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If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day, how many metric tonnes of coal will be needed for 8 engines, each running 10 hours a day, it begin given that 3 engines of the former type consume as much as 4 engines of the latter type ?

If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day, how many metric tonnes of coal will be needed for 8 engines, each running 10 hours a day, it begin given that 3 engines of the former type consume as much as 4 engines of the latter type ? Correct Answer 8

Let the required quantity of coal be x metric tonnes
More engines, More coal (Direct proportion)
More hours per day, More coal (Direct proportion)
More rate, More coal (Direct proportion)
\
$$\eqalign{ & \therefore \,\,\left( {5 \times 9 \times \frac{1}{3} \times x} \right) = \left( {8 \times 10 \times \frac{1}{4} \times 6} \right) \cr & \Leftrightarrow 15x = 120 \cr & \Leftrightarrow x = 8 \cr} $$

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