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Two coal loading machines, each working 12 h per day for 8 days, handle 9000 tonne of coal with an 1 efficiency of 90%, while 3 other coal loading machines at an efficiency of 80% are set to handle 12000 tonne of coal in 6 days. How many hours per day should each machine work?
According to the question,
time 1/amount of work done 1 = time 2/amount of work done 2
(12*8*.9):(9000/2)::(6*x*.8):(12000/3)
So,on solving , we get
x=16 h/day
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