Data Sufficiency

If x != -y, is (x-y)/(x+y) > 1?

1. x > 0
2. y < 0

OA: E

Here's, what I did, and apparently it's the wrong answer.

x - y > x + y --> 0 > 2y --> 0 > y. Once I saw that, I immediately chose B.

Can anybody show me what I did wrong, without using the picking number method. Thanks.

student22 wrote:If x != -y, is (x-y)/(x+y) > 1?

1. x > 0
2. y < 0

OA: E

Here's, what I did, and apparently it's the wrong answer.

x - y > x + y --> 0 > 2y --> 0 > y. Once I saw that, I immediately chose B.

Can anybody show me what I did wrong, without using the picking number method. Thanks.

You made one of the most common inequality errors: you cross-multiplied without considering whether the denominator could be 0 or negative.

Remember, when you multiply or divide both sides of an inequality by a negative, you have to flip the direction of the inequality.

You took:

(x-y)/(x+y) > 1

and multiplied both sides by (x+y). However, since (x+y) could be negative (or even 0; if x=1 and y=-1 that's what we get), it's not safe to do so.

Thanks for the reply, Stuart, I didn't consider that the denominator could be negative. It couldn't be 0 since the problem said that x != -y.

So, that being said is there a way to solve it without picking numbers?

student22 wrote:Thanks for the reply, Stuart, I didn't consider that the denominator could be negative. It couldn't be 0 since the problem said that x != -y.

So, that being said is there a way to solve it without picking numbers?

1! = 1

So if x = 1, then

y = -(1!) = -1

and

(x + y) = (1 + -1) = 0

My bad, by "!=" I mean not equal to. I don't know how to do the standard mathematical line through an equal sign, so I just used the programming version of "not equal to"

student22 wrote:If x != -y, is (x-y)/(x+y) > 1?

1. x > 0
2. y < 0

OA: E

Here's, what I did, and apparently it's the wrong answer.

x - y > x + y --> 0 > 2y --> 0 > y. Once I saw that, I immediately chose B.

Can anybody show me what I did wrong, without using the picking number method. Thanks.

I solved it in this way

(x-y)/(x+y)>1
multiplying both side by (x+y)^2 (this is positive always) i arrived
-(y^2)>xy
now y^2 is always positive so to hold the above either x or y has to be negative while the other positive and numerical value of x has to be greater than that of y

now neither of the two conditions confirms that |x|>|y|

hence ans E

but this took me more than 2 mins to reach...please let me kow if any other shorter method available.
[spoiler][/spoiler]

student22 wrote:My bad, by "!=" I mean not equal to. I don't know how to do the standard mathematical line through an equal sign, so I just used the programming version of "not equal to"

Even if it were a '!' (factorial) instead of '≠' (not equal to), there were many possibilities for x, y values apart from just 1 and -1. When x is 2, y is -2, x is 3, y is -6, x is 6, y is -720 etc. Nevertheless, you can copy the '≠' (not equal to) sign for future posts.

If x ≠ -y, then x + y ≠ 0, and hence (x - y)/(x + y) is defined, and it would be greater than 1 in the following two cases, otherwise NOT:

Case 1: When each of (x - y) and (x + y) is positive and (x - y) > (x + y). This is possible only when x >0, x > -y, and y < 0.

Case 2: When each of (x - y) and (x + y) is negative and (x - y) < (x + y). This is possible only when x < y, x < 0, and few considerations for y as well.

A bird-eye-view now over the question can make us feel that no statement alone is sufficient. If taken together, we could tempt to the case 1 above and could say, YES, [spoiler]C[/spoiler]; but NO! Case 1 also wants to corroborate whether x > -y or not. Hence, [spoiler]still insufficient

E[/spoiler]

Thanks sanju09 and liferocks for the replies, I like both of your methods for solving this. Sanju, I think your method might be a little bit faster though.