Pipes and cisterns

In short

Pipes and cisterns problems ask how long a tank takes to fill or empty when one or more pipes act on it together. Each pipe is described by the time it would take alone, which is converted into a rate (part of the tank per hour); inlet pipes add, outlet pipes and leaks subtract. It is the same reciprocal-rate algebra as work problems, with the one extra idea that a rate may be negative. The topic appears in almost every Indian placement aptitude test (TCS NQT, Infosys, Wipro, Accenture, Capgemini, Cognizant, AMCAT, eLitmus, CoCubes), in IBPS/SBI/SSC/RRB exams, in CAT/XAT/CMAT quant, and in first-year engineering/BBA foundation mathematics. The physical version of the same arithmetic is used daily in water-supply design, pump and overhead-tank sizing, and process-tank scheduling.

Explained

The basic model: time, rate and sign

• If a pipe fills a tank in x hours alone, its rate is 1/x tank per hour. In 1 hour it fills 1/x of the tank; in t hours, t/x.
• An inlet pipe has a positive rate (+1/x); an outlet pipe, waste pipe or leak has a negative rate (−1/y). This "negative work" is the only real difference from ordinary work problems.
• The tank's own size never matters unless the question gives a flow rate in litres per minute or asks for capacity — treat the full tank as 1 unit of work.
• Rates are additive: net rate = sum of all rates of pipes that are open. Time to fill = 1 ÷ net rate.
• Sign check first: if the outlet rates together exceed the inlet rates, the net rate is negative and the tank never fills — it empties. A positive answer in that situation is a wrong answer.
• Units must match: convert everything to hours or everything to minutes before adding (a rate of 1/20 per minute is 3 tanks per hour, not 1/20 per hour — mixing them is the commonest slip).

Standard results for pipes acting together

• Two inlets, x and y hours → together T = xy/(x + y) (the harmonic combination).
• One inlet x, one outlet y, with y > x → T = xy/(y − x) hours to fill.
• One inlet x, one outlet y, with x > y → the tank empties; a full tank empties in xy/(x − y) hours.
• Three pipes, two filling (x, y) and one emptying (z) → 1/T = 1/x + 1/y − 1/z.
• n identical pipes, each taking x hours → together x/n hours.
• Back-calculation: if A and B together take T and A alone takes x, then B alone takes Tx/(x − T).
• Pipe with a leak: if a pipe that should fill in x hours actually takes x′ hours (x′ > x), the leak alone empties a full tank in L = x·x′/(x′ − x).
• Rates are proportional to efficiency: "A is twice as fast as B" means rate(A) = 2 × rate(B), i.e. A takes half the time.

Two reliable solving methods

1. LCM (unit-capacity) method — preferred in exams.

• Take the tank's capacity as the LCM of all the given times, in arbitrary "units".
• Each pipe's rate in units/hour = capacity ÷ its own time (negative for outlets).
• Add rates, then time = capacity ÷ net rate. All arithmetic stays in whole numbers, no fractions.
• Example sketch: times 12, 18, 36 → capacity 36 units; rates +3, +2, −1 → net 4 → 36/4 = 9 hours.

2. Fraction / percentage method.

• Work in fractions of the tank (1/12 + 1/18 − 1/36) or in per-cent per hour (8.33% + 5.56% − 2.78% = 11.11% per hour → 100/11.11 = 9 hours).
• Useful when the question asks "what part of the tank is filled in 4 hours" or gives a part-full tank.
• Handy reciprocals to memorise: 1/3 = 33.33%, 1/6 = 16.67%, 1/7 ≈ 14.29%, 1/8 = 12.5%, 1/9 = 11.11%, 1/11 ≈ 9.09%, 1/12 = 8.33%, 1/16 = 6.25%.

