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 ?

Answer: Option B

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)
$$\begin{array}{c}\text{Engines 5}:8\text{hfill}\\ \text{Hours per day 9}:10\text{hfill}\\ \text{Rate }\frac{1}{3}:\frac{1}{4}\text{hfill}\end{array}\}::6:x$$
$$\begin{array}{rl}& \therefore \left(5\times 9\times \frac{1}{3}\times x\right)=\left(8\times 10\times \frac{1}{4}\times 6\right)\\ & \iff 15x=120\\ & \iff x=8\end{array}$$