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

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.

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, 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!

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.

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.

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 ?

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

Instead, you have
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?

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.

\(\frac{-v}{u(u+v)} - v < 0\)

\(\frac{1}{(u+v)} - \frac{1}{u}=\frac{u-(u+v)}{(u+v)u}=\frac{u-u-v}{(u+v)u}=\frac{-v}{(u+v)u}\).

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.

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.

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:

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