The standard question types

Pipes opened in turns, closed midway, and leaks found late

• Closed midway: if the tank is full in T hours overall and pipe B was shut after t hours, then (T/x) + (t/y) = 1 for the pipe that ran the whole time (x) and the one that ran t hours (y). One linear equation, one unknown.
• Alternate opening: compute the net gain per complete cycle, then find how many whole cycles fit *without overshooting*, and finish the remainder at the rate of whichever pipe is open next. The order matters: A-then-B and B-then-A can give different answers.
• Critical trap: never just divide capacity by net gain per cycle. The tank may become full *during* a cycle, in the middle of a filling spell. Always test the peak level reached inside each cycle.
• Leak discovered after some time: the work done before the leak is clean (t/x); the rest is done at the reduced net rate (1/x − 1/L). Equate the sum to 1.
• Emptying a part-full tank: if the tank is 3/4 full and the net rate is negative, time = (3/4) ÷ |net rate|.
• Tank overflow questions: if inlets are left open after the tank is full, the "extra" water simply overflows — the fill time is unaffected.

From the exam to the field: capacity, flow and diameter

• Real pipes are rated by discharge Q in litres per minute (LPM) or litres per second, not by "time to fill". Time to fill = volume ÷ net discharge; volume of a cuboidal tank in litres = length × breadth × height in metres × 1000.
• Rate from geometry: Q = A·v, so for the same water velocity the discharge is proportional to the cross-sectional area, i.e. to d². A 40 mm pipe carries about four times a 20 mm pipe, not twice — a favourite twist in questions that give pipe diameters.
• Real flow also depends on the head (height of the overhead tank) and on friction loss, so a tap's discharge falls as the sump level drops; aptitude problems assume a constant rate, which is an idealisation worth stating in an interview.
• Indian practice points: domestic water demand is conventionally taken as about 135 litres per capita per day for towns with full flushing systems (IS 1172 and the CPHEEO *Manual on Water Supply and Treatment*; editions are revised, so check the current one). Household overhead tanks are commonly 500-2,000 litres, sumps 1,000-5,000 litres; rotomoulded polyethylene water tanks are covered by IS 12701. Domestic pumps are rated in HP with a discharge curve in LPM against head.
• Limitations of the exam model: constant rates, instant switching, no pipe friction, no cavitation, no partial valve opening. It is arithmetic practice, not hydraulics — Bernoulli's equation and Darcy-Weisbach friction belong to fluid mechanics.

Speed technique for tests

• Spend five seconds deciding the sign and the method (LCM almost always) before writing anything.
• For two pipes, use the ratio shortcut: times in ratio a : b fill together in time ab/(a+b) × (common factor).
• Answer options are often "nice" numbers: estimate the net rate and eliminate. If two inlets take 12 and 18 hours, the joint time must lie between 6 and 12 — half the faster time and the faster time — and is in fact 7.2 h; anything outside that range is discardable instantly.
• Write the final answer in the units asked (hours and minutes, or minutes): 7.5 hours is 7 h 30 min, 42.75 h is 42 h 45 min.

Key points

• Rate - reciprocal of solo time; a pipe that fills in x hours does 1/x of the tank per hour.
• Negative work - an outlet pipe or leak has rate −1/y; it is subtracted from the inlet rates.
• Net rate - algebraic sum of the rates of all open pipes; time to fill = 1 ÷ net rate.
• Two inlets - together take xy/(x + y) hours, always less than the faster pipe alone.
• Inlet plus outlet - fill time xy/(y − x) when the inlet (x) is faster than the outlet (y).
• No-fill condition - if total outlet rate ≥ total inlet rate, the tank never fills.
• LCM method - take capacity = LCM of the times; all rates become integers.
• Leak formula - a leak that turns x hours into x′ hours empties a full tank in x·x′/(x′ − x).
• Partly full tank - remaining fraction ÷ net rate, never the full 1 ÷ net rate.
• Pipe closed early - form one equation: (time pipe ran)/(its solo time) summed over pipes = 1.
• Alternate opening - find the net gain per cycle, then check whether the tank fills *inside* a cycle.
• Order matters - starting with the inlet or with the outlet changes the answer in alternate-opening problems.
• Capacity in litres - only obtainable when one pipe's flow rate (L/min) is given; capacity = rate × that pipe's solo time.
• Discharge and diameter - at equal velocity, Q ∝ d², so doubling the diameter quadruples the flow.
• n identical pipes - take x/n hours; k of them take x/k.
• Efficiency language - "twice as fast" doubles the rate and halves the time, never the reverse.
• Overflow - water above the full mark is wasted; it does not shorten the fill time.
• Unit discipline - convert all times to one unit (minutes or hours) before adding rates.

