Discussion
Feed of \( 2000 \) kg/h contains \( 3 \) per cent foreign matter. If the cleaner removes \( 54 \) kg/h of it, the cleaning efficiency is
Answer: C
Correct answer: 90 per cent
Find the foreign matter present, then the fraction removed:
\[ 2000 \times 0.03 = 60\ \mathrm{kg/h\ foreign\ matter} \]
\[ \frac{54}{60} \times 100 = 90\ \mathrm{per\ cent} \]
The distractor \( 2.7 \) per cent would express the removal as a fraction of the whole feed, which measures nothing useful. Cleaning efficiency must always be referred to the impurity present, since the same 54 kg removed from a much dirtier feed would represent a far poorer performance.
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