Overview
Completion times cannot usually be added directly because each time describes a whole job performed by one worker. Convert to a rate—jobs per hour—so contributions over the same interval can be combined. If a worker’s speed changes, split the timeline into stages. We will use those ideas for cooperation, opposing flow, a delayed partner and the recovery of an individual rate from pairwise rates.
Convert a whole-job time into an hourly contribution
One machine finishes a batch in six hours and another in twelve. Their rates are 1/6 and 1/12 batch per hour. Working together, they complete 1/6 + 1/12 = 1/4 batch per hour, so one batch requires four hours.
The answer nine averages the completion times rather than combining productivity. A faster combined team should finish sooner than either machine alone, which already rules out nine. The reciprocal relationship matters twice: first to turn time into rate, and then to turn the combined rate back into time. Keep the unit “batch per hour” beside the sum so that 1/4 is not mistaken for the requested number of hours. This unit-based interpretation is more durable than memorizing a special two-worker shortcut.
Subtract rates when one process undoes the other
A pump fills a tank in six hours, while a drain empties it in nine. With both operating at constant rates, the net filling rate is 1/6 − 1/9 = 1/18 tank per hour. Starting empty, the tank therefore fills in eighteen hours.
The answer three subtracts the solo times, 9 − 6, instead of subtracting rates and then taking the reciprocal. Assign a sign based on contribution to the target quantity: filling is positive and draining is negative. The positive net rate confirms that filling is possible; its small size explains why the elapsed time exceeds the pump’s solo time. The stated constant-rate assumption lets the subtraction remain valid throughout the process. Do not introduce a changing-flow model when the problem explicitly supplies constant rates.
Write a separate work term for each stage
Machine A normally finishes a job in ten hours and B in fifteen. A begins alone. When B joins, A slows to half its normal rate; the total elapsed time is nine hours. Let t be the solo duration. A completes t/10 of the job during that stage. During the remaining 9 − t hours, the combined rate is 1/20 + 1/15 = 7/60.
The work equation is t/10 + (9 − t)·7/60 = 1. Multiplying by 60 gives 6t + 63 − 7t = 60, so t = 3. The remaining six hours form the joint stage. The alternative 7.5 comes from ignoring A’s slowdown, while six answers the wrong duration. A timeline makes both errors visible: each stage needs its own rate, and the unknown must be labeled as solo rather than joint time.
The joint rate is only slightly faster than A's original solo rate because A slows when B arrives. As a result, spending more time in the solo stage makes the total job take longer than it would under the joint rate throughout. This qualitative direction agrees with the equation and prevents importing the usual assumption that adding a second machine leaves the first machine's productivity unchanged.
Recover an individual rate from overlapping pairs
A and B together finish in six hours, A and C in eight, and B and C in twelve. Let their individual hourly rates be a, b and c. Then a + b = 1/6, a + c = 1/8 and b + c = 1/12. Add the first two equations and subtract the third to get 2a = 1/6 + 1/8 − 1/12 = 5/24. Thus a = 5/48 job per hour and A alone needs 48/5 hours.
The distractor 24/5 omits the division by two before taking a reciprocal; 5/48 reports the rate instead of time. For transfer, define one whole job, convert all solo or pair times to rates, and add work contributions over their actual durations. In practice, check that stage times are nonnegative and that all completed fractions total one. Those checks connect the equation back to the physical process.
Apply the method to a new case
Machine A finishes a job in 6 hours and machine B in 3 hours. They work together for 1 hour; then B stops. How many total hours does completion take? Their combined hourly rate is 1/6 + 1/3 = 1/2 of a job, so the first hour completes half. A alone needs (1/2)/(1/6) = 3 more hours, making 4 hours total. Adding the individual completion times would confuse time per job with jobs per hour. Using the combined rate for all four hours would ignore the stopped machine. Check the accumulated work: A runs for 4 hours and completes 4/6; B runs for 1 hour and completes 1/3. Together they complete exactly one job. Separate stages whenever the set of active workers changes, and report the total elapsed time rather than only the final stage.