Discussion
An ore contains 25 % of an alloy that has 90 % iron. Other than this, in the remaining 75% of the ore, there is no iron. How many kilograms of the ore are needed to obtain 60 kg of pure iron?
Answer: C
About 266.6 kg of ore is needed.
Only the alloy carries iron, and the alloy is 25% of the ore, of which 90% is iron. So the iron fraction of the ore is
\[ \begin{aligned} 0.25 \times 0.90 &= 0.225 \\ \text{ore} &= \frac{60}{0.225} \\ &= 266.67\ \text{kg} \end{aligned} \]
which the option gives as 266.6 kg. 300 kg would be right if the ore were 20% iron, and 270.6 kg and 250.67 kg do not correspond to the 22.5% iron content.
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