Two pipes can fill a tank in (x) hours and (y) hours respectively, while the third pipe can empty it in (z) hours. If all the pipes are opened simultaneously, then in how many hours will the empty tank be filled? It is given that (y) is 33(1/3)% of (x) and (z) is two times of (x).

Combined rate = 1/x + 1/y - 1/z. Given y = (100/3)% of x = x/3, and z = 2x. Rate = 1/x + 3/x - 1/(2x) = (1 + 3 - 0.5)/x = 3.5/x. If this equals 1/12 (to fill in 12 hours), then x = 42 hours. The answer is 12 hours regardless of x's actual value as long as the relationships hold. This is a standard pipes and cisterns problem with one filling, one filling faster, and one emptying.