Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full? (Campus Recruitment, 2003)

### Concept & Alternate Working When pipes are opened on alternate hours, calculate the work done in a full 2-hour cycle. Determine how many complete cycles fit into the total work, and then evaluate the remaining work sequentially. $$ \text{Work per Cycle} = \text{Efficiency}_A + \text{Efficiency}_B $$ ### Step-by-Step Solution 1. **Given**: Time for A = 6 hours, Time for B = 4 hours. A opens first. 2. **Calculation**: - Total capacity = LCM(6, 4) = 12 units. - Efficiency of A = 12 / 6 = 2 units/hr. - Efficiency of B = 12 / 4 = 3 units/hr. 3. Analyze the 2-hour cycle: - 1st hour (A works): fills 2 units. - 2nd hour (B works): fills 3 units. - Total work in 1 cycle (2 hours) = 2 + 3 = 5 units. 4. Calculate maximum full cycles: - Max cycles fitting in 12 units = 12 // 5 = 2 cycles. - Work done in 2 cycles (4 hours) = 2 * 5 = 10 units. 5. Process remaining work: - Remaining work = 12 - 10 = 2 units. - The 5th hour is Pipe A's turn. - Pipe A fills 2 units per hour. So it will take exactly 2 / 2 = 1 hour to finish the remaining 2 units. 6. Total time = 4 hours (from cycles) + 1 hour (from A) = 5 hours. ### Exam Strategy & Shortcut Mental math with LCM: Cap = 12. Cycle = 5 units/2hrs. Two cycles give 10 units in 4 hours. Remaining is 2 units. Next is A's turn, and A's capacity is exactly 2 units/hr, taking exactly 1 more hour. Total = 5. ### Common Pitfall A common mistake in alternate hour problems is dividing the total capacity directly by the combined efficiency without separating it into cycles. Doing $12 / 5 = 2.4 \times 2 = 4.8$ hours is incorrect because it ignores the discrete hourly turns. ### Final Answer Therefore, the correct answer is **5**.