Case 1: Workers have different rates
Work rate × Time to finish the job = 1 job done
Work rate = (1 job done) / (Time to finish the job)
Time of doing the job = (1 job done) / (Work rate)
Albert can finish a job in A days
Bryan can finish the same job in B days
Carlo can undo the job in C days
1/A = rate of Albert
1/B = rate of Bryan
-1/C = rate of Carlo
Albert and Bryan work together until the job is done: (1/A + 1/B)t = 1
Albert is doing the job while Carlo is undoing it until the job is done: (1/A - 1/C)t = 1
Lejon can finish a job in 6 hours while Romel can do the same job in 3 hours. Working together, how many hours can they finish the job?
Case 2: Workers have equal rates
Work load = no. of workers × time to finish the job
Work done = no. of workers × time of doing the job
To finish the job
Work done = Work load
If a job can be done by 10 workers in 5 hours, the work load is 10(5) = 50 man-hours. If 4 workers is doing the job for 6 hours, the work done is 4(6) = 24 man-hours. A remaining of 50 - 24 = 26 man-hours of work still needs to be done.
Eleven men could finish the job in 15 days. Five men started the job and four men were added at the beginning of the sixth day. How many days will it take them to finish the job?