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

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Senior Manager
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If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

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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]

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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.

(Edited)
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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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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.

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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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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.
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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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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?
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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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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.
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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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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
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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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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.

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]

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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.

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]

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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.

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]

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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.

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]

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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.

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

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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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If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

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11 Feb 2015, 17:49

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 ?
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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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11 Feb 2015, 20:21
23a2012 wrote:

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

u-(u+v) - uv( u+v)/(u+v)u<0
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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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12 Feb 2015, 05:46
VeritasPrepKarishma wrote:
23a2012 wrote:

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

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?
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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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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.

$$\frac{-v}{u(u+v)} - v < 0$$
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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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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?
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If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

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

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