Understanding the Milk Dilution Problem and Its Mathematical Solution

In this article, we explore a classic problem in mathematics and its practical application: the milk dilution problem. Through step-by-step calculations and explanations, you'll learn how to solve this challenging problem and how mathematical concepts can be applied in real-life scenarios.

Problem Statement

A container holds 80 litres of pure milk. If 4 litres of milk is taken out and replaced by water, this process is repeated twice more. The question is, how much pure milk now remains in the container?

Step-by-Step Solution

Solution 1

Let's start with the first replacement:

Initial amount of milk: 80 litres Amount of milk removed: 4 litres Amount of water added: 4 litres New quantity of milk: 80 - 4 76 litres New ratio of milk to total mixture: 76/80 0.95 Amount of milk removed in next replacement: 4 litres * 95% 3.8 litres (approximately 3.8 through calculation) New quantity of milk: 76 - 3.8 72.2 litres New ratio of milk to total mixture: 72.2/80 0.9025 Amount of milk removed in final replacement: 4 litres * 90.25% 3.61 litres (approximately 3.61 through calculation) New quantity of milk: 72.2 - 3.61 68.59 litres

Solution 2

Another approach to solve the problem is through the repeated substitution process:

Initial amount of milk: 80 litres Amount of milk removed: 4 litres Amount of water added: 4 litres New quantity of milk: 80 - 4 76 litres New ratio of milk to total mixture: 76/80 0.95 Amount of milk removed in next replacement: 4 litres * 95% 3.8 litres New quantity of milk: 76 - 3.8 72.2 litres New ratio of milk to total mixture: 72.2/80 0.9025 Amount of milk removed in final replacement: 4 litres * 90.25% 3.61 litres New quantity of milk: 72.2 - 3.61 68.59 litres

Solution 3

A third solution involves repeated calculations involving fractions:

Initial amount of milk: 80 litres Amount of milk removed: 4 litres Amount of water added: 4 litres New quantity of milk: 80 - 4 76 litres New ratio of milk to total mixture: 76/80 0.95 Amount of milk remaining in final replacement: 76 * 0.952 72.2 litres (using percentage calculation) Amount of milk removed in final replacement: 4 litres * 0.952 3.61 litres New quantity of milk: 76 - 3.61 68.59 litres

Conclusion

The problem of milk dilution involves repeated removal and replacement. Through the methodical application of mathematical concepts, we can determine the quantity of pure milk remaining after multiple replacements. This problem not only highlights the importance of understanding ratios and percentages but also the concept of exponential decay in a practical scenario.

Related Concepts

Understanding the milk dilution problem also helps in comprehending similar real-world applications, such as:

Mixture Problems: Calculating the composition of different mixtures. Water Replacement: Calculating the effect of adding water to a mixture. Percentage Calculation: Determining how much of a substance is left in a mixture after repeated additions.

By mastering these concepts, one can solve a wide range of problems related to mixtures and dilutions, making it a valuable tool in various fields including chemistry, biology, and even finance.