60 men can complete a work in 25 days. One man starts working on it, and one…

Concept: When a fixed number of workers works at a constant rate, total work equals workers × days, measured in man-days. If instead the workforce grows by one additional worker every day starting from a single worker, the cumulative man-days completed after n days is the sum of the first n natural numbers, n(n+1)/2 — the standard arithmetic-series identity for a linearly growing workforce.

Application:

1. Total work required = 60 men × 25 days = 1500 man-days.

2. One man works on day 1, two men on day 2, and so on, so after n days the cumulative work done is 1 + 2 + 3 + ... + n = n(n+1)/2 man-days.

3. Set the work done equal to the work required: n(n+1)/2 = 1500, i.e., n(n+1) = 3000.

4. Test integers near √3000 ≈ 54.8: for n = 54, n(n+1) = 54 × 55 = 2970 (below 3000); for n = 55, n(n+1) = 55 × 56 = 3080 (at or above 3000).

5. So the cumulative work first reaches 1500 man-days during the 55th day.

Cross-check: after 54 days the work done is 54 × 55 / 2 = 1485 man-days, 15 man-days short of the 1500 needed; on day 55 there are 55 men working, contributing 55 man-days that day alone — comfortably more than the 15 that remain — confirming the job finishes on day 55 and could not have finished a day earlier.