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Bunuel
If (x + y)/ z < 0 and z not equal to 0, then is the average of x, y, and z greater than zero?

(1) z = -5 and the range of x, y, and z is 6.

(2) x + y = 2 and the range of x, y, and z is 6.



We are looking at: \(\frac{x+y+z}{3}>0……..x+y+z>0\)
So, the question is whether sum of the three number is positive when x+y and z are of opposite sign.

(1) z = -5 and the range of x, y, and z is 6.
Since range is 6, the maximum value of x or y is -5+6 or 1.
Even when we take the largest possible alive of x and y, \(x+y+z=1+1-5=-3\).
Thus, the sum is always negative and answe is NO.
Sufficient

(2) x + y = 2 and the range of x, y, and z is 6.
Let us take x and y as extreme points so that z could be any value between them.
So \(x=y+6 or (y+6)+y=2…….y=-2…..x=-2+6=4\)
Now, z can be anything from -2 to 4.
When \(z=4……x+y+z=4-2+4=6,\) and answer is YES
When \(z=-2……x+y+z=4-2-2=0,\) and answer is NO
(You could take 1+1-5 too and you will get a NO.)
Insufficient


A
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