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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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New post 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.

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?
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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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New post 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.

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

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New post 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.

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?
If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? &nbs [#permalink] 27 Jul 2018, 14:15

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