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If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day. How much coal (in metric tonnes) will be needed for 8 engines, each running 10 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type?
A
B
C
D
Coal consumed by one engine of type 1 =6/(5*9) =2/15 metric tonnes per hour 3 engines of type 1 consume as much as 4 engines of type 2 Therefore, Coal consumed by one engine of type 2=(3/4)*(2/15) =1/10 MT/hour Coal consumed by 8 engines (of type 2),running 10 hours a day =(1/10)*8*10 =8 metric tonnes.
Hence, option 3 is the correct answer.
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