Is |x−z−y| > x−z+y?
(1) 0<x<z<y
(2) (x-z-y) is negative
Statement 1:
Case 1: x=1, z=2, y=3
Plugging this combination into |x−z−y| > x−z+y, we get:
|1−2−3| > 1−2+3
4 > 2
YES.
Case 2: x=1, z=1.1, y=1.2
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 1.1 − 1.2| > 1 − 1.1 + 1.2
1.3 > 1.1
YES.
Case 3: x=1, z=5, y=10
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 5 − 10| > 1 − 5 + 10
14 > 6
YES.
In every case -- whether x, y and z are close in value or far apart from each other -- the answer is YES.
SUFFICIENT.
Statement 2:
Cases 1, 2 and 3 also satisfy Statement 2.
In Cases 1, 2 and 3, the answer to the question stem is YES.
Case 4: x=0, z=0, y=1, with the result that x-z-y = 0-0-1 = -1.
Plugging this combination into |x−z−y| > x−z+y, we get:
|0 − 0 − 1| > 0 - 0 + 1
1 > 1
NO.
Since the answer is YES in Cases 1, 2 and 3 but NO in Case 4, INSUFFICIENT.
The correct answer is A.
