Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Aug 2016, 02:20

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 27 Jun 2012
Posts: 418
Concentration: Strategy, Finance
Followers: 71

Kudos [?]: 695 [3] , given: 183

If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]

### Show Tags

26 Dec 2012, 00:36
3
KUDOS
18
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

48% (02:27) correct 52% (01:55) wrong based on 311 sessions

### HideShow timer Statistics

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

(1) $$u+v >0$$
(2) $$v>0$$
[Reveal] Spoiler: OA

_________________

Thanks,
Prashant Ponde

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
VOTE GMAT Practice Tests: Vote Here
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6836
Location: Pune, India
Followers: 1928

Kudos [?]: 11973 [13] , given: 221

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

### Show Tags

25 Mar 2013, 21:24
13
KUDOS
Expert's post
5
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.

(Edited)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 34477 Followers: 6291 Kudos [?]: 79800 [7] , given: 10022 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 26 Dec 2012, 02:42 7 This post received KUDOS Expert's post 5 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. Answer: B. _________________ Manager Joined: 13 Oct 2012 Posts: 78 Concentration: General Management, Leadership Schools: IE '15 (A) GMAT 1: 760 Q49 V46 Followers: 1 Kudos [?]: -12 [1] , given: 0 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 03 Jan 2013, 22:21 1 This post received KUDOS The ques gets reduced to -> uv(u+v) + v > 0? if u and v both are + then the above is true , hence B Math Expert Joined: 02 Sep 2009 Posts: 34477 Followers: 6291 Kudos [?]: 79800 [1] , given: 10022 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 12 Jul 2016, 03:24 1 This post received KUDOS Expert's post smartguy595 wrote: $$\frac{1}{(u+v)} - \frac{1}{u}=\frac{u-(u+v)}{(u+v)u}=\frac{u-u-v}{(u+v)u}=\frac{-v}{(u+v)u}$$. Hi Bunuel, Can we cross multiply 'u+v' on left side even if we don't know the sign of 'u+v' We are concerned about the sign when dealing with inequalities: we should keep the sign if we multiply by a positive value and flip the sign when we multiply by a negative value. For equations we can multiply by u+v regardless of its sign. Hope it's clear. _________________ Senior Manager Joined: 17 Dec 2012 Posts: 442 Location: India Followers: 23 Kudos [?]: 361 [0], given: 14 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 26 Dec 2012, 06:40 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. _________________ Srinivasan Vaidyaraman Sravna http://www.sravnatestprep.com Classroom and Online Coaching Manager Joined: 26 Dec 2011 Posts: 117 Followers: 1 Kudos [?]: 28 [0], given: 17 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 01 Jan 2013, 07:41 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? Math Expert Joined: 02 Sep 2009 Posts: 34477 Followers: 6291 Kudos [?]: 79800 [0], given: 10022 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 02 Jan 2013, 04:20 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. Hope it helps. _________________ Manager Joined: 21 Jul 2012 Posts: 69 Followers: 0 Kudos [?]: 8 [0], given: 32 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 25 Mar 2013, 16:43 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! Manager Joined: 21 Jul 2012 Posts: 69 Followers: 0 Kudos [?]: 8 [0], given: 32 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 26 Mar 2013, 05:55 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6836 Location: Pune, India Followers: 1928 Kudos [?]: 11973 [0], given: 221 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 27 Mar 2013, 20:56 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11122
Followers: 511

Kudos [?]: 134 [0], given: 0

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

### Show Tags

13 May 2014, 09:08
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.
_________________
Intern
Joined: 12 Nov 2013
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

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

### Show Tags

05 Jun 2014, 00:04
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.

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?
Math Expert
Joined: 02 Sep 2009
Posts: 34477
Followers: 6291

Kudos [?]: 79800 [0], given: 10022

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

### Show Tags

05 Jun 2014, 01:40
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.

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?

$$\frac{1}{(u+v)} - \frac{1}{u}=\frac{u-(u+v)}{(u+v)u}=\frac{u-u-v}{(u+v)u}=\frac{-v}{(u+v)u}$$.

