jimmyjamesdonkey wrote:

3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in 9 hours. What was the ratio of the work performed by the fastest machine as compared to the slowest???

Answer to follow.

Really not sure about the answer, but here we go.

Machine A, B, C

A <-- Slowest

C <-- Fastest

Working:

3 hours, A+B+C

3 hours, B+C

3 hours, A+B+C

We know that A worked 6 hours where C worked 9 hours

Productivity A:B:C = 1:2:5

This means...

Machine: time: Job

A: t : 1 => A: 6 : 6/t

B: t : 2

C: t : 5 => C: 9: (5*(9/t))

Ratio = (45/t) / (6/t) = 15/2