Bunuel
If xyz ≠ 0, is xyz > 0?
(1) xz < 0
(2) x + y + z = 0
Important points:
1. xyz=/ 0. it means none of xyz are o.
we are looking for xyz>0.
statement 1: xz<0. we can conclude that one of z or x is negative. no information regarding y which could be negative or positive. Not sufficient.
statement 2: x+y+z=0. we know that none of the properties is 0. another point to be noted that all 3 can be negative. Thus it is a combo of both negative and positive.
case 1: x=-6
y=-6
z=12
x+y+z=0
-6-6+12=0
case 2: x=-6
y=12
z=-6
x+y+z=0
-6+12-6=0
case 3: x=12
y=-6
z=-6
x+y+z=0
12-6-6=0
try to analyze more: x=12
y=12
z=-24
12+12-24=0
so ultimately it is impossible to determine how many negative properties we have. It could be 2 or 1.
combining strategies: from 1 we can conclude that either x or z is negative. from 2 we know either any of the 2 or 1 properties has negative value. common thing in both statement is that one of the property could be negative. If only a property is negative xyz will not be greater that 0.
Thus, the best answer is C.