Let the time taken by pipe B to fill the tank alone be $t_B$ hours. From the problem, we know that pipe A takes 40% of the time taken by pipe B, meaning pipe A takes $0.4t_B$ hours to fill the tank.

Step 1: Determine the rates of the pipes

• The rate of pipe B (filling rate) is $\frac{1}{t_B}$ tanks per hour.
• The rate of pipe A is $\frac{1}{0.4t_B} = \frac{2.5}{t_B}$ tanks per hour.

When both pipes A and B are open together, their combined rate of filling the tank is: $\text{Rate of A and B together} = \frac{2.5}{t_B} + \frac{1}{t_B} = \frac{3.5}{t_B} \text{ tanks per hour}.$

Step 2: Filling time with pipes A, B, and C together

When pipe C is also opened, the time to fill the tank increases by $\frac{3}{4}$ more time compared to only A and B. Let the rate at which pipe C empties the tank be $\frac{1}{t_C}$ tanks per hour.

So, the combined rate of pipes A, B, and C is: $\text{Rate of A, B, and C} = \frac{3.5}{t_B} - \frac{1}{t_C}.$

Given that it takes 15 hours to fill the tank with all three pipes open, the rate at which the tank is filled when all three pipes are open is $\frac{1}{15}$ tanks per hour. Therefore, we have the equation: $\frac{3.5}{t_B} - \frac{1}{t_C} = \frac{1}{15}.$

Step 3: Filling time with only pipes A and B

The time taken to fill the tank with only pipes A and B is less, and the problem says that the time taken with pipe C open is $\frac{7}{4}$ times the time taken with only A and B. Therefore, the time to fill the tank with only pipes A and B is: $\frac{7}{4} \times 15 = 60 \div 7 = 8.57 \text{ hours}.$

Thus, the rate at which pipes A and B together fill the tank is: [ \frac{1}{8.57}. ] This forms thedifference of5minutes idealst