Pipes and Cisterns

Pipes and cisterns is time-and-work wearing a different costume: filling a tank is exactly the same maths as completing a job, with one twist. An inlet pipe does positive work, filling the tank, while an outlet pipe or a leak does negative work, emptying it. Add the positive rates and subtract the negative ones to get a net fill rate, and every question becomes the familiar total-over-rate division from the previous topic — no new arithmetic to learn.

Inlet and Outlet as Plus and Minus

Model the full tank as 1 unit, or better as the LCM of the given times. An inlet pipe that fills the tank in 6 hours contributes +1/6 per hour; an outlet pipe that empties it in 8 hours contributes −1/8 per hour. The signs are the only new idea here: filling is positive, draining is negative. Once every pipe is a signed rate, you simply add them all to find how fast the tank fills or empties on balance, exactly as you added workers' rates in the last topic.

Net rate = sum of inlet rates − sum of outlet rates Time = 1 / net rate

Net Fill Rate

When several pipes run at once, combine their signed rates into a single net rate and divide the tank size by it. Using the LCM approach, if filling takes 6 hours and emptying takes 8, set the tank at LCM(6, 8) = 24 units; the inlet adds 4 units an hour, the outlet removes 3, so the net is 1 unit an hour and the tank fills in 24 hours. If the net comes out negative, the tank is draining, which is the answer to a 'will it ever fill' type of question.

✗ Adding the two times 6 + 8 to combine the pipes  |  ✓ Netting the rates 4 − 3 = 1 unit/hr, so 24/1 = 24 hours

Leak Problems

A leak is just an unlabelled outlet. A classic sum says a pipe fills a tank in 5 hours, but with a leak it takes 6 hours; find the leak's emptying time. The pipe alone does 1/5 per hour; the pipe-with-leak does 1/6 per hour; the leak's rate is the difference, 1/5 − 1/6 = 1/30, so the leak empties a full tank in 30 hours. Any leak question reduces to subtracting the slower combined rate from the faster pipe-only rate, then flipping the result.

Leak rate = (pipe-only rate) − (pipe-with-leak rate)

Solved Examples

Example 1: Pipe A fills in 4 hours, pipe B in 6; with tank = 12 units, A = 3 and B = 2, net 5 units/hour, so 12/5 = 2.4 hours. Example 2: An inlet fills in 10 hours, an outlet empties in 15; tank = 30, net = 3 − 2 = 1 unit/hour, filling in 30 hours. Example 3: Fills in 8 hours, with leak 10 hours; leak = 1/8 − 1/10 = 1/40, so 40 hours.

Pipes Opened at Different Times

Harder SBI Clerk sums open pipes in stages: the inlet runs alone for two hours, then the outlet is opened as well. Handle these in segments — compute the units added during the inlet-only phase, then switch to the net rate for the combined phase and divide the remaining units by it. Keeping a running tally of the units filled at each stage keeps the bookkeeping honest. The arithmetic is never hard; the only risk is losing track of which pipes are open during which stretch of time, so note each changeover clearly.