GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 23 Jan 2019, 20:42

Close

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
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.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
  • Key Strategies to Master GMAT SC

     January 26, 2019

     January 26, 2019

     07:00 AM PST

     09:00 AM PST

    Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
  • Free GMAT Number Properties Webinar

     January 27, 2019

     January 27, 2019

     07:00 AM PST

     09:00 AM PST

    Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Senior Manager
Senior Manager
User avatar
Joined: 27 Jun 2012
Posts: 371
Concentration: Strategy, Finance
Schools: Haas EWMBA '17
If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 25 Dec 2012, 23:36
8
55
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

52% (01:36) correct 48% (02:14) wrong based on 819 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) \(v>0\)

_________________

Thanks,
Prashant Ponde

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
Reading Comprehension notes: Click here
VOTE GMAT Practice Tests: Vote Here
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here
Finance your Student loan through SoFi and get $100 referral bonus : Click here

Most Helpful Expert Reply
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 25 Mar 2013, 20:24
21
10
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.

Answer (B)

(Edited)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

General Discussion
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52433
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 26 Dec 2012, 01:42
15
12
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Director
Director
User avatar
S
Joined: 17 Dec 2012
Posts: 624
Location: India
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 26 Dec 2012, 05:40
1
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 Holistic Solutions
http://www.sravnatestprep.com

Holistic and Systematic Approach

Manager
Manager
avatar
Joined: 26 Dec 2011
Posts: 93
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 01 Jan 2013, 06:41
1
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
User avatar
V
Joined: 02 Sep 2009
Posts: 52433
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 02 Jan 2013, 03:20
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 13 Oct 2012
Posts: 48
Concentration: General Management, Leadership
Schools: IE '15 (A)
GMAT 1: 760 Q49 V46
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 03 Jan 2013, 21:21
The ques gets reduced to -> uv(u+v) + v > 0?
if u and v both are + then the above is true , hence B
Manager
Manager
avatar
Joined: 21 Jul 2012
Posts: 64
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 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!
Manager
Manager
avatar
Joined: 21 Jul 2012
Posts: 64
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 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?
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 27 Mar 2013, 19:56
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

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Intern
Intern
avatar
Joined: 12 Nov 2013
Posts: 2
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 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?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52433
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 05 Jun 2014, 00:40
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.

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?


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

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Director
Director
User avatar
Joined: 25 Apr 2012
Posts: 682
Location: India
GPA: 3.21
WE: Business Development (Other)
Premium Member Reviews Badge
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 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.”

Manager
Manager
User avatar
Status: Kitchener
Joined: 03 Oct 2013
Posts: 89
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)
Premium Member
If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 11 Feb 2015, 17:49
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 ?
_________________

Click +1 Kudos if my post helped

Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 11 Feb 2015, 20:21
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
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Manager
Manager
User avatar
Status: Kitchener
Joined: 03 Oct 2013
Posts: 89
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)
Premium Member
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 12 Feb 2015, 05:46
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
User avatar
D
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 12 Feb 2015, 20:27
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

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Manager
Manager
avatar
B
Joined: 14 Dec 2015
Posts: 50
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 16 Dec 2015, 06:03
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?
Senior Manager
Senior Manager
User avatar
Status: Always try to face your worst fear because nothing GOOD comes easy. You must be UNCOMFORTABLE to get to your COMFORT ZONE
Joined: 15 Aug 2014
Posts: 286
Concentration: Marketing, Technology
GMAT 1: 570 Q44 V25
GMAT 2: 600 Q48 V25
WE: Information Technology (Consulting)
GMAT ToolKit User Reviews Badge
If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 11 Jul 2016, 20:40
\(\frac{1}{(u+v)} - \frac{1}{u}=\frac{u-(u+v)}{(u+v)u}=\frac{u-u-v}{(u+v)u}=\frac{-v}{(u+v)u}\).

Hi Bunuel,
Can we cross multiply 'u+v' on left side even if we don't know the sign of 'u+v'
_________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

I need to work on timing badly!!

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52433
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

Show Tags

New post 12 Jul 2016, 02:24
1
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

GMAT Club Bot
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? &nbs [#permalink] 12 Jul 2016, 02:24

Go to page    1   2    Next  [ 26 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.