while trying to apply them. Let's try to think logically here.

\(u(u+v)\neq{0}\) implies that neither u nor (u + v) is 0 hence v is also not 0. \(u >0\)

Take statement 2 first since it is simpler: (2) \(v>0\)

Consider \(\frac{1}{(u+v)} < \frac{1}{u} + v\) If v is positive, \(\frac{1}{(u+v)}\) is less than \(\frac{1}{u}\) whereas \(\frac{1}{u} + v\) is greater than \(\frac{1}{u}\). Hence right hand side is always greater. Sufficient.

(1) \(u+v >0\)

Since u is positive, v can be either 'positive' or 'negative but smaller absolute value than u'. We have already seen the 'v is positive' case. In that case, right hand side is greater than left hand side. Let's focus on 'v is negative but smaller absolute value than u'. Say u is 4 and v is -3. Right hand side becomes negative while left hand side is positive. Hence right hand side is smaller. Not sufficient.

Answer (B)

u(u+v)\neq{0} implies that neither u nor (u + v) is 0 hence v is also not 0. I am confused why v cannot be 0? if u=5 and v=0, then 5(5+0) does not equal 0 and v=0 and it seems to meet the conditions?

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
27 Mar 2013, 19:56

Expert's post

jmuduke08 wrote:

VeritasPrepKarishma wrote:

PraPon wrote:

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

(1) \(u+v >0\) (2) \(v>0\)

Good Question. This is why GMAT is hard. The concepts are very simple but people get lost while trying to apply them. Let's try to think logically here.

\(u(u+v)\neq{0}\) implies that neither u nor (u + v) is 0 hence v is also not 0. \(u >0\)

Take statement 2 first since it is simpler: (2) \(v>0\)

Consider \(\frac{1}{(u+v)} < \frac{1}{u} + v\) If v is positive, \(\frac{1}{(u+v)}\) is less than \(\frac{1}{u}\) whereas \(\frac{1}{u} + v\) is greater than \(\frac{1}{u}\). Hence right hand side is always greater. Sufficient.

(1) \(u+v >0\)

Since u is positive, v can be either 'positive' or 'negative but smaller absolute value than u'. We have already seen the 'v is positive' case. In that case, right hand side is greater than left hand side. Let's focus on 'v is negative but smaller absolute value than u'. Say u is 4 and v is -3. Right hand side becomes negative while left hand side is positive. Hence right hand side is smaller. Not sufficient.

Answer (B)

u(u+v)\neq{0} implies that neither u nor (u + v) is 0 hence v is also not 0. I am confused why v cannot be 0? if u=5 and v=0, then 5(5+0) does not equal 0 and v=0 and it seems to meet the conditions?

Actually, there is no reason why v shouldn't be 0. The only thing is that (u + v) should not be 0. _________________

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
13 May 2014, 08:08

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. _________________

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
04 Jun 2014, 23:04

jmuduke08 wrote:

Bunuel wrote:

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

Is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)? --> is \(\frac{-v}{u(u+v)} <v\)?

(1) \(u+v >0\). Since \(u+v >0\) and \(u >0\), then \(u(u+v)>0\). Now, if \(v>0\), then \(\frac{-v}{u(u+v)}<0 <v\) but if \(v\leq{0}\), then \(\frac{-v}{u(u+v)}\geq{0}\geq{v}\). Not sufficient.

(2) \(v>0\). Since \(u >0\) and \(v>0\), then \(\frac{-v}{u(u+v)}<0<v\). Sufficient.

Answer: B.

Bunuel, how did you know to subtract 1/u first? For example, why didnt you add 1/u +v on the right hand side first? When I did this it became extremely messy and over two minutes so I had to guess, but I was hoping to be able to avoid this next time and easily see which form it needs to be in. Thanks for your help!

I have the same question too. I do not see where the -v is coming from.

1/u+v - 1/u = u - (u+v) / u(u+v) = +v / u(u+v). Where goes something wrong?

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
05 Jun 2014, 00:40

Expert's post

chrish06 wrote:

jmuduke08 wrote:

Bunuel wrote:

If \(u(u+v)\neq{0}\) and \(u>0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

Is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)? --> is \(\frac{-v}{u(u+v)} <v\)?

(1) \(u+v >0\). Since \(u+v >0\) and \(u >0\), then \(u(u+v)>0\). Now, if \(v>0\), then \(\frac{-v}{u(u+v)}<0 <v\) but if \(v\leq{0}\), then \(\frac{-v}{u(u+v)}\geq{0}\geq{v}\). Not sufficient.

(2) \(v>0\). Since \(u >0\) and \(v>0\), then \(\frac{-v}{u(u+v)}<0<v\). Sufficient.

Answer: B.

Bunuel, how did you know to subtract 1/u first? For example, why didnt you add 1/u +v on the right hand side first? When I did this it became extremely messy and over two minutes so I had to guess, but I was hoping to be able to avoid this next time and easily see which form it needs to be in. Thanks for your help!

I have the same question too. I do not see where the -v is coming from.

1/u+v - 1/u = u - (u+v) / u(u+v) = +v / u(u+v). Where goes something wrong?

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]
05 Jun 2014, 02:29

PraPon wrote:

If \(u(u+v)\neq{0}\) and \(u >0\), is \(\frac{1}{(u+v)} < \frac{1}{u} + v\)?

(1) \(u+v >0\) (2) \(v>0\)

Good question...I have done it quite a few times but get it wrong occasionally because of the long process of simplifying then inequality

The given expression can be written as

\(\frac{1}{u}\)\(+v\)\(-\) \(\frac{1}{(u+v)} >0\)

or \(\frac{[(u+v)+ u*(v+u) - u]}{u*(u+v)}\)Simplify and we get

\(\frac{v+uv(u+v)}{u(v+u)}\) >0

Or v*[\(\frac{1}{(u+v)}\) + 1] > 0

Now St 1 says u+v >0 and we know u>0 but we don't know whether v is greater than zero or not. Note that product of 2 nos is greater than zero if both the nos are of same sign. From St 1 we know that 1 term ie. [\(\frac{1}{(u+v)}\) + 1] > 0 but we don't know about v and hence not sufficient

St 2 says v > 0 and we know u>0 so u+v>0 and therefore the expression is greater than zero sufficient.

Ans is B _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

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...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...