Two supply tubes, M and N, can fill a large container in 60 minutes and 40 minutes, respectively. How long will it take to fill the empty container if tube N is active for the first half of the total time, and both tubes M and N are active together for the second half of the total time?

### Concept & Time-Splitting Equations When the problem splits the *total time* into fractions (e.g., halves) where different combinations of workers operate, set up an algebraic equation based on the total time rather than the volume. $$ \text{Total Work} = (\text{Rate}_1 \times \text{Time}_1) + (\text{Rate}_2 \times \text{Time}_2) $$ ### Step-by-Step Solution 1. **Determine Capacity and Rates:** * Let total capacity = LCM(60, 40) = 120 units. * Rate of Tube M = $120 / 60 = 2$ units/min. * Rate of Tube N = $120 / 40 = 3$ units/min. 2. **Set Up Time Variables:** * Let the total time taken to fill the container be $2t$ minutes. * First half of the time = $t$ minutes. * Second half of the time = $t$ minutes. 3. **Formulate the Equation:** * During the first half ($t$ mins), only N operates. Work done = $3 \times t$. * During the second half ($t$ mins), both M and N operate. Work done = $(2 + 3) \times t = 5 \times t$. * Total work equals the full capacity: $3t + 5t = 120$. 4. **Solve for Total Time:** * $8t = 120$ * $t = 120 / 8 = 15$ minutes. * The total time is $2t = 2 \times 15 = 30$ minutes. ### Exam Strategy & Shortcut Understand that the average efficiency across the whole period is just the average of the two halves. First half efficiency = 3. Second half efficiency = 5. Average efficiency = $(3 + 5) / 2 = 4$ units/min. Total time = $\frac{\text{Total Capacity}}{\text{Average Efficiency}} = \frac{120}{4} = 30$ minutes. This shortcut completely bypasses algebra. ### Common Pitfall Solving for $t$ and hastily selecting 15 minutes as the final answer. Remember that $t$ only represents *half* of the time. You must multiply by 2 to get the total time. ### Final Answer Therefore, the correct answer is **30 min**.