if x ≠-y, is (x-y)/(x+y) > 1?

[maribelsalazar02]

Hi, I am having a hard time understanding the following Gmat Prep Test data sufficiency question.

if x ≠ -y, is (x-y)/(x+y) > 1?

1) x> 0
2) y<0

I chose B because I tried to simplify the equation on the problem as much as I could and ended up with y < 0 (see below).

x-y > 1(x+y)
x-y > x + y
-y > y
0 > 2y
0 > y

Official Answer is E.

Thanks!!

[atul.prasad]

We can rephrase this question as :

is (x-y)/(x+y) -1 >0
on simplifying:

is -2y/(x+y) >0
or 2y/(x+y) < 0 (multiplying by -1, sign changes)

or 2 / (1+x/y) < 0 (dividing the numerator and denominator by y)

in this form it is easy to apply the options:
1) x > 0 .. 2/(1+x/y) is >0 if y >0 and < 0 if (x/y < -1)
Not sufficient
2) y < 0 ... 2/(1+x/y) is > 0 if x < 0 and < 0 if x > 0 and x/y < -1
Not Sufficient
together: (x/y) is always negative , but whether 2/(1+x/y) is < 0 or > 0 depends if |x/y| > 1

So E

The problem with your approach is that you cross multiplied.
You can safely cross multiply only if it is certain that the number multiplied is > 0

for example, here is a case where cross multiplying by a number < 0 changes sign..

(-1)/(-2) < 2/3
cross multiplying by -2
(-1) > -4/3 (note the sign change)

[jnelson0612]

Very nice work atul.

[agha79]

I am having hard time following Atuls approach and I am not sure what "If x <> -y" means.

Ron can you please help!!

[RonPurewal]

I am having hard time following Atuls approach and I am not sure what "If x <> -y" means.

Ron can you please help!!

apparently, "<>" is how computer programmers write "≠". don't feel bad, i didn't know this until ... about twenty seconds ago.
i'll go edit the original post.

--

an alternative approach to solving this problem is simply to plug in numbers -- this is certainly the approach i would take; treating fractional inequalities algebraically is fraught with issues.

let's assume that you can get rid of the individual statements (re-post if you don't know how).
let's take the statements together: x > 0 and y < 0.
let's try x = 1, y = -2 --> Is 3/-1 > 1? --> NO
let's try x = 2, y = -1 --> Is 3/1 > 1? --> YES

there you have it -- solid proof of insufficiency.

notice that this problem also underscores the importance of being familiar with both algebra and number-plugging methods: if a statement is sufficient, then that's probably easier to show with algebra -- but there's no doubt that number plugging is better at showing insufficiency.
be comfortable with both!

[supratim7]

Thank you Ron. Yes, testing numbers quite clearly shows insufficiency.

or 2 / (1+x/y) < 0 (dividing the numerator and denominator by y)

Going back to the older post, can we really divide by y here?? y is an unknown term, it could well be 0

Thanks

[tim]

Good catch. No, you should not divide anything by y unless you know for sure that it isn't 0.

[supratim7]

Appreciate it Tim :)

[jlucero]

Glad we can help.