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A diluted alcohol contains 16L of alcohol and rest is water. A new mixture in which the concentration of alcohol is 30% is to be prepared by replacing diluted alcohol. How many litres of the mixture shall be replaced with pure alcohol if there was initially 64L of water in the mixture.

A diluted alcohol contains 16L of alcohol and rest is water. A new mixture in which the concentration of alcohol is 30% is to be prepared by replacing diluted alcohol. How many litres of the mixture shall be replaced with pure alcohol if there was initially 64L of water in the mixture. Correct Answer 10L

Initial concentration of alcohol in the mixture = (16 × 100)/(64 + 16) = 20%

Suppose x litre of mixture is replaced:

∴ (80 – x) × 0.2 + x × 1 = 80 × 0.3

⇒ 16 – 0.2x + x = 24

⇒ 0.8x = 8

⇒ x = 10

∴ 10 litre of the mixture is replaced with the pure alcohol. 

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