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Magazine Chapter 4: Problem 38
Gerry Gundersen mixes different solutions with concentrations of \(25 \%, 40 \%,\) and \(50 \%\) to get 200 liters of a \(32 \%\) solution. If he uses twice as much of the \(25 \%\) solution as of the \(40 \%\) solution, find how many liters of each kind he uses.
Short Answer
Expert verified
Gerry uses 120 liters of 25%, 60 liters of 40%, and 20 liters of 50% solution.
Step by step solution
01
Establish Variables
Let the amount of the \(25\%\) solution be \(x\) liters, the amount of the \(40\%\) solution be \(y\) liters, and the amount of the \(50\%\) solution be \(z\) liters. According to the problem, \(x = 2y\) because he uses twice as much \(25\%\) solution as \(40\%\) solution.
02
Total Volume Equation
The total volume of the solution should be \(200\) liters. Therefore, we have the equation: \[ x + y + z = 200. \]
03
Concentration Equation
To satisfy the concentration requirement, the sum of the "active" part (concentration part) of each solution must equal the concentration of the final solution. Therefore, \[ 0.25x + 0.4y + 0.5z = 0.32 \times 200. \] Simplifying the right side, we get: \[ 0.25x + 0.4y + 0.5z = 64. \]
04
Substitute Variables
Substitute \(x = 2y\) (from Step 1) into both equations: \(x + y + z = 200\) becomes \[ 2y + y + z = 200, \] which simplifies to \[ 3y + z = 200. \] Similarly, substitute into the concentration equation \(0.25x + 0.4y + 0.5z = 64\), which becomes: \[ 0.25(2y) + 0.4y + 0.5z = 64. \] Simplify to get \[ 0.5y + 0.4y + 0.5z = 64, \] which further simplifies to \[ 0.9y + 0.5z = 64. \]
05
Solve the System of Equations
Now, solve the system of equations \[ 3y + z = 200 \] and \[ 0.9y + 0.5z = 64. \] Solve the first for \(z\): \[ z = 200 - 3y. \] Substitute this into the second equation: \[ 0.9y + 0.5(200 - 3y) = 64. \] Simplify to find \(y\): \[ 0.9y + 100 - 1.5y = 64, \] which simplifies to \[ -0.6y + 100 = 64. \] Solve for \(y\): \[ -0.6y = -36 \] \[ y = 60. \]
06
Find the Remaining Variables
With \(y = 60\), we substitute back to find \(x\) and \(z\): \[ x = 2y = 2 \times 60 = 120. \] And from \(z = 200 - 3y\), we have \[ z = 200 - 3 \times 60 = 200 - 180 = 20. \]
07
Verify the Solution
Verify by substituting \(x = 120\), \(y = 60\), and \(z = 20\) into the concentration equation: \[ 0.25(120) + 0.4(60) + 0.5(20) = 30 + 24 + 10 = 64. \] Since this checks out, our solution is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Algebraic Equations
Algebraic equations serve as a fundamental tool in solving many types of problems, including mixture problems. Here's why they are powerful:
- Represent Real-life Situations: Algebraic equations allow us to create mathematical models of real-world scenarios. In this case, we need equations to represent the volumes and concentrations of mixed solutions.
- Equations as Relationships: We use variables to represent unknown quantities. In the example, these variables are used to denote the liters of each solution.
- Balancing Equations: An equation works like a balance where both sides must be equal. This property helps ensure that our modeled situation correctly reflects what we are trying to find.
- Step-by-Step Breakdown: Problem solvers can break down complex problems into smaller, manageable equations. Each equation can be tackled methodically to progress toward a complete solution.
Concentration Calculations
Concentration calculations are crucial in chemical and solution mixtures. They help determine the precise makeup of a final solution:
- Understanding Concentration: Concentration refers to the amount of a substance within a mixture relative to the total amount of the mixture. It's often given as a percentage, indicating how much of a specific component is present.
- Calculating "Active" Parts: In the problem, each solution鈥檚 concentration (like 25%, 40%, and 50%) is an essential factor. Multiplying the volume of each solution by its concentration gives the active component.
- Balancing Concentrations: To ensure the final mixture has the correct concentration, sum up the active parts from each individual solution. This is compared to the desired concentration of the final volume.
- Practical Applications: Students can apply these calculations in real-life situations like cooking, chemistry experiments, and finding the right medication dosages.
System of Equations
A system of equations involves solving for multiple unknowns simultaneously. This is particularly useful in complex problems, like mixture scenarios:
- Multiple Variables: In the given problem, we have three variables representing different solution volumes. A single equation can't show all relationships, so a system is needed.
- Different Equations for Different Relationships: The problem sets up two main types of relationships: the total volume equation and the concentration equation. Each provides a piece of the solution puzzle.
- Substitution Method: One way to solve system equations is by substituting one equation into another. Here, substituting simplifies the solutions and reduces the number of unknowns step-by-step.
- Solution Verification: Once your calculations provide values for all variables, it's crucial to substitute them back into the initial equations to prove their accuracy.
