A container has a 100-litre mixture of acid and water in the ratio 4 : 1. 20 litres is removed and replaced with water. Then 20 litres is removed again and replaced with pure acid. What is the final percentage of acid?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 71.2%.
Total mixture = 100 L
Initial: Acid = 4/5 × 100 = 80 L, Water = 100 - 80 = 20 L
First operation: Remove 20 L mixture ⇒ Acid removed = (4/5) × 20 = 16 L
Water removed = 20 - 16 = 4 L
Remaining: Acid = 80 – 16 = 64 L, Water = 20 – 4 = 16 L
Add 20 L water ⇒ Acid = 64 L, Water = 16 + 20 = 36 L
Second operation: Remove 20 L mixture ⇒ Acid removed = (16/25) × 20 = 12.8 L
Water removed = (9/25) × 20 = 7.2 L
Remaining: Acid = 64 – 12.8 = 51.2 L, Water = 36 – 7.2 = 28.8 L
Add 20 L pure acid ⇒ Acid = 51.2 + 20 = 71.2 L, Water = 28.8 L
Total mixture = 71.2 + 28.8 = 100 L
Final acid percentage = (71.2/100) × 100 = 71.2%