Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in

### Concept & Continuous Plus Alternate Working When one pipe runs continuously while others alternate, construct a cycle based on the alternating pattern. Combine the fixed efficiency with the alternating efficiencies to find the work completed per cycle. ### Step-by-Step Solution 1. **Given**: Time for A = 12 hours, B = 15 hours, C = 20 hours. A is continuous; B and C alternate hourly. 2. **Calculation**: - Total capacity = LCM(12, 15, 20) = 60 units. - Efficiency of A = 60 / 12 = 5 units/hr. - Efficiency of B = 60 / 15 = 4 units/hr. - Efficiency of C = 60 / 20 = 3 units/hr. 3. Define the 2-hour cycle: - **1st hour**: A and B are open. Work = 5 + 4 = 9 units. - **2nd hour**: A and C are open. Work = 5 + 3 = 8 units. - Total work in 1 cycle (2 hours) = 9 + 8 = 17 units. 4. Calculate full cycles needed: - Divide total capacity by cycle work: 60 / 17. - We can fit 3 complete cycles: 3 * 17 = 51 units. - Time taken for 3 cycles = 3 * 2 = 6 hours. 5. Process remaining work: - Remaining work = 60 - 51 = 9 units. - The next (7th) hour will be worked by A and B. - A and B can fill exactly 9 units in 1 hour. 6. Total time = 6 hours + 1 hour = 7 hours. ### Exam Strategy & Shortcut Map the efficiencies quickly: A=5, B=4, C=3. Two hour block = (A+B) + (A+C) = 9 + 8 = 17 units. $17 \times 3 = 51$ units in 6 hours. Remaining 9 units matches exactly the first hour of the next cycle (A+B). Total time = $6+1=7$ hours. ### Common Pitfall A frequent error is assuming a 3-hour cycle (A, then B, then C) instead of reading carefully that A is open continuously while B and C alternate to join A. ### Final Answer Therefore, the correct answer is **7 hrs**.