DS-6

E?

E is correct, pls explain how?

I just took few numbers for x and y and I arrived at E.

Can someone please tell me what is my mistake... Im sure Im ignoring some fundamental number property!

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

(x-y) > (x+y)

0 > 2y OR y < 0 which is St 2, hence B

However when I substitute values for x and y I get E.

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

(x-y) > (x+y)

You can not multiply both sides of inequality if you don't know the sign of the number multiplied.

3 > 2 but if we multiply both sides with -2; -6 <4> 1 => (x-y) > (x+y) only if (x+y) > 0

Ah! Thank you very much!

is there any other method other than picking numbers for solving this? anybody wants to try? stuart, u wanna take a shot at this? TIA

Sure, I'll give it a whirl (although not all DS questions can be answered without picking #s).

Q: Is (x-y)/(x+y) > 1?

To get a "yes" answer, we need 2 things to happen:

First, the top and bottom have to both be + or both be -.

Second, the abs value of (x-y) has to be greater than the abs value of (x+y) (otherwise we'll have a fraction and it won't be > 1).

So, we need to know about the signs and relative values of x and y.

(1) x > 0

Doesn't mention y, but it still could be enough, we need to be careful.

If y is negative, then the top will be bigger than the bottom and we can get a "yes" by picking a y with smaller magnitude than x. However, we can also get a "no" by picking a y with greater magnitude than x (because we'll get a +/- which will be < 0).

Insufficient.

(2) y < 0

We've already done the work to show that negative y could lead to either a yes or a no (see above).

Insufficient.

Since we already considered y being negative along with statement 1, and got both a yes and a no, joining the statements together does us no good either.

(E) is correct.