On a number line distance between x and y is greater than

St1:
xyz --> negative.
We could have all x, y and z to the left of 0 on the number line and arranged such that z is not between x and y. We could also have all x, y and z to the left of 0 on the number line and arranged such that z is between x and y. Insufficient.

St2:
xy < 0 --> Either x or y is negative.

Same thing. I can arrange x, y or z to be between or to the left of x or y (whichever is negative). Insufficient.

Using St1 and St2:
xy < 0 and xyz < 0

Say x, is negative --> value of -5, y is positive, value of 5. The distance between x and y is 10. This satisfied the inequalitied xy < 0.

We know z must be positive, otherwise the inequality xyz < 0 cannot be satisfied, but we do not know which positive value z takes on the number line.

Ans E

Case 1 : Any number can be negative or positive. Not conclusive
Case 2.: z can be anywhere. Not conclusive again
Both : lets say x = -ve and y = +ve , z is obviously +ve. Ans distance bt x and y greater than distance bt x and z Hence
__________x______0__z___y_______________

Lets say x=+ve and y = -ve. Hence z can be anywhere.

__________y______0___z__x___z_______________

Therefore E

Great stuff guys! OA is E.