7.3 Work-Rate & Cooperative Labor
Key Takeaways
- Work = Rate × Time. If one job takes T time units, the worker's rate is 1/T of a job per unit time. Combined workers add rates: 1/A + 1/B + …
- Two workers with times A and B finish one job in AB/(A + B). That product-over-sum time is always less than either individual time.
- The LCM smart-number method assigns total work equal to LCM of the given times so that every rate is an integer; combined time is then (LCM units) / (sum of integer rates).
- Staggered shifts split the job into phases. Compute work completed in the solo phase, then finish the remainder at the combined rate. Total clock time is solo time plus together time.
- Pipes use signed rates: inlets are +1/T_fill and drains are −1/T_empty. Net rate is the signed sum; an empty tank never fills if the net rate is zero or negative.
Work Is Rate Times Time
Official Guide Math Review 3.3 treats work as another rate: output per unit time. The algebra is identical to uniform motion, which is why this chapter groups them. Replace distance with work W:
W = R × T, so R = W / T and T = W / R.
On almost every GMAT stem the "job" is unspecified — paint a house, assemble a machine, fill a tank. Assign W = 1 job. Then a person who finishes the job in T hours has rate 1/T jobs per hour. In 3 hours that person completes 3/T of the job. Fractions of a job are legal quantities; you do not need to know how many square feet the house has.
Units still matter. If one rate is in jobs per hour and another person's time is given in minutes, convert before you add. A machine that finishes in 90 minutes has rate 1/(3/2) = 2/3 job per hour, not 1/90 job per hour unless you have converted everything else to minutes.
Combined Rates and the Product-over-Sum Shortcut
Independent workers who do not block each other add rates:
R_combined = 1/A + 1/B + … + 1/N
Together-time for one job is the reciprocal, T_together = 1 / R_combined.
For two workers with individual times A and B,
1/T = 1/A + 1/B = (A + B)/(AB), so T = AB / (A + B).
That product-over-sum is always shorter than either A or B. If a choice is larger than the faster worker's solo time, eliminate it before you compute. Two people working together cannot take longer than the quicker of them working alone.
Worked example. Machine A finishes in 6 hours, Machine B in 12 hours. Product-over-sum: (6 × 12)/(6 + 12) = 72/18 = 4 hours. Fraction path: 1/6 + 1/12 = 1/4 job per hour, so time = 4 hours. Arithmetic mean (6+12)/2 = 9 hours is impossible: 9 is slower than A working alone.
LCM smart numbers (no-calculator default)
When three or more times appear, or when A and B are messy, assign the job a size equal to the LCM of the given times. Every rate becomes an integer, and you never hunt a common denominator.
Workers at 10 hours, 15 hours, and 30 hours. Set W = LCM(10, 15, 30) = 30 units. Rates: 30/10 = 3, 30/15 = 2, and 30/30 = 1 unit per hour. Combined 3 + 2 + 1 = 6 units per hour. Time = 30/6 = 5 hours. The unit-job path 1/10 + 1/15 + 1/30 = 3/30 + 2/30 + 1/30 = 6/30 = 1/5 gives the same 5 hours, but the integers are faster under 45-minute section timing.
Pick the LCM of all times that appear, including drain times and solo-phase times, not only the two numbers in a product-over-sum. If a drain takes 8 hours and inlets take 4 and 6, LCM(4, 6, 8) = 24 is the right tank size.
Staggered Shifts
When workers start or stop at different times, split the project into phases. Rates add only during the hours they actually overlap.
- Solo (or smaller-crew) phase: W₁ = R_solo × t_solo.
- Remaining work: W_rem = 1 − W₁ (or remaining LCM units).
- Together phase: T₂ = W_rem / R_combined.
- Total clock time = t_solo + T₂. The question usually wants this total, not T₂ alone.
Worked example. Painter X finishes a house in 10 hours; Painter Y finishes it in 15 hours. X works alone for 4 hours, then Y joins until the house is done. Solo work: 4 × (1/10) = 2/5 of the house. Remaining 3/5. Combined rate 1/10 + 1/15 = 1/6 house per hour. Together time (3/5) / (1/6) = 18/5 = 3.6 hours. Total time 4 + 3.6 = 7.6 hours. A common trap is to stop at 3.6, which is only the overlap, or to compute product-over-sum 10×15/(10+15) = 6 and ignore the staggered start.
If a worker leaves early, the second phase has a smaller combined rate, not a larger one. Compute the work already done while both were present, then finish at the remaining worker's solo rate.
Pipes Filling and Draining
A pipe is a worker with a signed rate.
- Inlet that fills an empty tank in T hours: +1/T
- Drain that empties a full tank in T hours: −1/T
- Net rate: sum of signed rates
- Time to fill an empty tank: 1 / R_net, provided R_net > 0
If drains dominate, R_net ≤ 0 and an empty tank never fills. If the tank starts full and net rate is negative, time to empty is (1) / |R_net|.
Worked example. Inlet A fills in 4 hours, inlet B fills in 6 hours, drain C empties in 8 hours. All three open on an empty pool. LCM(4, 6, 8) = 24 units. Rates +6, +4, and −3 units per hour. Net +7. Time = 24/7 hours. Fraction path: 1/4 + 1/6 − 1/8 = (6 + 4 − 3)/24 = 7/24, so T = 24/7 hours. Same number.
A stem that opens the drain only after the tank is half full is a two-phase pipe problem: fill at the inlet-only net rate to W = 1/2, then continue at the signed combined rate. Do not average the two net rates.
Group Labor Scaling
When workers are interchangeable, total work is proportional to the product of headcount, days, and hours per day:
W ∝ M × D × H
Two situations with the same efficiency satisfy
W₁ / (M₁ D₁ H₁) = W₂ / (M₂ D₂ H₂).
Worked example. Twelve workers build 4 houses in 15 days at 8 hours/day. How many days for 9 workers to build 6 houses at 10 hours/day? Then 4 / (12 × 15 × 8) = 6 / (9 × D × 10). Simplify the left side to 4/1,440 = 1/360. Right side 6/(90 D) = 1/(15 D). So 15 D = 360 and D = 24 days.
Efficiency changes — "each new worker is half as fast" — belong in the rate, not in M. Count effective workers, then scale.
| Problem type | Rate setup | Time or size formula |
|---|---|---|
| Two workers together | 1/A + 1/B | T = AB/(A + B) |
| Three or more together | sum of 1/Tᵢ or LCM units | T = 1 / R_combined |
| Staggered shift | phase 1 at R_solo, phase 2 at R_sum | t_total = t_solo + W_rem / R_sum |
| Inlet and drain | +1/T_in − 1/T_drain | T_fill = 1 / R_net |
| Crew scaling | W / (M D H) constant | cross-multiply the two situations |
Machine A can complete a job in 8 hours working alone, and Machine B can complete the same job in 24 hours working alone. If both machines work simultaneously at their constant rates, how many hours will they take to finish one job?
Working alone, Worker A can finish a project in 15 hours. Working alone, Worker B can finish the same project in 10 hours. Worker A works alone for 5 hours, then Worker B joins, and they finish the remaining work together. What is the total time, in hours, from when Worker A started until the project is complete?
Pipe A can fill an empty tank in 10 hours, Pipe B can fill the same empty tank in 15 hours, and a drain can empty the full tank in 30 hours. If the tank starts empty and all three pipes are opened at once, how many hours will it take to fill the tank?