Pipes A and B can fill a tank in $20$ hours and $30$ hours respectively and pipe C can empty the full tank in $40$ hours. If all the pipes are opened together, how much time will be needed to make the tank full?

### Concept & Multi-Pipe Net Efficiency When more than two pipes are active, algebraically sum their individual 1-hour work capacities. Treat filling pipes as positive work and emptying pipes as negative work. $$ \text{Net 1-hour Work} = \frac{1}{t_1} + \frac{1}{t_2} - \frac{1}{t_3} $$ ### Step-by-Step Solution * **Identify Individual Rates:** Pipe A (fill) $1$-hour work = $\frac{1}{20}$ Pipe B (fill) $1$-hour work = $\frac{1}{30}$ Pipe C (empty) $1$-hour work = $\frac{1}{40}$ * **Calculate Net Rate:** Net part filled in $1$ hour = $\frac{1}{20} + \frac{1}{30} - \frac{1}{40}$ * **Solve the Fractions:** Find the LCM of $20, 30, \text{ and } 40$, which is $120$. Convert fractions: $\frac{6}{120} + \frac{4}{120} - \frac{3}{120}$ Net rate = $\frac{6 + 4 - 3}{120} = \frac{7}{120}$ * **Find Total Time:** The total time is the reciprocal of the net rate. Time = $\frac{120}{7}$ hours. * **Convert to Mixed Fraction:** $120 \div 7 = 17$ with a remainder of $1$. Therefore, Time = $17\frac{1}{7}$ hours. ### Exam Strategy & Shortcut Use the LCM method (Total Work approach). Let Total Work = LCM($20, 30, 40$) = $120$ units. Efficiency of A = $120 / 20 = +6$ units/hr. Efficiency of B = $120 / 30 = +4$ units/hr. Efficiency of C = $120 / 40 = -3$ units/hr. Net Efficiency = $6 + 4 - 3 = 7$ units/hr. Total Time = $\frac{\text{Total Work}}{\text{Net Efficiency}} = \frac{120}{7} = 17\frac{1}{7}$ hours. ### Common Pitfall A frequent error is adding all three rates together ($\frac{1}{20} + \frac{1}{30} + \frac{1}{40}$) ignoring the fact that pipe C is an outlet pipe emptying the tank. ### Final Answer Therefore, the correct answer is **$17\frac{1}{7}$ hours**.