Work Rate: Two Faucets Filling a Tank Together

• Rates add; times do not. The whole problem turns on this one idea. You will see why adding 2 and 3 (or averaging them) cannot work, using only logic.
• How to convert "time to finish" into "fraction done per unit time" by taking a reciprocal, and how to convert back at the end.
• How to add fractions with unlike denominators in a context where the common denominator has a physical meaning (a convenient "job size").
• How to judge an answer before you finish: the combined time must fall between a certain minimum and maximum, and you can compute both bounds in your head.
• A pattern that reaches well beyond faucets: the same arithmetic governs parallel resistors, pumps, copy machines, and parallel computing.

Most people's first instinct on this problem is to do something with the numbers 2 and 3: add them, average them, multiply them. The better question is: what does each faucet accomplish in a single hour? That is a quantity we can add, because the two faucets are running during the same hour. Think of it as a ledger with one column per faucet.

Step 1 — Turn each time into a rate

If Faucet A fills the whole tank in 2 hours and pours at a steady pace, then in one hour it fills half the tank. Faucet B needs 3 hours for the whole tank, so in one hour it fills one third.

In general, a job that takes t hours has a rate of 1/t jobs per hour. This is the reciprocal relationship at the heart of every work-rate problem.

Step 2 — Add the rates

With both faucets open, each hour of running time adds A's contribution and B's contribution to the tank. To add 1/2 and 1/3 we need a common denominator. The smallest one is 6.

Together, the faucets fill five sixths of the tank every hour. Notice that this is already more than A manages alone (3/6), exactly as it should be.

Step 3 — Flip the rate to get the time

If the tank fills at 5/6 of a tank per hour, then the time for one full tank is the reciprocal. Dividing "1 tank" by "5/6 tank per hour" leaves a unit of hours.

Step 4 — Express the result in hours and minutes

The question asks for hours, so 1.2 hours (or 6/5 hours) is a complete answer. In everyday terms, 0.2 of an hour is 0.2 × 60 = 12 minutes, so the faucets need 1 hour and 12 minutes.

Fractions of a tank are fine, but there is a trick that removes the fractions altogether. The problem never tells us how big the tank is, so we are free to choose a size. Choose one that both times divide evenly. Since the times are 2 and 3 hours, a 6-gallon tank is a natural choice. (6 is the least common multiple of 2 and 3.)

Step 1 — Assign the tank a size

Let the tank hold 6 gallons. This is a modeling choice, not a fact from the problem. We will see in a moment why it cannot change the answer.

Step 2 — Find each faucet's rate in gallons per hour

Step 3 — Add the whole-number rates

Step 4 — Divide the job by the combined rate

Both methods are the same idea in different clothing: Method 1 keeps the tank as "1" and lets the rates be fractions, while Method 2 keeps the rates as whole numbers and lets the tank be a number of units. Many students find Method 2 easier to do under exam pressure, and Method 1 easier to generalize.

Substitute the answer into the situation. In 1.2 hours, each faucet runs at its own steady pace:

For a second, independent check, watch the tank fill in 12-minute blocks. In 12 minutes (one fifth of an hour) A fills 1/10 of the tank and B fills 1/15, so together they add 1/10 + 1/15 = 1/6 of a tank per block.

After exactly 6 blocks, which is 72 minutes, the tank is full. The 60-minute row also matches Step 2 of Method 1: after one hour the tank is 5/6 full.

You can bracket the answer before doing any fraction arithmetic. Adding a second faucet can only speed things up, so the combined time must be shorter than the faster faucet's solo time of 2 hours. And it cannot be shorter than 1 hour: if B were as fast as A, two 2-hour faucets would take exactly 1 hour, and B is slower than that.

The answer, 1.2 hours, sits just above the lower bound of 1 hour. That makes sense because A does most of the work (A supplies 3 of the 5 "shares" of the combined flow).

Let faucet A take a hours alone and faucet B take b hours alone. Their rates are 1/a and 1/b, and the combined time t satisfies:

With a = 2 and b = 3, this is t = 6/5. The "product over sum" form is a handy shortcut for exactly two workers. You can check it works here: product 6, sum 5, answer 6/5.

Notice that the same structure shows up when you add any quantities that combine through their reciprocals. Two resistors in parallel have 1/R = 1/R₁ + 1/R₂. Two lenses in contact combine their powers. Two pipes draining the same reservoir combine their flow rates. Mathematicians describe this as a "harmonic" combination. In fact, t is exactly half the harmonic mean of a and b, which explains why it is always smaller than the quicker worker's time.

• Electronics: two parallel resistors of 2 Ω and 3 Ω have a combined resistance of 6/5 = 1.2 Ω, the same arithmetic as the faucets, with current flowing instead of water.
• Computing: if one server processes a batch of requests in 2 hours and another in 3, splitting the work so that both finish together takes 1.2 hours. This is the logic behind load balancing and parallel processing.
• Operations and logistics: two crews unloading the same shipment, two printers working on one print order, or two pumps emptying the same flooded basement are all the same equation.

"Cistern" problems, in which several spouts or channels fill a basin at different speeds, are among the oldest word problems in mathematics. They appear in the writings of Heron of Alexandria and in the Greek Anthology, and similar pool-and-channel problems turn up in ancient Chinese mathematical texts. They endure in classrooms because they teach, in the simplest possible setting, the idea that a rate is a quantity that can be added, which is a first step toward the calculus of rates of change.

This problem was designed so that the numbers come out cleanly: 1.2 hours is a short decimal. Real problems often are not so kind. Suppose Faucet B took 5 hours instead of 3. Watch the first two steps. They look exactly the same:

Flip it: t = 10/7 ≈ 1.43 hours, or about 1 hour 26 minutes. The method has not changed at all. The only difference is that the final fraction does not convert to a short decimal. That is normal, and it should not make you doubt your work. Leave the answer as 10/7 if an exact value is wanted, or round sensibly if a decimal is acceptable. You can still use the sanity check from above: 1.43 is between 1 and 2, so it passes.

Once rates are comfortable, two natural extensions are waiting. The first is the "mixed team" problem, where workers join or leave at different moments (Push Further #4 is a first taste). The second is variable rates: what if a faucet's flow slowly decreases as the tank's water level rises? Then the rate is no longer a constant, and the total amount filled is found by integrating the rate over time. That is how this arithmetic turns into calculus.