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Alligation alternate is a useful mathematical method for solving mixture problems, especially when dealing with different concentrations or qualities. It simplifies the process of finding the ratio in which two or more ingredients should be mixed to achieve a desired concentration or quality. This article provides practice problems with solutions to help students master alligation alternate calculations.
Understanding Alligation Alternate
Alligation alternate involves comparing the quantities or qualities of two ingredients to find the ratio in which they should be mixed. It is particularly helpful when dealing with mixtures of solutions with different concentrations, such as medicines, chemicals, or food ingredients.
Practice Problems with Solutions
Problem 1: Mixture of Two Solutions
Two solutions of alcohol and water are mixed. The first solution contains 40% alcohol, and the second contains 80% alcohol. How should they be mixed to obtain 60% alcohol solution in a ratio of?
Solution:
- Alcohol in first solution = 40%
- Alcohol in second solution = 80%
- Desired alcohol concentration = 60%
Using alligation, we subtract the desired concentration from the concentrations of the two solutions:
- 80% – 60% = 20
- 60% – 40% = 20
The ratio of the two solutions is 20:20, which simplifies to 1:1. Therefore, equal parts of both solutions should be mixed to obtain a 60% alcohol solution.
Problem 2: Mixing Different Concentrations
A chemist has two solutions: one with 30% concentration and another with 70%. How much of each should be mixed to prepare 100 ml of a 50% solution?
Solution:
- Concentration of first solution = 30%
- Concentration of second solution = 70%
- Desired concentration = 50%
- Total volume = 100 ml
Calculate the parts using alligation:
- 70% – 50% = 20
- 50% – 30% = 20
The ratio of solutions is 20:20, which simplifies to 1:1. So, equal parts of both solutions are needed.
Since total volume is 100 ml, each part is 50 ml. Therefore, mix 50 ml of the 30% solution with 50 ml of the 70% solution.
Additional Practice Problems
Problem 3: Mixture of Different Grades
Two grades of sugar, one costing $2 per kg and the other $4 per kg, are mixed. To get a mixture costing $3 per kg, in what ratio should they be mixed?
Solution:
- Cost of first grade = $2/kg
- Cost of second grade = $4/kg
- Desired cost = $3/kg
Using alligation:
- 4 – 3 = 1
- 3 – 2 = 1
The ratio of the two grades is 1:1. Mix equal quantities of both grades to obtain the desired mixture.
Summary
Alligation alternate is a simple and effective method for solving mixture problems involving different qualities or concentrations. Practice with various problems helps in understanding its application and improving calculation speed.