Pipes and Cisterns: Fill, Drain and Net Rate Tricks for CSAT

Pipes and Cisterns is Time and Work in disguise: instead of workers building something, pipes fill a tank and drains empty it. A filling pipe (inlet) adds water at its rate, an emptying pipe (outlet or leak) removes water, and the tank fills only when the total inflow beats the total outflow. The single governing idea is net rate: inflow minus outflow per unit time. CSAT asks 1 or 2 questions on this almost every cycle, and every one of them yields to the same LCM method you use for Time and Work.

The net rate: filling against emptying

Think of a filling pipe as a worker doing positive work and an emptying pipe as a worker doing negative work. If pipe A fills a tank in 20 minutes and pipe B in 30 minutes, together they fill 1/20 + 1/30 = 5/60 = 1/12 of the tank per minute, so the tank fills in 12 minutes (CSAT 2015).

Worked example 1. A tap fills a cistern in 4 hours while another tap empties it in 9 hours. Both are opened together. After how much time is the cistern filled? Net rate = 1/4 - 1/9 = (9 - 4)/36 = 5/36 of the cistern per hour. Time = 36/5 = 7.2 hours. The minus sign is the whole concept: the emptying pipe fights the filling pipe, so the net rate is smaller and the time longer than either pipe alone.

Worked example 2. Pipes A and B fill a tank in 5 hours and 6 hours; pipe C empties it in 12 hours. All three opened together. Net rate = 1/5 + 1/6 - 1/12 = (12 + 10 - 5)/60 = 17/60 per hour. Time = 60/17 = 3.53 hours. Notice the tank still fills because the combined inflow beats the outflow.

The closed-early pattern: one pipe stops before the tank is full

CSAT loves the pattern where a pipe is switched off before the job is done. The method is two-phase: phase one runs at the combined rate, phase two runs at the remaining pipes' rate.

Worked example 3. Two taps C and D fill a tank in 36 and 48 minutes. Both are opened, but after some time tap D is closed and C alone finishes the tank, the whole process taking 24 minutes. After how many minutes was D closed? Let D run for t minutes. Work done = t x (1/36 + 1/48) + (24 - t) x 1/36 = 1. The combined rate is 7/144 per minute. t x 7/144 + (24 - t)/36 = 1 gives 7t/144 - 4t/144 = 1 - 24/36 = 1/3, so 3t/144 = 1/3 and t = 16 minutes. The slicker way: C ran all 24 minutes, filling 24/36 = 2/3 of the tank. D filled the remaining 1/3, at 1/48 per minute, so D ran for 48/3 = 16 minutes.

The half-tank pattern: more taps join mid-way

Worked example 4. A tap fills a tank in 4 hours. After half the tank is filled, three more similar taps are opened. Total time? One tap fills half the tank in 2 hours. Then 4 taps together fill at 4 times the rate, so the remaining half takes 2/4 = 0.5 hours. Total = 2.5 hours. The key phrase is 'similar taps': identical rates, so just multiply the rate by the number of taps.

Worked example 5. A tap fills a tank in 6 hours. After half is filled, three more similar taps open. One tap fills half in 3 hours; 4 taps fill the remaining half in 3/4 = 0.75 hours. Total = 3.75 hours.

Speed-ratio pipes and the single-pipe question

Worked example 6. One pipe fills a tank three times as fast as another. Together they fill it in 16 minutes. How long does the slower pipe take alone? Let the slower pipe take x minutes; the faster takes x/3. 1/x + 3/x = 4/x = 1/16, so x = 64 minutes. With a 'twice as fast' variant and 12 minutes together: 1/x + 2/x = 3/x = 1/12, so x = 36 minutes alone.

Traps that cost marks

First, forgetting the minus sign: an emptying pipe is subtracted, never added; a question where the outflow exceeds the inflow has the tank emptying, and CSAT sometimes offers 'cannot be determined' as a distractor, not the answer. Second, unit mixing: a stem giving one pipe in hours and another in minutes must be converted before any calculation. Third, 'opened simultaneously' versus 'opened one after another' changes the whole setup: simultaneous opening means adding rates from the start. Fourth, the classic emptying-vs-leak wording: a leak is simply an emptying pipe with an unstated rate, find it by subtraction exactly like the son-alone pattern in Time and Work.

Speed tips for the exam hall

Convert every pipe to a rate immediately, using the LCM method to keep numbers whole: for pipes of 20 and 30 minutes, treat the tank as 60 units and the rates as 3 and 2 units per minute. For closed-early questions, let the pipe that ran the whole time do its full share first, then divide the remainder by the other pipe's rate. For speed-ratio pipes, write the slower pipe's time as x and express the faster pipe as a fraction of x. And always sanity-check the sign: if a drain is open, the answer must be larger than the filling-only time.

Key Terms

• Inlet (filling) pipe is a pipe that adds water to a tank. Its rate is positive: a pipe filling a tank in 20 minutes adds 1/20 of the tank per minute.
• Outlet (emptying) pipe is a pipe that removes water from a tank, such as a drain or a leak. Its rate is negative in net-rate calculations.
• Net rate is the total inflow minus the total outflow per unit time. The tank fills only when the net rate is positive, and filling time equals tank capacity divided by net rate.
• Tank capacity in units is the LCM-method convention of treating the tank as a whole number of units (the LCM of the individual pipe times) so that each pipe's rate is a whole number.
• Closed-early pattern is a question type where one pipe is switched off before the tank is full. Solve it by giving the full-run pipe its whole share first, then handling the remainder.
• Similar taps are pipes with identical flow rates. If one similar tap fills a tank in t hours, n similar taps together fill it in t/n hours.