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This is the problem:

There is a job that has to be made in 60 days by a group of workers. After one day of work five workers more are added to the original group and they work one day. After this second day of work another 5 workers are added to the group, and then they work together and finish the job one day before the planned date. How many workers were there in the original group?

I tried to do like this: each worker of the original group of m workers did 1/(60m) of the whole job during the first day. Then the second day there were m+5 workers so they did $\frac{1-\frac{1}{60m}}{59(m+5)}$ of the job, and in the last day there already were m+10 workers who then worked 57 days (so the total ammount of days is 59...I think), so I tried to add the above quantities and came nothing close to the solution.

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The basic unit here is a dayload. Given that there were $m$ members in the original team the total number of dayloads is $N=60m$.

According to the story the $m$ people worked $59$ days, five more people worked $58$ days, and another $5$ people worked $57$ days. Therefore we have $$60 m=N= 59m+ 5\cdot 58+ 5\cdot 57\ ,$$ from which we immediately obtain $m=575$.

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One day for the whole group was saved by adding $5 \cdot 58 + 5 \cdot 57$ days by one worker. So there were originally $5 \cdot 58 + 5 \cdot 57 = 675$ workers.

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  • $\begingroup$ I can' understand the logic (and soundness) of this answer... $\endgroup$ – DonAntonio Oct 14 '13 at 12:46
  • $\begingroup$ @DonAntonio Assume that no extra workers were added. Then the workers that were present at the start had to do the work that the extra workers did in the last day. The extra workers did $5 \cdot 58 + 5 \cdot 57 = 675$ units of work, where one unit is how much one worker can do in one day. So the workers that were present at the start could do exactly $675$ units of work in one day, which means there were $675$ workers. $\endgroup$ – Arthur Oct 14 '13 at 17:52
  • $\begingroup$ As I understand it, there were $5$ extra workers the second day and $10$ extra workers from the third day on, @Arthur... $\endgroup$ – DonAntonio Oct 14 '13 at 18:36
  • $\begingroup$ @DonAntonio That is the way I understand the question too. The five workers that started on the second day did $58$ days of work each, and the five workers that started on the third day did $57$ days of work each, which gives a total of $675$ units of work done by the extra workers. $\endgroup$ – Arthur Oct 14 '13 at 18:42

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