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First of all, l love question like this. Thanks Bunuel! let x be the total volume of the original mixture of half alcohol and half water let y be the portion that was removed from the mixture x We must have in mind that the volume of the portion(y) removed is equal to the volume of solution that replaced it Hence, % of alcohol in final solution =(0.5x - 0.5y + 0.25y)/(x - y + y) =30% (0.5x - 0.25y)/(x) =0.3 0.5x - 0.25y =0.3x 0.2x =0.25y y = 0.8x or 80% of original mixture