Work Equivalence
Work equivalence is the "man-days" idea: a fixed job needs a fixed amount of total effort, so men × days is constant for the same job. If 10 workers finish a wall in 6 days, that wall is 60 man-days of work; 15 workers will finish it in 60/15 = 4 days. The same constant lets you solve "some workers leave or join" problems: compute the man-days already spent, subtract from the total, then divide the remainder by the new workforce. When hours per day also vary, extend it to man-day-hours: 10 men × 6 days × 8 hours = 480 man-hours. The crucial CAT caveat is that this assumes every worker is equally efficient; if efficiencies differ, convert everyone to a common unit first (e.g. in terms of "one man’s daily output"). Watch for inverse proportion: more workers ⇒ fewer days, so the relationship is a product staying constant, never a simple ratio.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Core rate & efficiency
| Work rate | Rate = Work ÷ Time (units per day) |
|---|---|
| Total work (LCM trick) | Total work = LCM of the given days |
| Efficiency from days | Efficiency = Total work ÷ days |
| Combined time | Time = Total work ÷ (sum of efficiencies) |
| Efficiency ∝ 1/Time | If A is k× as fast as B ⇒ A takes 1/k of B’s time |
CAT power-tools
| Two together (classic) | Time = (a × b) ÷ (a + b) days |
|---|---|
| Man-days (work equivalence) | M1 × D1 = M2 × D2 (same job) |
| Full chained formula | M1·D1·H1 / W1 = M2·D2·H2 / W2 |
| Wages share | Wages split in the ratio of work done (= ratio of efficiencies × days worked) |
| Leak / negative work | Effective rate = inflow rate − leak rate |