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crackgmat007
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(i) xyz < 0 impilies that x,y, and z are non-zero and either one of the three values is negative with other two positive, or all three values are negative.
Case (a) x < 0
Then x < z < y so z is between x and y.
Case (b) z < 0.
Choose z = -1 , x = 1, y = 5. then z < x < y
So (i) is insufficient on it's own.
(ii) xy < 0 implies x, and y are non-zero, and one value is positive while the other value is negative.
Case (c) x > 0, y < 0
Choose y = -5, x = 1, z = 2. Then y < x < z
Case (d) x < 0, y > 0
Choose x = -1, y= 5, z = 1. Then x < z < y.
So (ii) is insufficient on it's own.
(i) and (ii) together implies that z > 0. So we have two cases.
Case (e) x > 0, y < 0
y < x < z
Case (f) x < 0, y > 0
x < y < z
So (i) and (ii) aren't sufficient together. Answer is E.
