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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # 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 Retired Moderator Joined: 04 Aug 2016 Posts: 508 Location: India Concentration: Leadership, Strategy GPA: 4 WE: Engineering (Telecommunications) Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 01 Jul 2017, 01:03 1 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. When I first looked at this problem, I did not get how to use given condition u(u+v) not equal to 0. I took numbers and build test cases to check. Is that a valid approach in this case or did I just get lucky? Math Expert Joined: 02 Sep 2009 Posts: 50544 Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 01 Jul 2017, 01:49 1 warriorguy 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. When I first looked at this problem, I did not get how to use given condition u(u+v) not equal to 0. I took numbers and build test cases to check. Is that a valid approach in this case or did I just get lucky? 1. $$u(u+v)\neq{0}$$ means that neither u nor u + v is 0. This is given to ensure that the denominators in $$\frac{1}{(u+v)}$$ and $$\frac{1}{u}$$ are not 0 and thus the fractions are defined. 2. For DS questions testing values is good to get insufficiency (one value gives a NO answer, another gives an YES answer, which means that the statement is not sufficient). Getting that a statement is sufficient by testing values is a bit trickier. You can get say YES answer with several values and conclude that the statement is sufficient but there might be some value which is giving a NO answer and you just missed it, thus incorrect conclusion. So, you should test properly and be careful. _________________ Intern Joined: 07 May 2015 Posts: 35 If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 27 Jul 2018, 13:15 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. I'm curious why you knew to take those steps to simplify the equation. When I first saw that u > 0, my first inclination was to multiply both sides by u, to get u/(u+v) < vu --> u/u + u/v < vu --> 1 + u/v < vu -->stuck. What made you think to first subtract the fraction? EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12841 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] ### Show Tags 11 Sep 2018, 12:04 Hi All, We're told that U(U+V) is not equal to 0 and U > 0. We're asked if 1/(U+V) is less than the sum of 1/U and V. This is a YES/NO question. While it might look complex, it can be handled rather easily by TESTing VALUES and a bit of Number Property knowledge. 1) (U+V) > 0 IF.... U = 1, V = 1, then 1/(1+1) = 1/2 and 1/1 + 1 = 2... 1/2 IS less than 2 and the answer to the question is YES. U = 1, V = 0, then 1/(1+0) = 1/1 and 1/1 + 0 = 1... 1 is NOT less than 1 and the answer to the question is NO. Fact 1 is INSUFFICIENT. 2) V > 0 With the information in Fact 2, we know that BOTH U and V are POSITIVE. This allows us to use a particular Number Property rule to our advantage. Regardless of whether you are using positive fractions or positive integers, when BOTH variables are POSITIVE, 1/(U+V) will ALWAYS be LESS than 1/U. As either (or both) of the values of U or V INCREASE, the value of 1/(U+V) will DECREASE. Obviously 1/U = 1/U, but when you 'add in' a positive value for V to that first fraction, the value of that fraction will DECREASE. Try any positive values for U and V and you'll see. Thus, 1/(U+V) will be LESS than 1/U. Adding V to 1/U simply increases the difference. Thus, the answer to the question is ALWAYS YES. Fact 2 is SUFFICIENT. Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

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17 Sep 2018, 12:16
Hi,

Approach of VeritasKarishma, is brilliant. The approach is very much logical. Its like calculations in brain.

However what if i needed to solve algebrically

So here it is
we are given that u(u=v)$$\neq{0}$$
means u$$\neq{0}$$, (u+v) $$\neq{0}$$ and u$$\neq{-v}$$
we are also given that u>0
we are asked it $$\frac{1}{u+v}< \frac{1}{u}+ v$$
Lets try to rephrase out question
Because u>0
$$\frac{1}{u+v}< \frac{1+ uv}{u}$$
$$\frac{1}{u+v}- \frac{1+ uv}{u}$$ <0
$$\frac{u- (1+uv)(u+v)}{(u+v)}$$ <0
$$\frac{u- (1+uv)(u+v)}{(u+v)}$$ <0
$$\frac{u- u -v -uv(u+v)}{(u+v)}$$ <0
$$\frac{ -v -uv(u+v)}{(u+v)}$$ <0
multiplying both sides by -1 and flipping the sign
$$\frac{ v +uv(u+v)}{(u+v)}$$ >0

if you want to identify what the question is really asking simplify further (though in above step you should have identified what is being asked)
Splitting the numerator and cancelling the common terms we get
$$\frac{ v}{(u+v)}$$ +uv>0
So we see above inequality holds if we are told that v>0
So question really is IS v>0

So Statement 2 answers our question

This method is lengthy and not sure you should employ on the test day but if we start thinking what is being asked in DS question we can mark the answer confidently.

Probus
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Posts: 529
Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

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21 Oct 2018, 23:15
Top Contributor
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)

1. No individual information of u. Insuffient.

2. v>0 and u>0.
Let v=1, u=1.
1/2<1/1
Ans. B
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? &nbs [#permalink] 21 Oct 2018, 23:15

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