gmatcracker2010 wrote:
If x ≠ -y is \(\frac{x-y}{x+y}>1\)?
(1) x > 0
(2) y < 0
I'm actually going to go straight to plugging in numbers here, without rephrasing. That's different from usual, but there are a few reasons I'm going to approach it like that:
- I can't just cross-multiply, because I don't know whether the denominator of the fraction is positive or negative. So the easiest way to rephrase won't work at all.
- Also, this looks like an algebra problem, not a positive/negative type problem. That's because the inequality is "greater than 1", not "greater than 0" (positive) or "less than 0" (negative). So, we actually care about what the specific numbers are, and we can't necessarily just use logical reasoning and figure out whether they're positive or negative.
So, let's start with the statements!
Statement 1 x is positive.
Let's try x = 1 to make things easy.
is \(\frac{1-y}{1+y}>1\)?
Let's see what happens when we use different values for y...
if y = 1, then \(\frac{1-y}{1+y}=\frac{1-1}{1+1}\)= 0, which is NOT greater than 1.
How can we make the fraction bigger than 1? We'd like the top to be bigger than the bottom. Well, the only way for that to happen is by subtracting a NEGATIVE number on the top. Let's try making y negative:
if y = -2, then \(\frac{1-y}{1+y}=\frac{1-(-2)}{1+(-2)} = \frac{3}{-1}\)= -3, which is NOT greater than 1.
Interesting. To make it greater than 1, we need to keep the denominator from becoming negative. Let's make y a smaller negative number:
if y = -0.5, then \(\frac{1-y}{1+y}=\frac{1-(-0.5)}{1+(-0.5)} = \frac{1.5}{0.5}\)= 3, which IS greater than 1.
Since it can go either way, the statement is insufficient.
Statement 2: We can reuse the two cases we just tested! We found a case where y = -2, and the result was NOT greater than 1. We also found a case where y = -0.5, and the result WAS greater than 1. This statement is insufficient as well.
Statements 1 and 2 together Since we used the exact same cases to prove both statements insufficient individually, we can also use those same cases when we put the statements together. Therefore, the statements are insufficient together and the answer is E.
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