- Let W be the total work required to build the wall. - Let 'w' be the amount of work one worker does in one day. Assume all workers work at the same rate. - The total work done per day is the number of workers multiplied by 'w'. - The work is completed in 4 days with a changing number of workers: - Day 1: 6 workers work. Work done = 6w. - Day 2: 5 workers work. Work done = 5w. - Day 3: 4 workers work. Work done = 4w. - Day 4: 3 workers work. Work done = 3w. - Total work W = (Work done on Day 1) + (Work done on Day 2) + (Work done on Day 3) + (Work done on Day 4). - W = 6w + 5w + 4w + 3w = (6 + 5 + 4 + 3)w = 18w. - Now, consider the scenario where none of the workers leave. All 6 workers work every day. - Let 'd' be the number of days it would take the 6 workers to complete the work if they all worked together every day. - In this case, the total work W = (Number of workers) (Work per worker per day) (Number of days) = 6 w d. - Equate the total work from both scenarios: 18w = 6wd. - Since 'w' is the work done by a worker per day (and is non-zero), we can cancel 'w' from both sides: 18 = 6d. - Solve for d: d = 18 / 6 = 3.
6 workers start to work in the construction of a wall. They all work together for 1st day. Thereafter, from second day one worker leaves the work every day. Thus, the total job is completed in 4 days. If none of the workers would have leave the job, how many days they would have taken to complete the work?
Asked in GPSC Prelims CSP 1 2019 · Topic: Mechanics (Newton's Laws of Motion and their applications)
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