| If 5 engines consume 6 metric tonnes of coal when each is running 9 h a day, how many metric tonnes of coal will be needed for 8 engines, each running 10 h 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) \[6\frac{12}{25}\]
Correct Answer: B
Solution :
| 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] |
| \[\left. \begin{matrix} \text{Engines} & 5:8 \\ \text{Hours}\,\text{per}\,\text{day} & 9:10 \\ \text{Rate} & \frac{1}{3}:\frac{1}{4} \\ \end{matrix}\, \right\}::6:x\] |
| \[\therefore \] \[\left( 5\times 9\times \frac{1}{3}\times x \right)=\left( 8\times 10\times \frac{1}{4}\times 6 \right)\] |
| \[\Rightarrow \] \[15x=120\,\,\,\Rightarrow \,\,\,x=8\] |
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