question : Is |x-y| > |x| - |y| ?buckrich wrote:Is |x-y| > |x| - |y| ?
(1) y < X
(2) xy < 0
Statement 1:
y < x
if x=5, y = 2
|x-y| = |5-2| = 3
|x| - |y| = |5| - |2| = 3, so |x-y| = |x|-|y|
if x=5, y=-2
|x-y| = 7
|x|-|y| = 3, so |x-y| > |x|-|y| ... So insufficient
Statement 2:
xy < 0 => x and y have opposite signs
if x = 5, y= -2
|x-y| = 7
|x|-|y| = 3, so |x-y| > |x| - |y|
if x =-2, y=5
|x-y| = |-2-5| = 7
|x|-|y|=|-2|-|5| = 2-5 = -3 , still |x-y| > |x| - |y| --- Sufficient
Hence, B
Using graph :
y<x => means the area below the line y=x, means it contains half of the 3rd quadrant, full 2nd quadrant and half 1st quadrant. So, clearly there are some areas (like the 1st quadrant), which is outside the graph of |x-y| > |x|-|y| as drawn below.
Also, you can see that xy > 0 lies in 2nd and 4th quadrant:
when x>0, y<0 (Means 4th quadrant)
when x<0, y>0 (Means the 2nd quadrant)
The graph for |x-y| > |x| -|y| is shown below (clearly all the areas in 2nd and 4th quadrant lies inside this).
