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

 It is currently 08 Mar 2014, 13:51

### 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:
Senior Manager
Joined: 27 Jun 2012
Posts: 416
Followers: 30

Kudos [?]: 266 [0], given: 179

If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]  25 Dec 2012, 23:36
00:00

Difficulty:

35% (medium)

Question Stats:

62% (02:38) correct 37% (01:42) wrong based on 54 sessions
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,
PraPon

VOTE: vote-best-gmat-practice-tests-excluding-gmatprep-144859.html
Tough RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 4028
Location: Pune, India
Followers: 857

Kudos [?]: 3612 [5] , given: 144

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink]  25 Mar 2013, 20:24
5
KUDOS
Expert's post
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

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 16817 Followers: 2771 Kudos [?]: 17572 [3] , given: 2225 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] 26 Dec 2012, 01:42 3 This post received KUDOS Expert's post 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 [?]: -16 [0], given: 0 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] 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 Senior Manager Joined: 17 Dec 2012 Posts: 350 Location: India Followers: 9 Kudos [?]: 125 [0], given: 8 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} +vor 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 andu>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@gmail.com Sravna Test Prep http://www.sravna.com Online courses and 1-on-1 Online Tutoring for the GMAT and the GRE Manager Joined: 26 Dec 2011 Posts: 117 Followers: 1 Kudos [?]: 8 [0], given: 17 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? Math Expert Joined: 02 Sep 2009 Posts: 16817 Followers: 2771 Kudos [?]: 17572 [0], given: 2225 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. Hope it helps. _________________ Manager Joined: 21 Jul 2012 Posts: 61 Followers: 0 Kudos [?]: 6 [0], given: 32 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! Manager Joined: 21 Jul 2012 Posts: 61 Followers: 0 Kudos [?]: 6 [0], given: 32 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 857 Kudos [?]: 3612 [0], given: 144 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?   [#permalink] 27 Mar 2013, 19:56
Similar topics Replies Last post
Similar
Topics:
If mv<pv<0 is v>0? 4 24 Feb 2007, 16:35
4 If mv < pv < 0, is v > 0? 10 03 Sep 2008, 10:44
2 If u and v are positive real numbers, is u>v? 1. u^3/v 4 16 Jul 2011, 06:17
2 If mv < pv< 0, is v > 0? 3 20 May 2012, 13:14
14 If mv < pv < 0, is v > 0? 6 20 Jun 2012, 01:53
Display posts from previous: Sort by