Pipes X and Z fill a tank, while pipe Y drains it. Pipe Y can empty a full tank in 2 hours less than the time pipe X takes to fill an empty tank. When pipes X, Y, and Z are opened together, an empty tank is filled in 3 hours. If pipes Y and Z are opened together with the tank initially empty, and pipe Y is closed after 1 hour, then pipe Z alone takes 1 additional hour to fill the tank. If pipe X can fill the tank in less than 8 hours, how many minutes does pipe Z take to fill the empty tank by itself?

Let the times for pipes X, Y, and Z be x, y, and z respectively; given Y empties two hours faster than X fills, y = x - 2. The first condition gives 1/x - 1/(x-2) + 1/z = 1/3, and the second condition where Y and Z work for one hour followed by Z working alone for one hour gives 1/z - 1/(x-2) + 2/z = 1. Solving the second equation yields z = (x^2 - 2x)/(x^2 - 2x + 3), and substituting this into the first equation with the constraint x < 8 yields the consistent value x = 6. Plugging x = 6 back into the formula for z gives z = 24/24 = 1 hour. The result is 60 minutes.

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