Pipes & Cisterns

This is time-and-work with the tank as the "work". An inlet pipe has a positive rate; an outlet pipe or a leak has a negative one.

• A pipe filling in a hours has rate +1/a of the tank per hour.
• A pipe emptying in b hours has rate −1/b.
• Add all the rates that are open, then time = 1 ÷ net rate.

Four types:

1. Part filled in t hours = t/a; the empty part is 1 − t/a.
2. Two inlets together → 1/a + 1/b, so time = ab/(a+b).
3. A leak. The tank normally fills in c hours but takes d hours with a leak. The leak's rate is 1/c − 1/d, so it empties a full tank in cd/(d − c) hours.
4. One pipe opened first. Work out what the first pipe filled alone, then divide the remaining fraction by the combined rate.

The single most common mistake is type 3: subtracting the hours — writing "9 − 6 = 3 hours" for a tank that fills in 6 but takes 9 with a leak. Rates subtract, times do not. Guard: the leak must be slower than the filling pipe, otherwise the tank would never fill at all — so its emptying time must always come out larger than the 6 hours the pipe takes on its own.

If the net rate ever comes out negative or zero, the tank never fills — that is the answer, not an arithmetic error.

inlet = +1/a, outlet = −1/b · net rate = sum · time = 1/net rate · leak: 1/c − 1/d → cd/(d−c) · two inlets → ab/(a+b)

A cistern fills in 6 hours, but with a leak in the bottom it takes 9 hours. In how long will the leak empty a full cistern? → Leak rate = 1/6 − 1/9 = 3/18 − 2/18 = 1/18, so it empties the tank in 18 hours. Sanity check: with both acting, the net rate is 1/6 − 1/18 = 3/18 − 1/18 = 2/18 = 1/9, i.e. 9 hours — exactly as the question stated. The tempting 9 − 6 = 3 hours fails this check completely.