Plugging in is the most efficient approach right?? you can work this out under 1 min.

tonebeeze wrote:I am interested in learning different approaches to solving this problem.

If x ≠ -y, is (x-y)/(x+y) > 1?

1. x >0
2. y< 0

OA = E

You can pick numbers but you can also use some reasoning. What really helps here is knowing that in DS inequality questions, GMAT is often testing your knowledge of the following rule:

when you multiply or divide an inequality by a negative, the inequality sign "flips."

In this question, if you were to solve for the numerator (x-y), you notice you obviously have to multiply both sides by the denominator (x+y). But if (x+y) were negative, that inequality sign would flip...and without knowing whether the inequality sign flips, you can't answer the question.

So, the question's really asking you to determine whether the sign will flip. Which requires knowing the sign of the denominator (i.e., (x+y)).

Because the denominator has both x and y in it, (1) and (2) are insufficient by themselves.

In combo, for the denominator you have:

(pos) + (neg)

which may be positive or negative depending on how "big" x and y are without considering their sign (i.e., their absolute value).

If x ≠ -y, is (x-y)/(x+y) > 1?

1. x > 0
Don't know anything about y

2. y < 0
Don't know anything about x

3. Combined:
x - y = will always yield a positive value
x + y = could yield a positive value (if |y| > |x|) or negative value (if |y|<|x|)
So it should be (E)