Three filling pipes, R, S and T, together, can fill an empty tank in two hours. S can fill the tank four times as fast as T. Initially, R alone is opened and after exactly x hours, it is closed and immediately S and T are opened together. The tank is full after exactly another y hours. If the tank was filled in a total of 4 hours, and x ≠ y, find the time (in hours) that T alone would take to fill the tank.6201224

Let's denote the rates at which pipes R, S, and T can fill the tank as r, s, and t respectively. We know that these rates are in tanks/hour.

From the problem, we know that:

1. R, S, and T together can fill the tank in 2 hours. So, r + s + t = 1/2.
2. S can fill the tank four times as fast as T. So, s = 4t.
3. The tank was filled in a total of 4 hours, with R filling for x hours and S and T filling for y hours. So, x + y = 4.
4. In x hours, R can fill xr of the tank, and in y hours, S and T can fill ys + yt of the tank. Since the tank is full, xr + ys + yt = 1.

We can substitute s = 4t into the first equation to get r + 4t + t = 1/2, which simplifies to r + 5t = 1/2.

We also know that R filled the tank for x hours and S and T filled the tank for y hours. So, we can write the equation xr + (4t)y + ty = 1. Substituting r = 1/2 - 5t into this equation gives us x(1/2 - 5t) + 5ty = 1.

We now have two equations, x + y = 4 and x(1/2 - 5t) + 5ty = 1, and two unknowns, x and t. We can solve this system of equations to find the values of x and t.

Solving these equations, we find that t = 1/12. Therefore, T alone would take 12 hours to fill the tank.