Carcass wrote:
If xy = 30, yz = 20, and zx = 150, what is a possible value of xyz?
(A) 20
(B) 45
(C) 120
(D) 240
(E) 300
Given: xy = 30, yz = 20, and zx = 150
Multiply all three equations to get: (xy)(yz)(zx) = (30)(20)(150)
Simplify: x²y²z² = 90,000
Rewrite as follows: (xyz)² = 90,000
So, EITHER xyz = 300 OR xyz = -300
Answer: E