GMAT PREP QUESTION

[subbiah1.an]

To answer this qs, you need to know the |x| and |Y|.
Individually you cannot answer this. When you combine and substitute,
ex : X = 3, Y = -1 => (3+1)/ 2 => 2 > 1. "Yes"
but when x = 1, Y = -3 => (1+3)/(-2) => -2 >1. "No"
So you need to |x| and |Y| to answer this qs. hence 'E'.

Please correct me, if I am wrong.

[tomslawsky]

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

(1) x>0
(2) y<0

if i re-arrange the equation in question i get the following:

x-y>x+y, further implying that -y>y, therefore -2y>0 or y<0

So the question after re-phrasing it is asking is y<0?

and using the above i answered the question with B

However the Official answer per the GMAT prep is E

Could someone please explain the methodology so that I can understand as to where I could be going wrong.

thanks in advance.

Best,

M.

I actually like this question, it highlights exactly how inequalities can be "hard" compared to straight algebra questions. The answer to the question is illuminated when you take the final inequality in A and multiply it by -1. The inequaluity sign reverses and you have contradictory statements: Y is greater than 0 and Y is less than 0.

[ayshaw.asif]

The way you have simplified the inequality is not correct because the denominator can be negative or positive and hence when you multiply it by 1 the sign can either be greater than or less than.
Hope this helps

[tomslawsky]

Since dividing or multiplying by a negative number flips the inequality sign, you cannot multiply or divide by an unknown (i.e., a variable), as it could be negative. This is a common trap. For example:

xz < 10z cannot be solved by dividing both sides by z to get x < 10.
If z were negative, the inequality would end up as x > 10. Unless a problem states that a variable is positive or negative, both sides cannot be divided or multiplied by an unknown as you cannot be certain whether to flip the inequality sign.

It is extremely important that you cement this property of inequalities into your mind as the test makers like to trick people on this topic (more information about this important topic within multiplying & dividing inequalities).

More Information About Inequalities
http://www.platinumgmat.com/gmat_study_ ... equalities

I'm NOT affiliated in any way with the website, I just came across it.

[RonPurewal]

this is a difficult problem, because it resists simple algebra. you CANNOT multiply through by the denominator (x + y), because the sign of that denominator is unknown.

therefore, you have to leave the problem as written (ugly as it may be).

since there's no simple algebraic solution, a fallback is to recognize the types of numbers that are important in the problem, and try numbers across those categories.

there are two things that matter in this problem (as may be deduced from an inspection of the problem + experience with these sorts of things):
1. positive vs. negative
2. the relative magnitudes of x and y

let's try numbers across both of these categories.

statement (1)
x must be positive, but y could be positive or negative, and smaller or bigger (or the same) in magnitude.
if x = 1 and y = 2 --> answer = NO
if x = 2 and y = 1 --> answer = NO
if x = 1 and y = -2 --> answer = NO
if x = 2 and y = -1 --> answer = YES
insufficient
[editor: at this point you could notice that the last two examples also satisfy statement 2, and therefore satisfy statements 1 and 2 together. this fact proves that the answer is E right now, and you're done. if you don't notice this (most students won't), then go on.]

statement (2)
y must be negative, but x could be positive or negative, and smaller or bigger (or the same) in magnitude.
if x = -1 and y = -2 --> answer = NO
if x = -2 and y = -1 --> answer = NO
if x = 1 and y = -2 --> answer = NO
if x = 2 and y = -1 --> answer = YES
insufficient

together
x must be positive and y must be negative, but the relative magnitudes can go any way (bigger/smaller/same)
if x = 1 and y = -2 --> answer = NO
if x = 2 and y = -1 --> answer = YES
insufficient

ans (e)