A container contains a mixture of three liquids A, B,

Solution

Let the initial total volume of the mixture be x liters. So, the initial quantities of A, B, and C are 4x/9, 3x/9, and 2x/9 liters, respectively. After removing 18 liters, the quantities of A, B, and C remaining are: A: 4x/9 - (4/9) × 18 = 4x/9 - 8 B: 3x/9 - (3/9) × 18 = 3x/9 - 6 C: 2x/9 - (2/9) × 18 = 2x/9 - 4 Adding 9 liters of liquid C, the new quantity of C becomes: New C = (2x/9 - 4) + 12 = 2x/9 + 8 The new ratio is given as 7:4:2. Therefore: (4x/9 - 8): (3x/9 - 6): (2x/9 - 4) = 4: 3: 5 Let's set up the equations using the last two parts of the ratio: (3x/9 - 6) / (2x/9 – 4) = 3/5 Cross-multiplying: 5(3x/9 - 6) = 3(2x/9 - 4) 15x/9 - 30 = 6x/9 – 12 x = 30 – 12 x = 18 liters