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 ?
Answer with explanationAnswer: Option A
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.