Pipes & Cisterns

1Inlets add, outlets subtract

A pipe that fills a tank in x hours does +\frac{1}{x} of the tank per hour. A pipe or leak that empties the same tank in y hours does -\frac{1}{y} per hour. There is no separate "emptying" logic — outlets are negative workers, and the net rate is just the sum of the signed rates.

The tank fills only when that sum is positive. If the outlet is as fast as the inlet, or faster, the net rate is zero or negative and the tank never fills, no matter how long you wait.

Figure. On a capacity of 30 units, the inlet contributes +3 units an hour and the outlet removes 2. The net bar is their algebraic sum, not a third pipe — and because it sits above the dashed empty line, the tank fills. Swap the magnitudes so the outlet is -3 and the inlet +2, and the net bar flips below that line: same figure, opposite verdict.

How it works

1. Write every rate with a signFill in x hours becomes +1/x per hour; empty in y hours becomes -1/y.
2. Add the signed ratesHelpers and inlets add; drains and leaks subtract. One number remains.
3. Invert only if the net is positiveTime to fill is 1 divided by the net rate. A zero or negative net means the tank does not fill.

2Net fill time from one inlet and one outlet

When a single inlet of x hours runs against a single outlet of y hours and y > x, the net rate is \frac{1}{x}-\frac{1}{y} and the time to fill is \frac{xy}{y-x}. The denominator is the difference of the times, not their sum — swapping that sign is the standard wrong answer on this formula.

The condition y > x is not decoration. If the outlet is faster (y < x), the same expression goes negative and correctly reports that the tank is draining. Two filling pipes use the sibling formula \frac{xy}{x+y}, where the denominator adds because both rates are positive.

Figure. Concrete case of the fill formula xy/(y-x) with x=5, y=20: on a 20-unit tank the inlet adds 4 an hour and the outlet removes 1, so the net bar is +3 and the tank fills in 20/3 hours. The outlet bar sits below the dashed empty line because its rate is negative; if it were taller than the inlet the net would flip below that line and the tank would drain.

How it works

1. Confirm the outlet is the slower pipeNeed y > x. Otherwise the tank drains and "time to fill" is the wrong question.
2. Write \frac{xy}{y-x}Product of the two times over the outlet-minus-inlet difference.
3. Cross-check with signed ratesNet = 1/x - 1/y; invert. The two routes must agree.

3Take capacity as the LCM of the pipe times

Fractions slow pipes problems the same way they slow ordinary work problems. Call the tank's capacity the LCM of every pipe's time — inlet and outlet alike — and each hourly flow becomes a whole number of units. Filling pipes get a positive integer; emptying pipes get a negative one.

The choice of capacity never appears in the answer, because you divide by the net rate at the end. What you gain is that two inlets and an outlet, or a pipe closed halfway, stay in integer units until that final division.

The LCM route

1. Capacity = LCM of all timesInclude every inlet and every outlet mentioned in the question.
2. Rate = ± capacity ÷ own timePlus for a filling pipe, minus for an emptying pipe or leak.
3. Net, then divideAdd the signed rates; time is capacity divided by that net, provided the net is positive.

4Finding how long a leak takes to empty

If a pipe alone fills a tank in a hours, but with a leak the same fill takes b hours and b > a, the leak is the missing negative rate. Its emptying time for a full tank is \frac{ab}{b-a} hours — the same shape as the net-fill formula, read for the leak instead of for the tank.

The leak's rate is fill rate minus effective rate: \frac{1}{a}-\frac{1}{b}. Inverting that difference is exactly \frac{ab}{b-a}. Use the shortcut directly rather than setting up two rate equations from scratch every time.

Figure. On a 30-unit tank the clean inlet is +3 an hour and the slowed fill is only +2, so the leak is the missing -1. Inverting that rate empties 30 units in 30 hours — the same \frac{ab}{b-a} the ledger used. The figure's point is the subtraction of rates, not of times: the gap between the first two bars is the third bar's magnitude.

How it works

1. Name a and ba is the clean fill time; b is the slowed fill time with the leak. Need b > a.
2. Write \frac{ab}{b-a}That is the leak's time to empty a full tank on its own.
3. Check with LCM unitsCapacity LCM(a,b); leak rate = clean rate − slowed rate; invert.

5Pipes open for only part of the time

When a pipe is opened late or closed early, do not invent a single combined rate for the whole question. Split the timeline into intervals, compute the units each pipe contributes over the hours it is actually open, and sum to one full tank.

The LCM capacity makes this bookkeeping mechanical: each open hour of a pipe adds or subtracts a fixed integer. Work interval by interval until the remaining units are a clean multiple of the rate still running, then finish with one division.

Figure. Work done against time for the worked example: A alone for the first 3 hours (shallower slope), then A and B together (steeper). The kink is the only geometric fact that matters — each segment's slope is the rate in force during that interval, and the horizontal position of the top point is the total clock time 8.4 hours. Axes are the renderer's default frame; both series ends sit on them: empty at the origin, full at the top of the y-axis.

How it works

1. Fix capacity as the LCMEvery pipe then has an integer units-per-hour rate.
2. Credit each interval separatelyUnits = rate × hours open in that stretch. Sum what has been filled so far.
3. Finish the remainder at the final rateRemaining units divided by whoever is still open — that is the last stretch of time.