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?

• 89.28 Kg.
  
  Initially , let there was 'x' kg of iron in 500 kg. ore.
  Iron in the 200 kg of removed =200*12.5/100= 25 kg.
  So weight of the ore remained= 500-200= 300 kg and iron in the ore= x-25 kg.
  Given, % of iron in an ore after removing slag is 20% more than the original 500 kg ore.
  So (x-25)/300 = (120/100)*x/500
  => 5x - 125= 3.6x
  => 1.4x= 125
  => x= 125/1.4
  => x=89.28
  
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