zachlebo wrote:if X+Y/Z > 0, is X<0?
1) X<Y
2) Z<0

my answer was E. both the numerator and the denominator have to either be pos. or neg. to be > 0. statement 1 does not give sufficient info on z, only saying that x is less than y.
statement 2, again only gives info on z, but not on x and y.

the OA is C, and i am confused why. i do not know why. with z being negative and x being less than y, how does that make it negative?

I assume the inequality in the question has x+y in the numerator - that is, that it should read (x+y)/z > 0. Neither statement could be sufficient alone since we certainly need information about all of our unknowns here.

Taking the two statements together, if we know z < 0, then for (x+y)/z > 0 to be true, the numerator will also need to be negative, so x+y < 0. Rewriting Statement 1, we also know that x-y < 0. So we have:

x + y < 0
x - y < 0

and if you add your two inequalities, you find 2x < 0, so x < 0. In general, when you have two inequalities that would look like a 2 equations/2 unknowns problem if you had equals signs, it will often be useful to add the inequalities - just make sure they face the same way. And also be aware that you cannot subtract inequalities (something we often do with equations).

Alternatively, rather than add the inequalities, you can look at them conceptually. If x+y is negative, then at least one of x or y must be negative, so certainly the smaller of the two letters must be negative. So if x < y, x must be negative.

So the answer is C.