1. **Problem statement:** Pipe A fills a tank in 3 hours, pipe B fills the same tank in 6 hours, and pipe C empties the tank in 8 hours. Pipes A and B start filling the empty tank together, and after 1 hour, pipe C is also opened. We need to find the total time taken to fill the tank. 2. **Formula and rates:** - Rate of pipe A = $\frac{1}{3}$ tank/hour - Rate of pipe B = $\frac{1}{6}$ tank/hour - Rate of pipe C (emptying) = $-\frac{1}{8}$ tank/hour (negative because it empties) 3. **Step 1: Calculate amount filled in the first hour by pipes A and B only:** $$\text{Amount filled in 1 hour} = \left(\frac{1}{3} + \frac{1}{6}\right) \times 1 = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$ So, after 1 hour, half the tank is filled. 4. **Step 2: Calculate the combined rate when pipes A, B, and C are open:** $$\text{Combined rate} = \frac{1}{3} + \frac{1}{6} - \frac{1}{8} = \frac{8}{24} + \frac{4}{24} - \frac{3}{24} = \frac{9}{24} = \frac{3}{8}$$ So, the tank fills at $\frac{3}{8}$ tank/hour when all three pipes are open. 5. **Step 3: Calculate remaining amount to fill:** $$1 - \frac{1}{2} = \frac{1}{2}$$ 6. **Step 4: Calculate time to fill the remaining half with all three pipes open:** $$\text{Time} = \frac{\text{Remaining amount}}{\text{Rate}} = \frac{\frac{1}{2}}{\frac{3}{8}} = \frac{1}{2} \times \frac{8}{3} = \frac{8}{6} = \frac{4}{3} \text{ hours} = 1 \text{ hour } 20 \text{ minutes}$$ 7. **Step 5: Calculate total time to fill the tank:** $$1 \text{ hour (first phase)} + \frac{4}{3} \text{ hours (second phase)} = \frac{3}{3} + \frac{4}{3} = \frac{7}{3} \text{ hours} = 2 \text{ hours } 20 \text{ minutes}$$ **Final answer:** The total time taken to fill the tank is $\boxed{2 \text{ hours } 20 \text{ minutes}}$.