Pipes and cisterns problems are time and work problems in disguise: inlet pipes do positive work and outlets or leaks do negative work. Enter the hours each pipe needs to see the net rate and the time to fill the tank.
Enter the filling time of each inlet pipe.
Enter the emptying time of the outlet or leak.
Read the time to fill.
Net rate: 1/T = 1/a + 1/b − 1/c
Pipes and cisterns questions ask how long a tank takes to fill or empty when several pipes run together. Inlet pipes add water and outlets or leaks remove it. This calculator handles two inlet pipes and one outlet or leak, and reports the net hourly rate and the time to fill.
The topic appears in SSC, bank PO and clerk, railway and state PSC aptitude sections. It is really the time and work model with a twist: an outlet does negative work. The same reasoning applies to overhead tanks in housing societies, where a leak slows filling.
Enter the hours each inlet needs alone, and the hours the outlet needs to empty a full tank. Use 0 for a pipe that is not present.
1. For each inlet that fills the tank in a hours, write its rate as +1/a tank per hour.
2. For each outlet or leak that empties the tank in c hours, write its rate as −1/c tank per hour.
3. Add all rates to get the net rate: R = 1/a + 1/b − 1/c.
4. If R is positive, time to fill = 1 ÷ R.
5. If R is zero or negative, the tank never fills; with water already in it, it would stay level or drain.
6. For LCM-style working, set tank capacity = LCM of all the times, find litres per hour for each pipe, and divide.
Treating outlets as negative work lets one formula cover every case. Each pipe contributes a rate, positive for filling and negative for emptying, and the net rate is their sum. The time is the reciprocal of the net rate, provided it is positive. With inlets of 4 and 12 hours and an outlet of 8 hours, the net rate is 1/4 + 1/12 − 1/8 = 5/24 tank per hour, so the tank fills in 24/5 = 4.8 hours. Without the outlet it would take 3 hours, so the outlet adds 1.8 hours.
A common question gives the normal filling time and the longer time with a leak, and asks how long the leak alone would take to empty the tank. If the pipe fills in t hours normally and in T hours with the leak, then 1/t − 1/L = 1/T, so 1/L = 1/t − 1/T. For t = 6 and T = 8, 1/L = 1/24 and the leak empties a full tank in 24 hours. The same equation can be rearranged to find any one missing time.
If the outlet empties faster than the inlets fill, the net rate is negative and the reciprocal would be a negative time, which has no physical meaning. The calculator reports that the tank does not fill. If the rates are exactly equal, the level stays constant. Questions that open and close pipes at different times are solved phase by phase: compute the fraction filled in each phase with the pipes open in that phase, then add the fractions until they reach 1.
A housing society's overhead tank has a main inlet that fills it in 5 hours and a borewell pipe that fills it in 10 hours, but a cracked valve would empty the full tank in 15 hours.
Filling per hour: 1/5 + 1/10 = 0.3
Emptying per hour: 1/15 = 0.066667
Net filling per hour: 0.3 − 0.066667 = 0.233333
Time to fill: 1 ÷ 0.233333 = 4.2857 hours
Answer: Time to fill the tank 4.2857 hours; Net filled per hour 23.33%
Adding the outlet's rate instead of subtracting it.
Adding times instead of rates.
Reporting a negative time instead of recognising the tank never fills.
Ignoring that pipes in multi-phase questions are open only for part of the time.
Mixing minutes and hours across pipes.
Solving aptitude questions in SSC, banking and railway exams.
Estimating how long a water tank takes to fill with a leak.
Planning pump schedules for overhead and underground tanks.
Explaining signed rates in a classroom setting.
What if the leak is faster than the pipes?
The net rate is negative, so the tank never fills; it would empty instead.
How do I find the leak time if I know the delayed fill time?
1/leak = 1/normal time − 1/delayed time.