A mixture contains alcohol and water in the ratio 3:2. If 10 liters of water are added, the ratio becomes 3:4. What was the initial volume of the mixture?
["How to Solve Mixture Problems: A Case Study with a 3:2 Alcohol-to-Water Mixture", "When dealing with evaporation or dilution problems involving mixtures, understanding ratios and algebra is key. One common type of problem involves a mixture containing alcohol and water in a 3:2 ratio. In this article, we explore a specific scenario: when 10 liters of water are added to such a mixture, the new ratio changes to 3:4. We walk through the math step-by-step to determine the initial volume of the mixture.", "---", "### The Problem Setup", "Initially, the mixture contains alcohol and water in a 3:2 ratio.", "Let the initial volume of the mixture be ( V ) liters.", "Let the amount of alcohol be ( 3x ) liters and water be ( 2x ) liters.\nSo,\n[\n3x + 2x = V \Rightarrow V = 5x\n]", "Now, 10 liters of water are added, so the new amount of water becomes:\n[\n2x + 10\n]\nAlcohol remains unchanged at ( 3x ).", "The new ratio becomes 3:4, so:\n[\n\frac{3x}{2x + 10} = \frac{3}{4}\n]", "---", "### Solving the Equation", "Cross-multiplying:\n[\n4 \cdot 3x = 3 \cdot (2x + 10)\n]", "Simplify both sides:\n[\n12x = 6x + 30\n]", "Subtract ( 6x ) from both sides:\n[\n6x = 30\n]", "Divide by 6:\n[\nx = 5\n]", "Now substitute back to find the initial mixture volume:\n[\nV = 5x = 5 \ imes 5 = 25 \ ext{ liters}\n]", "---", "### Conclusion", "The initial volume of the alcohol-and-water mixture was 25 liters. By adding 10 liters of water, the water portion increased from 10 liters (2x when x=5) to 20 liters, shifting the ratio from 3:2 to 3:4 as confirmed.", "This problem illustrates how ratios change with proportional additions — a crucial skill in chemistry, beverage formulation, and chemical engineering.", "For similar mixture challenges, always:", "- Define variables based on the given ratio\n- Express quantities in terms of a common variable\n- Set up proportional equations based on changes\n- Solve algebraically for unknowns", "Understanding these patterns helps solve real-world problems involving dilution, concentration, and formulation with precision."]
