rommysingh wrote:If xyz ≠0, is x (y + z) >= 0?
(1) ¦y + z¦ = ¦y¦ + ¦z¦
(2) ¦x + y¦ = ¦x¦ + ¦y¦
Statement 1: |y + z| = |y| + |z|
Test the following cases:
y=1, z=1
y=-1, z=1
y=1, z=-1
y=-1, z=-1.
|y + z| = |y| + |z| holds true only for the cases in red.
Implication:
y and z have the SAME SIGN.
If x=1, y=1 and z=1, then x(y+z) > 0.
If x=-1, y=1 and z=1, then x(y+z) < 0.
INSUFFICIENT.
Statement 2: |x + y| = |x| + |y|
As illustrated by our work in Statement 1, the equation here will hold true only if x and y have the same sign.
If x=1, y=1 and z=1, then x(y+z) > 0.
If x=-1, y=-1 and z=2, then x(y+z) < 0.
INSUFFICIENT.
Statements combined:
Statement 2 requires that x and y have the same sign.
Statement 1 requires that y and z have the same sign.
Implication:
x, y and z ALL have the same sign.
If x, y and z are all POSITIVE, then x(y+z) = (positive)(positive + positive) = (positive)(positive) = positive.
If x, y and z are all NEGATIVE, then x(y+z) = (negative)(negative + negative) = (negative)(negative) = positive.
Since x(y+z) > 0 in each case, SUFFICIENT.
The correct answer is
C.
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