Work-Rate Problems: Combined Time for Workers, Pipes, and Machines
October 29, 2026
Work-rate problems ask how long a job takes when several workers, pipes, or machines work together. The key idea is to think in rates: if a worker finishes a job in t hours, the worker completes 1/t of the job per hour. Rates add when workers work together, so the combined rate is 1/t₁ + 1/t₂ + ..., and the combined time is the reciprocal of that sum.
If Anna can paint a room in 6 hours and Ben can paint it in 4 hours, their rates are 1/6 and 1/4 of a room per hour. Together, 1/6 + 1/4 = 2/12 + 3/12 = 5/12 of a room per hour, so the job takes 12/5 = 2.4 hours (2 hours 24 minutes).
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Three machines: one finishes an order in 3 hours, one in 6, one in 12. Combined rate = 1/3 + 1/6 + 1/12 = 4/12 + 2/12 + 1/12 = 7/12 per hour, so the time is 12/7 = 1.71 hours. Two pipes: A fills a tank in 10 hours and B in 15 hours; together, 1/10 + 1/15 = 5/30 = 1/6, so 6 hours. Filling and draining: a tank fills in 5 hours with the inlet pipe, and a drain empties a full tank in 8 hours. With both open, the net rate = 1/5 − 1/8 = 8/40 − 5/40 = 3/40 per hour, so the tank fills in 40/3 = 13.3 hours. Subtract rates for work that undoes progress.
A shortcut for two workers: combined time = (t₁ × t₂) ÷ (t₁ + t₂). For 6 and 4: 24 ÷ 10 = 2.4 hours.
| Problem | Rates (jobs per hour) | Combined time |
|---|---|---|
| Workers taking 6 h and 4 h | 1/6 + 1/4 = 5/12 | 2.4 hours |
| Machines taking 3, 6, and 12 h | 1/3 + 1/6 + 1/12 = 7/12 | 1.71 hours |
| Pipes taking 10 h and 15 h | 1/10 + 1/15 = 1/6 | 6 hours |
| Fill in 5 h, drain in 8 h | 1/5 − 1/8 = 3/40 | 13.3 hours |
Partial Work and Different Start Times
If Anna works alone for 2 hours before Ben joins, she completes 2/6 = 1/3 of the job. The remaining 2/3 is done by both at 5/12 per hour: (2/3) ÷ (5/12) = 1.6 hours. The total time is 3.6 hours. Set up an equation: Anna's work + Ben's work = 1. If Anna works for t hours and Ben for t − 2 hours (starting later): t/6 + (t − 2)/4 = 1. Multiply by 12: 2t + 3(t − 2) = 12, so 5t = 18 and t = 3.6 hours. The equation method handles workers who join or leave at different times.
Sanity Checks
The combined time must be less than the time of the fastest individual: 2.4 is less than 4. If the answer is longer than one of the individual times, you probably subtracted instead of added or inverted wrongly. For two identical workers, the time halves. With more workers, the time falls, but real teams have coordination costs, so rates may not add perfectly: nine women cannot make a baby in one month. Work-rate math models ideal conditions and is useful for first estimates in scheduling, staffing, and manufacturing.
Related Problems
The same logic applies to travel problems with upstream and downstream rates, to taps and baths, to combining machines in a plant, and to parallel resistors in electronics, where the combined resistance is 1/(1/R₁ + 1/R₂). Memorize the approach: convert times to rates, add, and invert.
Frequently Asked Questions
How do you solve a work-rate problem?
Convert each worker's time to a rate (1/time), add the rates, and take the reciprocal to get the combined time.
How long do workers taking 6 and 4 hours need together?
2.4 hours.
What is the shortcut formula for two workers?
(t₁ × t₂) ÷ (t₁ + t₂).
How do I handle a drain or leak?
Subtract its rate from the filling rate.
Do real teams always combine rates perfectly?
No. Coordination costs and limits on parallel work can reduce the gain.