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If 6 engines consume 24 metric tonnes of coal, when each is working 8 hours day, how much coal will be required for 9 engines, each running 13hours a day, it being given that 2 engines of former type consume as much as 3 engines of latter type ?

Correct Answer: 45 metric tonnes

Explanation:

2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton 


i.e. 3 engines of latter type consumes 1 ton for one hour


Hence 9 engines consumes 3 tons for one hour


For 15 hours it is 15 x 3 = 45 metric tonnes.


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