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[email protected]
- Newbie | Next Rank: 10 Posts
- Posts: 1
- Joined: Fri Feb 28, 2014 11:10 pm
Use the following formula:[email protected] wrote:Two coal loading trucks handle 9000 tonnes of coal at an efficiency of
90% working 12 hours per day for 8 days. How many hours a day, 3 coal
loading trucks should work at an efficiency of 80% so as to load 12000 tonnes
of coal in 6 days
(trucks)(efficiency)(days)(hours per day) / work = (trucks)(efficiency)(days)(hours per day) / work.
Original values:
Trucks = 2.
Efficiency = 90.
Days = 8.
Hours per day = 12.
Work = 9000.
New values:
Trucks = 3.
Efficiency = 80.
Days = 6.
Hours per day = x.
Work = 12000.
Plugging these values into the formula above, we get:
(2 trucks)(90% efficiency)(8 days)(12 hours per day) / (9000 tons) = (3 trucks)(80% efficiency)(6 days)(x) / (12000 tons)
(2*90*8*12)/9000 = (3*80*6*x)/12000
(2*9*12)/3 = (3*6*x)/4
4 = x/4
x = 16 hours.
