If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours a day; how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type ?
Answer
Answer (as printed):
Explanation
Let 3 engines of former type consume 1 unit in 1 hour. Then, 4 engines of latter type consume 1 unit in 1 hour. $\therefore$ 1 engine of former type consumes $\frac{1}{3}$ unit in 1 hour. 1 engine of latter type consumes $\frac{1}{4}$ unit in 1 hour. Let the required consumption of coal be $x$ units. Less engines, Less coal consumed (Direct Proportion). More working hours, More coal consumed (Direct Proportion). Less rate of consumption, Less coal consumed (Direct Proportion). Number of engines $9:8$, Working hours $8:13$, Rate of consumption $\frac{1}{3}:\frac{1}{4}$ :: $24:x$. $\therefore \left(9 \times 8 \times \frac{1}{3} \times x\right) = \left(8 \times 13 \times \frac{1}{4} \times 24\right) \Leftrightarrow 24x=624 \Leftrightarrow x=26$. Hence, the required consumption of coal = 26 metric tonnes.
Explanation as extracted from the printed page; notation may be imperfect.