Formulas and examples

Example 1 — three pipes with a part-full tank (LCM method). Pipe A fills a tank in 8 min, B in 12 min, and C empties it in 6 min. The tank is already 1/3 full and all three are opened. When is it full?

• Capacity = LCM(8, 12, 6) = 24 units. Rates: A = +3, B = +2, C = −4 units/min.
• Net rate = 3 + 2 − 4 = +1 unit/min (positive, so it will fill).
• Already filled = 1/3 × 24 = 8 units; remaining = 24 − 8 = 16 units.
• Time = 16 ÷ 1 = 16 minutes.

Example 2 — a pipe closed before the tank is full. Two pipes fill a cistern in 15 h and 20 h. Both are opened together, but the second is closed after some time; the cistern is full in 12 h. When was the second pipe closed?

• First pipe runs all 12 h → 12/15 = 4/5 of the tank.
• Remaining = 1 − 4/5 = 1/5, done by the second pipe at 1/20 per hour.
• t = (1/5) ÷ (1/20) = 4 hours after the start.
• Check: 12/15 + 4/20 = 0.8 + 0.2 = 1. ✓

Example 3 — leak discovered after the pipe has run a while. A pipe can fill a tank in 15 h. After it has run for 5 h a leak develops, and the tank is full only 20 h after the start. How long would the leak alone take to empty a full tank?

• Clean stage: 5 h × 1/15 = 1/3 filled. Remaining = 2/3, taken in 20 − 5 = 15 h.
• Net rate in stage 2 = (2/3)/15 = 2/45 tank per hour.
• 1/15 − 1/L = 2/45 → 3/45 − 1/L = 2/45 → 1/L = 1/45.
• Leak alone empties the full tank in 45 hours.

Example 4 — capacity in litres from a waste pipe's flow rate. Two taps fill a tank in 20 min and 30 min. A waste pipe can empty 3 litres per minute. With all three open the tank fills in 18 min. Find the capacity.

• Inlets together: 1/20 + 1/30 = (3 + 2)/60 = 1/12 per min.
• 1/12 − 1/W = 1/18 → 1/W = 1/12 − 1/18 = (3 − 2)/36 = 1/36 → W = 36 min to empty the full tank.
• Capacity = 3 L/min × 36 min = 108 litres.

Example 5 — pipes opened alternately (the overshoot trap). Pipe A fills a tank in 6 h; outlet B empties it in 8 h. They are opened alternately for one hour each, starting with A. When is the tank full?

• Capacity = LCM(6, 8) = 24 units. A = +4 units/h, B = −3 units/h.
• Net gain per 2-hour cycle = +1 unit. Naive answer 24 × 2 = 48 h — wrong.
• After 2n hours the level is n units. During the next A-hour the level peaks at n + 4.
• Full (24 units) first becomes reachable when n + 4 = 24 → n = 20, i.e. after 40 h the level is 20 units, and the 41st hour (A's turn) raises it 20 → 24.
• Tank is full in exactly 41 hours.

