Nine engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type?

Let required amount of coal be y metric tonnes
9 engines consumes 24 metric tonnes of coal in 8 hours a day.
More engines, more amount of coal (direct proportion)
If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption
More rate of consumption, more amount of coal (direct proportion)
More hours, more amount of coal(direct proportion)
Now we have , 9 × ( 1 / 3 ) × 8 × y = 8 × ( 1 / 4 ) × 13 × 24
⇒ 3 × 8 × y = 8 × 6 × 13
⇒ 3 × y = 6 × 13 ⇒ y = 2 x 13 = 26 metric tonnes