Data Sufficiency

GMATGuruNY wrote:

massi2884 wrote:If u(u+v) ≠ 0 and u > 0, is 1/(u+v) < 1/u + v?
1) u+v >0
2) v>0

Statement 1: u+v > 0
Plugging u=1 and v=1 into 1/(u+v) < 1/u + v, we get:
1/(1+1) < 1/1 + 1
1/2 < 2.
YES.

Plugging u=2 and v=-1 into 1/(u+v) < 1/u + v, we get:
1/(2-1) < 1/2 - 1
1 < -1/2.
NO.

Since in the first case the answer is YES, and in the second case the answer is NO, INSUFFICIENT.

Statement 2: v>0
Since all of the values are positive, we can rephrase the questions stem.
Is 1/(u+v) < 1/u + v?

Put the right side over a common denominator:
1/(u+v) < (1+uv)/u

Cross-multiply:
u < (u+v)(1+uv)

Distribute on the right-hand side:
u < u + u²v + v + uv².

Subtract u from each side:
0 < u²v + v + uv².

Question stem, rephrased:
Is u²v + v + uv² > 0?
Since all of the values on the left-hand side are positive, the answer is YES.
SUFFICIENT.

The correct answer is B.

Two take-aways:
1. When we're plugging values into a Yes/No DS question, we need to determine what types of values could change the answer from YES to NO or vice versa. Since statement 2 requires that v>0, the sign of v seems to be an issue here. Thus, we should consider NEGATIVE values when we plug into statement 1.

2. When an inequality problem is restricted to positive values, algebra becomes much easier to implement, since we don't have to worry about the direction of the inequality.

Asma77 wrote:
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 ?

The steps in red contain algebraic errors and yield an incorrect rephrase of the question stem.
I suggest that you check your work.