If xy ≠ 0, is the reciprocal of x/y greater than x/y?

Question: Is y/x > x/y?
When will y/x be greater than x/y?
- If the fraction is positive (x and y are both positive or both negative), then |x| should be less than |y|.
- If the fraction is negative (exactly one of x and y is negative), then |x| should be more than |y|.


(1) –(1)x > y
If x is positive, y is certainly negative and |x| is less than |y|. In this case y/x < x/y.
If x is negative, y could be negative or positive and we cannot say anything about the relative absolute values of x and y because they depend on whether y is +ve or -ve.
This statement alone is not sufficient.

(2) xy > 0
The fraction is positive. Either both x and y are positive or both are negative. Is |x| < |y|? We do not know. So this statement alone is also not sufficient.

Using both statements together,
If x is positive, y is negative - this is not possible since fraction is positive.
x must be negative and y must be negative. So -x will be positive and will be greater than y irrespective of whether its absolute value of x is greater than or less than the absolute value of y. So we still do not know if |x| is less than |y|.
For example,
x = -2, y = -3 satisfy all conditions. Here y/x > x/y.
x = -3, y = -2 satisfy all conditions. Here y/x < x/y.

Both statements together are not sufficient.

Answer (E)

The question asks whether y/x>x/y or y^2-x^2>0 (taking it all in one side) or (y-x)* (y+x)>0

Indirectly it wants to check, if (y-x) & (y+x) have same sign or not (than is, are both +ve or -ve)

given xy not equal to zero.

1) -x>y or y+x<0 here we have no information about y-x, hence insufficient

2) xy>0 or x & y have same signs (both +ve or both -ve). Here we can't comment upon the sign of x+y (either +ve or -ve) and of y-x since we do not know y>x or y<x. hence insufficient

combining 1) & 2) we have y+x as -ve or both x & y are negative (since they have same signs). But still we do not know the sign of of y-x, as we do not know y>x or y<x. hence E

is x/y < y/x or, x^2<y^2 or, |x| < |y|
(1) if x = 2, y = -3, x^2 < y^2. again, if x=-4, y = 2, then x^2 >y^2.not sufficient.
(2) when x is positive, y is also positive, and vice versa. not sufficient.
Together, suppose, x = -2, y =-1, then |x| > |y|. again, x = -4, y = -6, then |x|< |y|. not sufficient.
E is the answer IMHO.