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500 kg of ore contained a certain amount of iron. After the first blast furnace process, 200 kg of slag containing 12.5% of iron was removed. The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore. How many kg of iron were there in the original 500 kg ore?
90.5
89.2
87.3
81.2
Initially ‘x’ kg of iron in 500 kg ore. Iron in the 200 kg of removed =200*12.5/100= 25 kg. The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore So (x-25)/300 = (120/100)*x/500 => x – 25 = 18x/25 => 7x = 625 => x = 89.2
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Noorbir Singh
pls explain the formula applied
Arshdeep Sidhu
20% more than the quantity. Not percentage. That would give x to be 212.5 kg
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