Two faucets that fill a tank in 3 h and in 6 h take 2 h together, not 4.5 h. You add what each faucet fills in one hour and invert the total. You will see three solved cases, with two faucets, with a drain and with a missing time, and how to check each answer.
| Situation | What you do with the rates | Example |
|---|---|---|
| Two or more faucets open at once | Add them and invert the total | 3 h and 6 h → 2 h |
| A faucet and a drain | Subtract the drain's rate | 4 h and 12 h → 6 h |
| You know the total and one part | Subtract the known part from the total | Together 4 h, one alone 6 h → the other 12 h |
The essentials
- The rate of a faucet is the reciprocal of its time; if it fills the tank in 5 h, it contributes 1/5 per hour.
- Rates are added because they measure the same thing, tanks per hour. Times are not added.
- Check the result by multiplying the total rate by the time, which must give 1 tank.
Worked example: 3 h and 6 h faucets
One faucet fills a tank in 3 hours and another fills it in 6 hours. How long do they take if both are opened at once?
- Turn each time into a rate. The first contributes 1/3 of a tank per hour and the second, 1/6.
- Find the common denominator. The least common multiple (LCM) of 3 and 6 is 6, so 1/3 = 2/6.
- Add the rates. 2/6 + 1/6 = 3/6 = 1/2 of a tank per hour.
- Invert the total. 1 ÷ 1/2 = 2 hours.
They take 2 hours.
Check. In 2 hours the first faucet fills 2 of 3 parts and the second fills 1 of 3, which add up to one full tank.
Why you add rates and not times
A rate says how much work gets done in one hour, and the work of two faucets open at once builds up hour by hour. A time, on the other hand, says how long a whole task lasts, and two times for simultaneous tasks do not pile up. Adding 3 + 6 = 9 h would describe doing one task and then the other, not working at the same time.
Averaging does not work either. The average of 3 h and 6 h is 4.5 h, but the 3 h faucet contributes twice as much as the 6 h one, and together they take less time than either one alone. That is why an answer larger than the shorter of the times gives away a mistake.
The last step uses the reciprocal, which Wikipedia describes as the number that gives 1 when multiplied by x; for a fraction a/b it is b/a (accessed October 17, 2026). If you want to review how to get it, see the guide on the reciprocal of a fraction. For adding with different denominators, use the adding fractions calculator; the guide on why denominators are not added explains step 2 of the example.
A faucet with a drain
A drain works against the faucet. If a faucet fills the tank in 4 hours and a drain empties it in 12 hours, the net rate is 1/4 − 1/12 = 3/12 − 1/12 = 1/6 of a tank per hour. The time is 6 hours.
With the drain open, the tank takes longer than with the faucet alone, and 6 h is more than 4 h. If you get a shorter time, you added the drain’s rate. The subtracting fractions calculator handles that subtraction.
When one participant’s time is missing
Sometimes you know the combined time and the time of one alone. Two workers together finish a job in 4 hours, and the second, working alone, would take 6 hours. The combined rate is 1/4 and the second worker’s is 1/6, so the first worker’s rate is the difference, 3/12 − 2/12 = 1/12 of the job per hour. The first worker takes 12 hours.
Here the operation is a subtraction because you know the total and one part. The missing time must be longer than the combined time, and 12 h is. If you prefer to set it up as an equation, the fraction equations tool solves an equality with an unknown.
How to check any result
Multiply the total rate by the time and confirm that it gives exactly 1. For the drain, the net rate times 6 hours gives 1; for the worker, the rate times 12 hours also gives 1. If it does not give 1, a rate was calculated wrongly or an addition should have been a subtraction. The guide on verifying a fraction answer gathers more checks.
The second check is about order of magnitude. With several faucets, the combined time is less than the smallest of the individual times; with a drain, it is greater than the time of the faucet alone. And if the times come in different units, convert them to a single unit before finding the rates.
Limits of the method
The method assumes that each faucet, worker or printer keeps its pace from start to finish. If the flow changes, for example because the pressure drops, the rates are no longer constant and a different setup is needed. Distance, speed and time problems follow another relationship, covered in the guide on speed with fractions.
The same arithmetic shows up in physics: resistors connected in parallel combine by adding reciprocals, as the article on parallel resistors shows. For more problems set in everyday situations, visit the fraction word problems.
Practice
- One faucet fills a tank in 5 h and another in 20 h. How long do they take together?
- Two faucets fill in 4 h and 6 h, and a drain empties in 12 h. How long does the tank take to fill?
- A 2 h faucet runs alone for 1 h and then a 6 h faucet is opened. How long does it take in total, in hours and minutes?
See the answers
- 1/5 + 1/20 = 4/20 + 1/20 = 5/20 = 1/4, so they take 4 h.
- The rates are 3/12 + 2/12 − 1/12 = 4/12, or 1/3 per hour: it takes 3 h.
- In the first hour it fills half, and half is left. Together they fill 1/2 + 1/6 = 2/3 per hour, so the second phase lasts 1/2 ÷ 2/3 = 3/4 h = 45 min, and the total is 1 h 45 min.
Source: Wikipedia, article Multiplicative inverse (accessed October 17, 2026).