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? 500 किलोग्राम अयस्क में
लोहे की एक निश्चित मात्रा होती है। पहले ब्लास्ट
फर्नेस प्रक्रिया के बाद 200 किलोग्राम स्लैग जिसमें
12.5% लोहे शामिल हैं को हटा दिया गया.लोहे का
प्रतिशत शेष अयस्क में मूल प्रतिशत से 20%
अधिक है.मूल 500 किलो अयस्क में कितने किलो
लोहा था.
A)54.2B)46.3 C)58.5D)42.4 E)89.2
Answers
Hii Dear,
◆ Answer -
Total iron = 89.29 kg
◆ Explanation -
Let x be the original percent of iron in the ore. Then rotal iron content would be -
Total iron = 500 × x/100
Total iron = 5x
Iron that was removed in the slag is -
Iron removed = 200 × 12.5/100
Iron removed = 25 kg
Now remaining iron would be 5x-25 kg in 300 kg ore. Given that this is 20% more than original iron.
x/100 × (120/100) = (5x-25)/300
3.6x = 5x - 25
1.4x = 25
x = 17.85 %
Total iron in the original ore was thus -
Total iron = 5x
Total iron = 5 × 17.85
Total iron = 89.29 kg
Therefore, total iron in the original ore was 89.25 kg.
Hope this helps...
Answer:
212.5 kg
Step-by-step explanation:
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
Let say % Of Iron in Origin = F %
Iron in 500 kg = (F/100)*500 = 5F kg
200 kg of slag containing 12.5% of iron was removed.
Iron = (12.5/100) * 200 = 25 kg
Iron remained = 5F - 25 kg
Total ore remained = 500 - 200 = 300 kg
% of iron = ((5F -25)/300 )* 100 = (5F -25)/3 %
(5F -25)/3 = F + 20
=> 5F - 25 = 3F + 60
=> 2F = 85
=> F = 42.5 %
Iron in original ore = 5F = 5*42.5 = 212.5 kg