Discussion
Grain enters a cleaner at \( 5000 \) kg/h with \( 4 \) per cent impurity and leaves at \( 0.5 \) per cent impurity in a clean stream of \( 4750 \) kg/h. The impurity removed is
Answer: A
Correct answer: 176 kg/h
Compute the impurity in each stream and take the difference:
\[ \text{in} = 5000 \times 0.04 = 200\ \mathrm{kg/h} \]
\[ \text{out} = 4750 \times 0.005 = 23.75\ \mathrm{kg/h} \]
\[ \text{removed} = 200 - 23.75 = 176\ \mathrm{kg/h} \]
The distractor \( 200 \) assumes complete removal, ignoring the impurity still present in the clean stream. Note also that the reject stream is \( 250 \) kg/h, so about \( 74 \) kg/h of good grain leaves with the impurity - a real loss worth measuring.
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