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. 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.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 27 Jun 2012
Posts: 371
Concentration: Strategy, Finance

If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?
[#permalink]
Show Tags
25 Dec 2012, 23:36
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) \(v>0\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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




Veritas Prep GMAT Instructor
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
25 Mar 2013, 20:24
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 >




Math Expert
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
26 Dec 2012, 01:42



Director
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
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.
_________________
Srinivasan Vaidyaraman Sravna Holistic Solutions http://www.sravnatestprep.com
Holistic and Systematic Approach



Manager
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
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?



Math Expert
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
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? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 13 Oct 2012
Posts: 48
Concentration: General Management, Leadership

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?
[#permalink]
Show Tags
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
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
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
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
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
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
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
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
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
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
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? \(\frac{1}{(u+v)}  \frac{1}{u}=\frac{u(u+v)}{(u+v)u}=\frac{uuv}{(u+v)u}=\frac{v}{(u+v)u}\). 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? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 25 Apr 2012
Posts: 682
Location: India
GPA: 3.21
WE: Business Development (Other)

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

If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?
[#permalink]
Show Tags
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)  (1uv/u) <0 u(u+v)  uv( u+v)/(u+v)u<0 uuuvuv/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: 8810
Location: Pune, India

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

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?
[#permalink]
Show Tags
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)  (1uv/u) <0
u(u+v)  uv( u+v)/(u+v)u<0 uuuvuv/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)  (1uv/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 uuv+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: 8810
Location: Pune, India

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?
[#permalink]
Show Tags
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
uuv+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
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
16 Dec 2015, 06:03
SravnaTestPrepSravnaTestPrep 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
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)

If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?
[#permalink]
Show Tags
11 Jul 2016, 20:40
\(\frac{1}{(u+v)}  \frac{1}{u}=\frac{u(u+v)}{(u+v)u}=\frac{uuv}{(u+v)u}=\frac{v}{(u+v)u}\). 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
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
12 Jul 2016, 02:24




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 ]



