Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is

This problem can be solved by setting up equations based on the information given.

1. From 2 pm to 10 pm is 8 hours. During this time, pipe A was open for 8 hours and pipe B was open for 7 hours. Since pipe A fills the tank and pipe B drains it, we can say that the amount of water added by pipe A in 8 hours minus the amount of water drained by pipe B in 7 hours is equal to one full tank. We can write this as: 8A - 7B = 1.

2. From 2 pm to 6 pm is 4 hours. During this time, pipe A was open for 4 hours and pipe B was open for 2 hours. Using the same logic as above, we can write this as: 4A - 2B = 1.

We now have a system of two equations, and we can solve for A and B.

First, multiply the second equation by 2 to make the coefficients of B the same in both equations: 8A - 4B = 2.

Subtract this equation from the first to eliminate B: 8A - 7B - (8A - 4B) = 1 - 2, which simplifies to 3B = -1. Therefore, B = -1/3. This means that pipe B drains 1/3 of the tank per hour.

Substitute B = -1/3 into the first equation to find A: 8A - 7(-1/3) = 1, which simplifies to 8A + 7/3 = 1. Solving for A gives A = (1 - 7/3) / 8 = -1/6. This means that pipe A fills 1/6 of the tank per hour.

If pipe B is not opened at all, then the time taken to fill the tank is the reciprocal of the rate at which pipe A fills the tank, or 1 / (1/6) = 6 hours. Since there are 60 minutes in an hour, this is equivalent to 6 * 60 = 360 minutes.