Three pipes can fill a reservoir in 10, 15 and 20 hours respectively. If the three taps are opened one after another in the given order, with a certain fixed time gap between them, the reservoir fills in 5 hours. The time gap is

### Concept & Staggered Openings with Uniform Gaps When pipes are opened at fixed intervals, the total time each pipe operates can be expressed algebraically relative to the first pipe. The sum of the work done by all pipes must equal the total capacity. $$ \text{Work} = (\text{Rate}_A \times \text{Time}_A) + (\text{Rate}_B \times \text{Time}_B) + (\text{Rate}_C \times \text{Time}_C) $$ ### Step-by-Step Solution 1. **Given**: Times are 10, 15, and 20 hours. Total fill time is 5 hours. 2. **Calculation**: - Total capacity = LCM(10, 15, 20) = 60 units. - Efficiency A = 60 / 10 = 6 units/hr. - Efficiency B = 60 / 15 = 4 units/hr. - Efficiency C = 60 / 20 = 3 units/hr. 3. Define time active for each pipe: - Let the fixed time gap be $t$ hours. - Pipe A is open for the full 5 hours. Work = $5 \times 6 = 30$ units. - Pipe B is open for $(5 - t)$ hours. Work = $4 \times (5 - t)$ units. - Pipe C is open for $(5 - 2t)$ hours. Work = $3 \times (5 - 2t)$ units. 4. Set up the total work equation: - $30 + 4(5 - t) + 3(5 - 2t) = 60$ - $30 + 20 - 4t + 15 - 6t = 60$ - $65 - 10t = 60$ - $10t = 5$ - $t = 0.5$ hours. 5. Convert to minutes: - $0.5 \text{ hours} = 30 \text{ minutes}$. ### Exam Strategy & Shortcut Pipe A does half the work ($5/10$) instantly, leaving 30 units (out of 60). If there were no gap, B and C would work 5 hours too, doing $20 + 15 = 35$ units. The "extra" 5 units of work ($35 - 30$) comes from the time they were *closed*. B loses $4t$ and C loses $6t$ ($2t \times 3$). So $10t = 5$, giving $t = 0.5$ hours. ### Common Pitfall A standard error is assigning the time gap as $t$ for B and $t$ for C instead of accumulating the gap (i.e., making C wait $2t$). ### Final Answer Therefore, the correct answer is **30 min**.