If u(u+v) 0 and u 0 is 1/(u + v)

Question

If  u(u+v)
eq{0}  and  u >0 , is  \frac{1}{(u+v)} < \frac{1}{u} + v ?

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

KarishmaB · Accepted Answer

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)
eq{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)

Bunuel · Answer

If u(u+v) eq{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)} 0 . Since u+v >0 and u >0 , then u(u+v)>0 . Now, if v>0 , then \frac{-v}{u(u+v)}<0 0 . Since u >0 and v>0 , then \frac{-v}{u(u+v)}<0

SVaidyaraman · Answer

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.

pavanpuneet · Answer

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?

Bunuel · Answer

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)0 you should write -1/u(u+v)<1 BUT if v<0 you should write -1/u(u+v) >1. Hope it helps.

jmuduke08 · Answer

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!

jmuduke08 · Answer

u(u+v)
eq{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?

KarishmaB · Answer

Actually, there is no reason why v shouldn't be 0. The only thing is that (u + v) should not be 0.

chrish06 · Answer

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?

Bunuel · Answer

\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.

23a2012 · Answer

I answered the above equestion as follow please correct me if I am wrong

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 ?

KarishmaB · Answer

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

+ uv(u + v) ]/u(u + v) < 0

Instead, you have 
u-(u+v)  - uv( u+v) /(u+v)u<0

23a2012 · Answer

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?

KarishmaB · Answer

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

sairam595 · Answer

\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'

Bunuel · Answer

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.

warriorguy · Answer

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?

Bunuel · Answer

1.  u(u+v)
eq{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.

EMPOWERgmatRichC · Answer

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:  B

GMAT assassins aren't born, they're made, 
Rich

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

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