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# If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/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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01 Jul 2017, 02:03
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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.

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?
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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 Jul 2017, 02:49
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.

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

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27 Jul 2018, 14:15
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.

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?
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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 Sep 2018, 13:04
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.

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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, 13:16
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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Re: If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

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22 Oct 2018, 00:15
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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)

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 ?  [#permalink]

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11 Jun 2019, 02:18
PrashantPonde wrote:
If $$u(u+v)\neq{0}$$ and $$u >0$$, is $$\frac{1}{(u+v)} < \frac{1}{u} + v$$?

(1) $$u+v >0$$
(2) $$v>0$$

Simple solution to the qn:

Given that U>0
Option-1 says U+V>0 THAT IMPLIES 1/(U+V)>0,that means to proof the qn the value of (1/U) +V must atleast be greater than 0,but e have only information regarding U >0 ,BUT dont have value or range of V.SO insufficient.
Option-2 says V>0
so ,dont get tempted to combine,rather try individually.
bkz its already says U>0 and now its saying V>0 ,SO WHAT EVER COMBINATIONS AND ARITHMATIC OPERATION WE WILL DO OF POSITIVE KIND.so the answer to the qn can either be YES or NO ,but cant have both,bkz its only 2 positive numbers we have to operate with.so B is 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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23 Jun 2019, 04:47
1
Bunuel wrote:
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.

Bunuel
Shouldn't the highlighted part be LESS than (<)?
Thanks__
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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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23 Jun 2019, 05:14
Bunuel wrote:
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.

Bunuel
Shouldn't the highlighted part be LESS than (<)?
Thanks__

Right. Edited the typo. Thank you for noticing.
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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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24 Jun 2019, 01:10
PrashantPonde wrote:
If $$u(u+v)\neq{0}$$ and $$u >0$$, is $$\frac{1}{(u+v)} < \frac{1}{u} + v$$?

(1) $$u+v >0$$
(2) $$v>0$$

best way to solve this question would be to simplify the given expression
$$\frac{1}{(u+v)} < \frac{1}{u} + v$$

we get
1<v+uv(u+v)
also given $$u(u+v)\neq{0}$$ and $$u >0$$
#1
$$u+v >0$$
valid only if both u & v are +ve or either of u or v is -ve and smaller than the +ve sign
insufficeint
#2
$$v>0$$
and also given u>0 so 1<v+uv(u+v)
will be true
IMO 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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26 Jun 2019, 04:14
Probus wrote:
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

Hi Probus
Shouldn't you put $$u$$ in the denominator of the highlighted part?
Also, there is a typo in the first highlighted part.
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If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?  [#permalink]

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18 Sep 2019, 22:42
PrashantPonde wrote:
If $$u(u+v)\neq{0}$$ and $$u >0$$, is $$\frac{1}{(u+v)} < \frac{1}{u} + v$$?

(1) $$u+v >0$$
(2) $$v>0$$

If $$u(u+v)\neq{0}$$ and $$u >0$$, is $$\frac{1}{(u+v)} < \frac{1}{u} + v$$?
Q. $$\frac{u}{u+v} < 1 + uv ?$$

(1) $$u+v >0$$
Q. $$u < (u+v)(1 + uv) ?$$
Q. $$u < u+v + uv(u+v) ?$$
Q. $$-v < uv(u+v) ?$$
Since sign of v is unknown
NOT SUFFICIENT

(2) $$v>0$$
v>0; u>0; uv>0;u+v>0
Q. $$u < (u+v)(1 + uv) ?$$
Q. $$u < u+v + uv(u+v) ?$$
Q. $$-v < uv(u+v) ?$$
Since uv(u+v)>0 but -v<0
uv(u+v)>-v
SUFFICIENT

IMO B
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If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ?   [#permalink] 18 Sep 2019, 22:42

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