Comparison — same two pipes, three different schedules

Where it is used

• Municipal and building water supply: sizing sumps, overhead tanks and pumps so that the daily demand (litres per capita per day) is met within the hours of supply — exactly the net-rate calculation, done by civil and public-health engineers and by plumbing consultants.
• Pump and tank manufacturers (Kirloskar, Crompton, Texmo, Sintex, Ashirvad, Supreme): matching pump discharge in LPM to tank volume and head; dealer sizing charts are built on fill-time arithmetic.
• Process industries: filling and draining reactors, blending tanks and ETP sump pits in chemical, dairy (Amul-type plants), brewing and pharmaceutical units, where simultaneous inflow and outflow is the normal operating state.
• Irrigation and canal operation: reservoir inflow against release and seepage loss — the "leak" term is real and measured.
• Recruitment tests and government exams: 1-3 questions per paper in TCS NQT, Infosys, Wipro, Capgemini, Accenture, AMCAT, eLitmus, CoCubes, IBPS/SBI PO and Clerk, SSC CGL/CHSL, RRB NTPC, CAT/XAT/CMAT.
• Analyst and operations roles: the identical model describes production inflow versus despatch in warehouses, ticket arrival versus resolution in BPO/IT support queues, and token-bucket or buffer fill-and-drain rates in systems design.

Interview questions

• Q: Why do you add reciprocals of the times instead of the times themselves? A: Because the quantities that add are rates (tank per hour), not durations; adding times would wrongly suggest two pipes take longer than one.
• Q: Two pipes fill a tank in 12 and 24 hours. How long together? A: 1/12 + 1/24 = 3/24 = 1/8, so 8 hours.
• Q: A pipe fills a tank in 10 h; a leak empties it in 15 h. When will the tank fill with both acting? A: 1/10 − 1/15 = 1/30, so 30 hours.
• Q: What if the leak emptied it in 8 hours instead? A: Net rate = 1/10 − 1/8 = −1/40, which is negative, so the tank never fills; a full tank would empty in 40 hours.
• Q: A pipe fills a tank in 6 h, but with a leak it takes 9 h. How long does the leak take to empty the tank? A: 1/6 − 1/9 = 1/18, so the leak alone empties it in 18 hours.
• Q: Does the capacity of the tank matter? A: No, unless a flow rate in litres per minute is given or the capacity is asked; otherwise the tank is one unit of work.
• Q: Three pipes A, B, C together fill a tank in 5 h; A is twice as fast as B and B twice as fast as C. How long does C alone take? A: Rates k, 2k, 4k sum to 7k = 1/5, so k = 1/35 and C alone takes 35 hours (B 17.5 h, A 8.75 h).
• Q: Why can't you divide total capacity by the net gain per cycle in alternate-opening problems? A: The tank often becomes full in the middle of a filling spell, so the last cycle is incomplete; dividing assumes every cycle is completed and overstates the time.
• Q: Two pipes have diameters 20 mm and 40 mm and the same water velocity. How do their fill times compare? A: Discharge goes as d², so the 40 mm pipe delivers four times as much and fills the tank in one-quarter of the time.
• Q: A tank is 2,400 litres and a pump delivers 40 litres per minute while a tap draws 10 litres per minute. Fill time? A: Net 30 L/min, so 2,400/30 = 80 minutes.
• Q: Where does this model break down in a real water system? A: Real discharge falls with reduced head and rises with pipe diameter, varies with friction loss and pump wear, and valves are rarely switched instantaneously — the constant-rate assumption is an idealisation.
• Q: Two inlets take x and y hours; can their combined time ever exceed the faster pipe's time? A: Never — adding a positive rate can only increase the net rate, so the joint time is always less than min(x, y).

Common mistakes

• Adding or averaging the times instead of the rates, giving a joint time longer than a single pipe.
• Forgetting the minus sign for an outlet pipe or leak, or subtracting an inlet by mistake when the question wording is reversed.
• Mixing minutes and hours in the same rate equation without converting.
• Using 1 ÷ net rate for a tank that is already partly full, instead of the remaining fraction ÷ net rate.
• In alternate-opening problems, dividing capacity by the gain per cycle and ignoring that the tank can fill mid-cycle — and ignoring which pipe starts.
• Computing capacity in litres when no flow rate is given, or multiplying the flow rate by the *combined* time instead of that pipe's own solo time.
• Reading "A is twice as fast as B" as "A takes twice the time", which inverts the rates.
• Reporting a decimal answer in the wrong form — writing 7.5 hours as "7 hours 50 minutes".

Related topics

Time and work · Ratio, proportion and partnership · Time, speed and distance · Simplification and approximation · Mensuration: volumes and surface areas · Percentages