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 ?

Explanation:-

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)
$$\left. \begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Engines 5}}:8 \hfill \\ {\text{Hours per day 9}}:10 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Rate }}\frac{1}{3}:\frac{1}{4} \hfill \\ \end{gathered} \right\}::6:x$$
$$\eqalign{ & \therefore \,\,\left( {5 \times 9 \times \frac{1}{3} \times x} \right) = \left( {8 \times 10 \times \frac{1}{4} \times 6} \right) \cr & \Leftrightarrow 15x = 120 \cr & \Leftrightarrow x = 8 \cr}$$