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 being given that 3 engines of the former type consume as much as 4 engines of the latter type?
(a)$3 \frac{1}{8}$
(b)8
(c)$8 \frac{8}{9}$
(d)$6 \frac{12}{25}$
Answer
Answer (as printed): B
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). Engines $5:8$, Hours per day $9:10$, Rate $\frac{1}{3}:\frac{1}{4}$ :: $6:x$. $\therefore \left(5 \times 9 \times \frac{1}{3} \times x\right) = \left(8 \times 10 \times \frac{1}{4} \times 6\right) \Leftrightarrow 15x = 120 \Leftrightarrow x = 8$.
Explanation as extracted from the printed page; notation may be imperfect.