Pipe A alone takes only 40% of the time taken by pipe B alone to fill an empty tank. When an outlet pipe C is also opened simultaneously with pipe A and pipe B, it takes 3/4 more time to fill the empty tank than it takes, when only pipe A and pipe B are opened together. If it takes 15 hours to fill when all the three pipes are opened simultaneously, then in how much time (in hours) can pipe C empty the full tank operating alone

Let's denote the time taken by pipe B alone to fill the tank as 't'.

According to the problem, pipe A alone takes only 40% of the time taken by pipe B. So, the time taken by pipe A to fill the tank is 0.4t.

The rate of work done is the reciprocal of the time taken to complete the work. So, the rates of work done by pipe A and pipe B are 1/(0.4t) and 1/t respectively.

When pipe A and pipe B are opened together, their combined rate of work done is the sum of their individual rates, which is 1/(0.4t) + 1/t = 1/(0.4t) + 2.5/(0.4t) = 3.5/(0.4t).

When an outlet pipe C is also opened simultaneously with pipe A and pipe B, it takes 3/4 more time to fill the tank than it takes when only pipe A and pipe B are opened together. So, the time taken is 1.75 times the time taken by pipe A and pipe B together, which is 1.75/(3.5/(0.4t)) = 0.4t.

According to the problem, this time is 15 hours. So, we have 0.4t = 15, which gives t = 15/0.4 = 37.5 hours.

The rate of work done by pipe C is the difference between the combined rate of pipe A and pipe B and the rate when all three pipes are opened together, which is 3.5/(0.4t) - 1/(0.4t) = 2.5/(0.4t).

So, the time taken by pipe C to empty the full tank operating alone is the reciprocal of its rate, which is 0.4t/2.5 = 0.16t.

Substituting t = 37.5 hours, we get the time taken by pipe C to empty the full tank operating alone as 0.16 * 37.5 = 6 hours.