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

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If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 25 Dec 2012, 23:36
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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
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 26 Dec 2012, 01:42
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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.
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 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.
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 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?
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 02 Jan 2013, 03:20
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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.
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 03 Jan 2013, 21:21
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The ques gets reduced to -> uv(u+v) + v > 0?
if u and v both are + then the above is true , hence B
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] 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!
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 25 Mar 2013, 20:24
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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)
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] 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?
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 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.
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 13 May 2014, 08:08
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] 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?
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] New post 05 Jun 2014, 00:40
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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.
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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.


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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? [#permalink] 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
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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?   [#permalink] 05 Jun 2014, 02:29
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