Let’s solve this problem step by step.

1. Let the capacity of the tank be 1 unit (for simplicity).

2. Rate of filling/emptying:

   • Pipe A can fill the tank in 2 hours, so its filling rate is: $\frac{1}{2} \text{ units per hour}$
   • Pipe B can empty the tank in 6 hours, so its emptying rate is: $-\frac{1}{6} \text{ units per hour}$
   • Pipe C can empty the tank in $x$ hours, so its emptying rate is: $-\frac{1}{x} \text{ units per hour}$

3. Combined rate when all pipes are open: Since the tank is full in 6 hours when all pipes are open, the net rate of filling with all three pipes is: $\frac{1}{6} \text{ units per hour}$

   Therefore, the combined rate of A, B, and C must equal $\frac{1}{6}$: $\frac{1}{2} - \frac{1}{6} - \frac{1}{x} = \frac{1}{6}$

4. Solve for $x$: Rearranging the equation to isolate $\frac{1}{x}$: $\frac{1}{2} - \frac{1}{6} - \frac{1}{x} = \frac{1}{6}$ First, find a common denominator to simplify $\frac{1}{2} - \frac{1}{6}$: $\frac{1}{2} = \frac{3}{6} \quad \text{and} \quad \frac{1}{2} - \frac{1}{6} = \frac{3 - 1}{6} = \frac{2}{6} = \frac{1}{3}$ Substituting back: $\frac{1}{3} - \frac{1}{x} = \frac{1}{6}$ Now, isolate $\frac{1}{x}$: $\frac{1}{3} - \frac{1}{6} = \frac{1}{x}$ Finding a common denominator again: $\frac{1}{3} = \frac{2}{6}, \quad \frac{2}{6} - \frac{1}{6} = \frac{1}{6}$ So, $\frac{1}{x} = \frac{1}{6}$ Thus, $x = 6$

Answer:

The value of $x$ is 6 hours.

Do you want further details on any step?

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1. What would happen if pipe C took longer to empty the tank?
2. How does changing the filling rate of pipe A affect the overall time?
3. Can this problem be solved using another method?
4. What if only pipes A and B were operating?
5. How would you solve if all pipes had the same filling/emptying rate?

Tip: Always double-check each rate, especially if signs differ, as they affect the overall solution.