Bunuel wrote:

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.