anandnk wrote:

(A+B) + (B+C) + (C+A) = 6/5+3/2+2 = 47/10

In this time 3 times work is done

2A+2B+2C will do one job in 47/30 hours

so

A+B+C will need 47/60 hrs to do one job, working together

Thus they need 47 minutes to complete one job.

Hey Anand,

I do not see any difference between yours and calnhob's way of interpreting this problem.

Looking at your logic, I believe that you also considered that

A and B together took 6/5 hours to complete the ENTIRE job.

B and C together took 3/2 hours to complete the ENTIRE job.

A and C together took 2 hours to complete the ENTIRE job.

This is the same way in which calnhob interpreted the problem. But I think he applied the right formula.

So your statement "In 47/10 hours the work is completed 3 times is correct" (If each pair works once and finishes the entire job)

However, I think it is not correct to say that

If 2A + 2B + 2C can complete 3 times a job in 47/10 minutes, it will take 47/30 minutes to complete the job once. The proportionality can not be applied here because everybody has different rate of working. Here the effeciency factor comes in to play. (consider a simple example where all the players work with a different rates and it will be more clear)

In my opinion, the work related problems can not be solved by applying the straight formula. Because the longer one takes to finish a job, he lesser is the amount of work that can be finished in unit time.