arithmetic - in steel plant if 9 engines consume 24 metric tonnes of coal when each is working 8 hours day how much coal will be required for 8 engines each running 13 hours a day it being given that 3 engines of former type consume as much as 4 engines of - skillgun

As 9 engines consumes 24 metric tonnes of coal, when each working for 8 hours.
Therefore, 1 engines consume in 1 hour = 24/(8*9) = 3
3 engines of former type = 4 engines of latter type.
6 engine of former type = 8 engine of latter type.
Then, Coal required for 8 engines each running for 13 hours a day = 24*6*13/(8*9)
= 26 metric tonnes