How to Solve Work-Equivalence (Men-Days) Problems
Solve men-days work-equivalence aptitude problems using the M×D×H invariant, efficiency weighting and worked examples with practice questions.
Expected Interview Answer
Work-equivalence problems are solved with the man-days (or man-hours) invariant: Men × Days × Hours per day is constant for a fixed amount of work, so M1×D1×H1 = M2×D2×H2 whenever the total job size stays the same.
The core idea is that "work" has a fixed total size, measured in man-days: if 10 men take 6 days, the job is 60 man-days of work, regardless of how the workforce is later reorganized. Doubling the men halves the days, tripling the hours per day cuts the days to a third, and these proportional relationships fall directly out of keeping the man-days product constant. When workers have different efficiencies, weight each worker by their efficiency multiplier before multiplying by days — a worker twice as efficient contributes two "standard" man-days per day worked. The same invariant extends naturally to man-days-per-unit-of-work problems, such as comparing how many men are needed to build walls of different lengths in different times.
- One invariant (M×D×H = constant) replaces case-by-case reasoning
- Handles scaling workforce, days, and hours per day uniformly
- Extends cleanly to jobs of different sizes via men-days-per-unit-work
AI Mentor Explanation
If a ground crew of 10 workers takes 6 days to prepare a pitch, the job represents 60 worker-days of labor, and that total stays fixed no matter how the crew is reorganized. Double the crew to 20 workers and the same pitch needs only 3 days, because 20×3 = 60 preserves the invariant. Work-equivalence problems are entirely about holding this men×days product constant while men, days, or hours per day change.
Worked example (different work sizes and hours)
Base rate
- 12×10×8 / 100 = 9.6
Target equation
- 9.6 = M2×20×6 / 150
Solve M2
- M2 = 1440/120 = 12 men
Step-by-Step Explanation
Step 1
Compute total man-days
For a fixed job, Men × Days (× Hours/day) is constant.
Step 2
Handle different work sizes
Divide by the amount of work: (M×D×H)/W stays constant across scenarios.
Step 3
Weight by efficiency if given
A worker k times as efficient contributes k standard man-days per day.
Step 4
Solve the proportion
Set up M1D1H1/W1 = M2D2H2/W2 and solve for the unknown quantity.
What Interviewer Expects
- Correct statement of the men-days invariant
- Correct handling of differing hours per day
- Adjustment for different work sizes (e.g. wall lengths) via the /W term
- Correct weighting when workers have different efficiencies
Common Mistakes
- Forgetting to divide by the amount of work when job sizes differ
- Ignoring a change in hours per day between the two scenarios
- Treating all workers as equally efficient when the problem states otherwise
- Setting up the proportion inverted (multiplying instead of dividing correctly)
Best Answer (HR Friendly)
“I rely on the men-days invariant: for a fixed job, the product of men, days, and hours per day stays constant, so I can set up M1×D1×H1 equals M2×D2×H2 whenever the workforce or schedule changes. If the two scenarios involve different amounts of work, like walls of different lengths, I divide both sides by the work size before equating them. If workers have different efficiencies, I weight each one by their efficiency multiplier before multiplying by days.”
Follow-up Questions
- How do you adjust the invariant when the work itself is a different size in each scenario?
- How would you incorporate workers who join partway through the job?
- How does efficiency (a worker twice as fast) modify the men-days formula?
- How do you extend this to three interacting variables like men, machines, and days?
MCQ Practice
1. 15 men can finish a job in 12 days. How many men are needed to finish it in 9 days?
15×12 = 180 man-days; 180/9 = 20 men.
2. 8 men working 6 hours a day build a wall in 10 days. How many days would 6 men working 8 hours a day take?
8×6×10 = 480 man-hours; 480/(6×8) = 10 days.
3. If 20 men can complete a task in 15 days, working at double efficiency how many equivalent standard-men-days does the task represent?
Task size = 20×15 = 300 standard man-days; efficiency doubling changes the men needed, not the fixed task size.
Flash Cards
Core work-equivalence invariant? — Men × Days × Hours/day is constant for a fixed job.
How to compare different work sizes? — Divide by the work amount: (M×D×H)/W stays constant.
How to handle unequal efficiency? — Weight each worker by their efficiency multiplier before multiplying by days.
What does 15 men × 12 days represent? — 180 man-days — the fixed total work of the job.