A cistern can be filled by two pipes filling separately in 12 and 16 minutes separately. Both the pipes are opened together for a certain time but being clogged, only $\frac{7}{8}$ of the full quantity of water flows through the former and only $\frac{5}{6}$ through the latter pipe. The obstructions, however, being suddenly removed, the cistern is filled in 3 minutes from that moment. How long was it before the full flow began? (M.A.T., 2006)

### Concept & Variable Efficiency When pipes operate at reduced capacity, calculate their new working efficiencies based on the given fractions. Determine the work done during the full capacity phase to find the remaining work that was completed during the clogged phase. $$ \text{Time} = \frac{\text{Work Completed}}{\text{Working Efficiency}} $$ ### Step-by-Step Solution 1. **Given**: Pipe 1 fills in 12 min, Pipe 2 fills in 16 min. Last 3 minutes are at full capacity. 2. **Calculation**: - Let total capacity = LCM(12, 16) = 48 units. - Normal efficiency of Pipe 1 = 48 / 12 = 4 units/min. - Normal efficiency of Pipe 2 = 48 / 16 = 3 units/min. - Normal combined efficiency = 4 + 3 = 7 units/min. 3. Calculate clogged efficiencies: - Clogged Pipe 1 = $4 \times \frac{7}{8} = 3.5$ units/min. - Clogged Pipe 2 = $3 \times \frac{5}{6} = 2.5$ units/min. - Clogged combined efficiency = 3.5 + 2.5 = 6 units/min. 4. Work backward from the end: - In the last 3 minutes at full flow, work done = $3 \times 7 = 21$ units. - Remaining work that was done while clogged = $48 - 21 = 27$ units. 5. Calculate time spent clogged: - Time clogged = Work done clogged / Clogged efficiency - Time clogged = $27 / 6 = 4.5$ minutes, or $4\frac{1}{2}$ minutes. ### Exam Strategy & Shortcut Using the LCM of 48 gives clean numbers. Normal rate is 7. Clogged rate is $4(7/8) + 3(5/6) = 3.5 + 2.5 = 6$. The final 3 mins at rate 7 fill 21 units. Thus, $48 - 21 = 27$ units were filled at the clogged rate of 6. $27 / 6 = 4.5$ mins. ### Common Pitfall A common mistake is trying to set up an algebraic equation for the total time $T$, taking $T-3$ as the clogged time, which often leads to algebraic errors compared to the straightforward reverse work method. ### Final Answer Therefore, the correct answer is **$4\frac{1}{2}$ min**.