To solve this problem, we can use the concept of work. Work is equal to the rate at which a task is completed multiplied by the time it takes to complete the task. In this case, the task is filling the tank, and the work required is the volume of water in the tank.
Let's assume that the capacity of the tank is T, and let's denote the rate at which each pipe fills the tank as R. Since all the pipes are the same size, we can assume that each pipe fills the tank at the same rate R.
When 4 pipes are on, the rate at which the tank is filled is 4R. Therefore, in 6 hours, the work done is 6 * 4R.
When 3 pipes are on, the rate at which the tank is filled is 3R. Therefore, in 10 hours, the work done is 10 * 3R.
Since the work done is the same in both cases (filling the tank completely), we can set up the equation:
6 * 4R = 10 * 3R
Now, let's solve for R:
24R = 30R
Subtracting 24R from both sides, we get:
6R = 0
Dividing both sides by 6, we find:
R = 0
This means that each pipe is not filling the tank at all! There seems to be an error or missing information in the problem statement.
However, if we assume that there is a positive rate at which each pipe fills the tank, then the problem can be solved. But in this case, we can't determine the exact number of pipes required to fill the tank in 3 hours without knowing the value of R.
Please provide additional information or clarify the problem statement to get a definitive answer.