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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?

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? Correct Answer 266.66 kg

Let there is 100 kg of ore.25% ore contains 90% off Iron that means 25 kg contains; $$\frac{{25 \times 90}}{{100}} = 22.5\,{\text{kg}}\,{\text{iron}}$$22.5 kg Iron contains 100 kg of ore.Then, 1 kg of iron contains = $$\frac{{25}}{{100}}{\text{kg}}\,{\text{ore}}$$Hence, 60 kg iron contains
= $$\frac{{100 \times 60}}{{22.5}}$$
= 266.66 kg ore

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