gmat prep -II ; If x is not equal to -y,

[bblast]

If x is not equal to -y, is (x-y)/(x+y) > 1 ?


1> x>0

2> y<0

I could solve this by looking at the statements. But I would like to see whether we can algebraically reduce a solution.
E

[Anurag@Gurome]

If x is not equal to -y, is (x-y)/(x+y) > 1 ?
1> x>0
2> y<0

Consider the following two cases,

• 1. x = 2 and y = -1 --> (x - y)/(x + y) = 3/1 = 3 > 1
  2. x = 2 and y = -3 --> (x - y)/(x + y) = 5/(-1) = -5 < 1

Both of the above cases satisfy both the statements but in one case the answer is to the main question is YES in other case it is NO. Hence, both statements together is not sufficient to answer the question.

The correct answer is E.

...I would like to see whether we can algebraically reduce a solution.

The problem can be solved algebraically too, but in that case the solution will be a bit more time consuming.

[amit2k9]

you can condition this as per x and y as follows

add 1 to sides and you obtain

x/(x+y)> 1

subtract 1 from both sides and you obtain

y/(x+y) < 0

a. for x= 2 and y = -1|1 you have different results. Not sufficient.

b. for x= +|- 2 and y= -1 you have different results.not sufficient.

a+b

x=2 and y=-1 satisfies both the equations.
thus C it is.

[bblast]

Thanks Anurag and Amit. Sorry I had erred in pasting the OA. {I was under the impression that I had got this one right-but realized after reading anurag's post and reviewing in detail}


Anurag- Could you please share with me an algebraic solution for this as well ?

[Anurag@Gurome]

Anurag- Could you please share with me an algebraic solution for this as well ?

For x > 0 and y < 0, (x - y) will be always positive.
But (x + y) may be positive or negative and accordingly (x - y)/(x + y) will be greater than 1 or less than 1.

If x > |y|, then (x - y)/(x + y) = (x + |y|)/(x - |y|) > 1
If x < |y|, then (x - y)/(x + y) = (x + |y|)/(x - |y|) < 0 < 1

Hence, both statements together is not sufficient.

[bblast]

yup, the solution makes sense, but you are right, one would be better picking numbers for this.

Thanks a ton !!