Let the initial quantity of alcohol in the mixture = x liter So, the initial quantity of water in the mixture = (20 – x) liters According to the question ∶ (x + 1) = (20 – x) × 50/100 (x + 1) = (20 – x) × 1/2 2x + 2 = 20 – x 3x = 18 x = 6 The initial quantity of alcohol in the mixture = 6 liters The initial quantity of water in the mixture = 20 – 6 = 14 liters Required ratio = 6 ∶ 14 = 3 ∶ 7