kuttingchai wrote:
Voted for E
Solved it following way
(A) |Y| > |Z|
Case 1 :
---[-X =-4]-------[-Y = -3]------[-Z = -2]------[0]-------[Z = 2]-------[Y=3]-------[X=4]------
Case 2 :
---[-Y = -3]------[-Z = -2]------[-X =-1]-------[0]-------[X=1]-------[Z = 2]-------[Y=3]------
when x>0
|X-Y| = |1-3| = 2 and |X-Z| = |1-2| = 1
therefore |X-Y| > |X-Z|
|X-Y| = |4-3| = 1 and |X-Z| = |4-2| = 2
therefore |X-Y| < |X-Z|
when x<0
|X-Y| = |-1+3| = 2 and |X-Z| = |-1+2| = 1
therefore |X-Y| > |X-Z|
|X-Y| = |-4+3| = 1 and |X-Z| = |-4+2| = 2
therefore |X-Y| < |X-Z|
not sufficient
(B) X<0 dont know about Y and Z, therefore nit sufficient
(C) X < 0 and |Y| > |Z|
when x<0
|X-Y| = |-1+3| = 2 and |X-Z| = |-1+2| = 1
therefore |X-Y| > |X-Z|
|X-Y| = |-4+3| = 1 and |X-Z| = |-4+2| = 2
therefore |X-Y| < |X-Z|
[not sufficent]
therefore E
First, remember what do we use the absolute value for ? We use it to express the distance between two real numbers on the number line.
So, |x - y| means the distance between x and y, and |x| = |x - 0| means the distance between x and 0. Also, |x + 3| = |x - (-3)| expresses the distance between x and -3. Obviously, |x - y| = |y - x| (it is the same distance).
So, the question can be rephrased as "is the distance between x and y greater that the distance between x and z ?" or "is x closer to z than to y?"
(1) |y| > |z| means that the distance from y to 0 is greater than the distance from z to 0. This in fact is not important in this case. But certainly y and z are distinct. Regardless of wheather y > z or y < z (both cases are possible, try to draw the number line and visualize the points), we can place x closer to either y or z. So, (1) is not sufficient.
(2) Is evidently not sufficient. Take a point x on the number line at the left of 0, then you can place y and z as you please, each one can be closer or farther from x. Also, now you can consider y = z, in which case the two distances are equal.
Evidently, considering (1) and (2) together won't help either. For example, put x at the left of 0, y and z on either side of x such that y < x and z > x, both negative. You can play with each distance and put either y or z closer to x.
Therefore, answer, E.
_________________
PhD in Applied Mathematics
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