Mixture and alligation Advance Concepts

Presents the Rule of Constant as an example. In Example 4, it asks how many liters of pure water need to be added to a 40-liter, 30% milk solution in order to obtain a 15% milk solution. There are three methods that can be used to solve this problem, including the rule of Mixture and alligation Advance Concepts, which will be explained below.

Method I:

The standard school textbook approach is used in this method. Let’s assume that the quantity of water added is x liters. The 40-liter solution initially contains 30% milk, which is equal to 12 liters. With the addition of water, the quantity of the new solution becomes (40 + x) liters. According to the problem, the percentage of milk in the new solution should be 15%. Therefore, we can set up the equation (\frac{{12}}{{40 + x}} \times 100 = 15). Solving for x, we find that x = 40 liters.

Method II:

The rule of alligation is applied in this method. We assume that two solutions, milk and water, are mixed together to obtain the new solution. Taking milk as the common element in both solutions, the first solution has 30% milk and the second solution (pure water) has 0% milk. When mixed, the final solution has 15% milk. Therefore, the ratio of the quantity of the first solution to the second solution should be 15:15, which simplifies to 1:1. As a result, 40 liters of pure water should be mixed to obtain the desired new solution.

Method III:

The rule of constant is used in this method. In this rule, we focus on an element in the mixture whose amount does not change but its percentage changes due to the change in the total amount of the mixture. As calculated earlier, the quantity of milk in the first solution is 12 liters, and it will remain the same in the new solution as well. This means that 12 liters is equal to 30% of the first solution and 15% of the new solution. Therefore, the quantity of the new solution can be calculated as (\frac{{100}}{{15}} \times 12) = 80 liters. Since the increase in the quantity of the new mixture is due to adding extra water, the quantity of water added is 80 – 40 = 40 liters.

In addition to the above methods, there are also variations in problems based on replacement. Two types are mentioned:

Type I:

When the quantity withdrawn and quantity replaced are the same, the volume of milk in the final solution can be calculated using the formula (x{\left( {1 – \frac{a}{x}} \right)^n}), where x is the initial volume, a is the volume of milk replaced each time, and n is the number of times the operation is repeated.

Example 5 is provided to illustrate this type. From a 40-liter solution of pure milk, 5 liters of milk is replaced with an equal volume of water three times. Applying the formula, we find that the final volume of milk is equal to (40{\left( {1 – \frac{5}{{40}}} \right)^3}), which is approximately 26.8 liters.

Type II:

When the quantity withdrawn and quantity replaced are not the same, a different approach is required. Example 6 demonstrates this type. From a 40-liter solution of pure milk, 5 liters of milk is replaced with 6 liters of water, and then 6 liters of the mixture is replaced with 7 liters of water. The volume of milk in the final solution can be determined by considering the ratio of milk to water after each operation.

Overall, understanding and practicing these Mixture and alligation Advance Concepts can be helpful in solving various problems involving mixtures and solutions.