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A mixture of alcohol and water comprises 60% alcohol. First, 20% of the mixture is replaced with water and then the volume of the resultant mixture is increased by 20% by adding only alcohol. What is approx. percentage of alcohol in the final mixture?

A mixture of alcohol and water comprises 60% alcohol. First, 20% of the mixture is replaced with water and then the volume of the resultant mixture is increased by 20% by adding only alcohol. What is approx. percentage of alcohol in the final mixture? Correct Answer 57%

Let the Initial Volume of Alcohol and Water is 6000 and 4000 respectively

20% of the mixture is removed and replaced with water

⇒ Alcohol = 6000 - 1200 = 4800

⇒ Water = 4000 - 800 + 2000 = 5200

Volume of resultant mixture increased by 20% by adding alcohol.

New Volume of Resultant mixture 12, 000

⇒ Alcohol = 4800 + 2000 = 6800

⇒ Water = 5200

⇒ Percentage of Alcohol = (6800 / 12000) × 100 = 56.67 % ≈ 57%

∴ the required percentage of alcohol is 57%

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