Hope it's clear.
_________________
Moderator
Joined: 25 Apr 2012
Posts: 728
Location: India
GPA: 3.21
Followers: 42

Kudos [?]: 615 [0], given: 723

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

### Show Tags

05 Jun 2014, 03:29
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.”

Manager
Status: Kitchener
Joined: 03 Oct 2013
Posts: 98
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)
Followers: 0

Kudos [?]: 21 [0], given: 144

If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]

### Show Tags

11 Feb 2015, 18:49

1/(u+v)<1/u+v

1/(u+v) -1/u -v <0

1/(u+v) - (1-uv/u) <0

u-(u+v) - uv( u+v)/(u+v)u<0

u-u-uv-uv/u<0

-2uv/u <0

-2v<0 or does v positive?

can I devided both sides by -2 and it will be does v>o ?
_________________

Click +1 Kudos if my post helped

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6836
Location: Pune, India
Followers: 1928

Kudos [?]: 11973 [0], given: 221

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

### Show Tags

11 Feb 2015, 21:21
23a2012 wrote:

1/(u+v)<1/u+v

1/(u+v) -1/u -v <0

1/(u+v) - (1-uv/u) <0

u-(u+v) - uv( u+v)/(u+v)u<0

u-u-uv-uv/u<0

-2uv/u <0

-2v<0 or does v positive?

can I devided both sides by -2 and it will be does v>o ?

You have messed up the calculations a bit:

1/(u+v) - (1-uv/u) <0

After this step,
1/(u+v) - 1/u + uv/u < 0

[u - (u + v) + uv(u + v)]/u(u + v) < 0

u-(u+v) - uv( u+v)/(u+v)u<0
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: Kitchener Joined: 03 Oct 2013 Posts: 98 Location: Canada Concentration: Finance, Finance GPA: 2.9 WE: Education (Education) Followers: 0 Kudos [?]: 21 [0], given: 144 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 12 Feb 2015, 06:46 VeritasPrepKarishma wrote: 23a2012 wrote: I answered the above equestion as follow please correct me if I am wrong 1/(u+v)<1/u+v 1/(u+v) -1/u -v <0 1/(u+v) - (1-uv/u) <0 u-(u+v) - uv( u+v)/(u+v)u<0 u-u-uv-uv/u<0 -2uv/u <0 -2v<0 or does v positive? can I devided both sides by -2 and it will be does v>o ? You have messed up the calculations a bit: 1/(u+v) - (1-uv/u) <0 After this step, 1/(u+v) - 1/u + uv/u < 0 [u - (u + v) + uv(u + v)]/u(u + v) < 0 Instead, you have u-(u+v) - uv( u+v)/(u+v)u<0 OK, I see my mistake here so it will be u-(u+v)+uv(u+v)/u(u+v)<o u-u-v+uv(u+v)/u(u+v)<0 -v+uv/u<0 v(-1+u)/u<0 is that correct? _________________ Click +1 Kudos if my post helped Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6836 Location: Pune, India Followers: 1928 Kudos [?]: 11973 [0], given: 221 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 12 Feb 2015, 21:27 23a2012 wrote: OK, I see my mistake here so it will be u-(u+v)+uv(u+v)/u(u+v)<o u-u-v+uv(u+v)/u(u+v)<0 -v+uv/u<0 v(-1+u)/u<0 is that correct? Let me show you the entire calculation since I think we messed up. $$\frac{1}{(u+v)} < 1/u + v$$ $$\frac{1}{(u+v)} - \frac{1}{u} - v < 0$$ $$\frac{u - (u + v) - vu(u + v)}{u(u + v)} < 0$$ $$\frac{u - u - v - vu(u + v)}{u(u + v)} < 0$$ $$\frac{-v - vu(u + v)}{u(u + v)} < 0$$ You cannot simply it further except if you want to separate out the terms. $$\frac{-v}{u(u+v)} - v < 0$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Intern
Joined: 15 Dec 2015
Posts: 43
Followers: 0

Kudos [?]: 5 [0], given: 30

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

### Show Tags

16 Dec 2015, 07:03
SravnaTestPrep

SravnaTestPrep wrote:

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

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

Is this legit unless we know the signs of LHS and RHS?
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?   [#permalink] 16 Dec 2015, 07:03

Go to page    1   2    Next  [ 21 posts ]

Similar topics Replies Last post
Similar
Topics:
1 If u, v, and w are integers, is u > 0 ? 1 24 Jul 2016, 09:45
95 If mv < pv < 0, is v > 0? 9 20 Jun 2012, 02:53
6 If mv < pv< 0, is v > 0? 6 20 May 2012, 14:14
7 If u and v are positive real numbers, is u>v? 1. u^3/v 7 16 Jul 2011, 07:17
8 If mv < pv < 0, is v > 0? 10 03 Sep 2008, 11:44
Display posts from previous: Sort by