- Identify variables \(x\) and \(y\) for the volumes of the 80% and 40% solutions, respectively.
- Create a system of equations based on total volume (\(x + y = 12\)) and total alcohol content (\(0.8x + 0.4y = 8.4\)).
- Solve the system by substitution to find \(x = 9\) and \(y = 3\).
- Final Answer: \(\boxed{9 \text{ liters of 80\% solution and 3 liters of 40\% solution}}\)
Explanation
- Analyzing and Modeling the Mixture Problem
To solve for the volume of each alcohol solution, we define the variables: let \(x\) be the number of liters of the \(80\%\) alcohol solution and \(y\) be the number of liters of the \(40\%\) alcohol solution. We know the total volume of the final mixture is \(12\) liters, giving us the first equation:
\[x + y = 12\]
Next, we track the amount of pure alcohol. The \(80\%\) solution contains \(0.80x\) liters of pure alcohol, the \(40\%\) solution contains \(0.40y\) liters of pure alcohol, and the final \(12\)-liter mixture at \(70\%\) concentration must contain \(0.70 \times 12 = 8.4\) liters of pure alcohol. This gives us our second equation:
\[0.80x + 0.40y = 8.4\]
- Completing the Data Table
We can organize the data in a table as requested:
| | \(80\%\) solution | \(40\%\) solution | Final solution |
| :--- | :--- | :--- | :--- |
| Number of Liters | \(x\) | \(y\) | \(12\) |
| Liters of Alcohol | \(0.8x\) | \(0.4y\) | \(8.4\) |
- Solving the System of Equations
We have the system of linear equations:
1) \(x + y = 12\)
2) \(0.8x + 0.4y = 8.4\)
From the first equation, we can express \(y\) in terms of \(x\): \(y = 12 - x\). Substituting this into the second equation gives:
\[0.8x + 0.4(12 - x) = 8.4\]
Distributing the \(0.4\):
\[0.8x + 4.8 - 0.4x = 8.4\]
Combining like terms:
\[0.4x + 4.8 = 8.4\]
Subtracting \(4.8\) from both sides:
\[0.4x = 3.6\]
Dividing by \(0.4\):
\[x = \frac{3.6}{0.4} = 9\]
Now, substitute \(x = 9\) back into the expression for \(y\):
\[y = 12 - 9 = 3\]
Thus, we need \(9\) liters of the \(80\%\) solution and \(3\) liters of the \(40\%\) solution.
Examples
This method is widely used in chemistry and pharmacology to prepare solutions of specific concentrations from existing stock solutions. For example, a nurse preparing a specific intravenous medication concentration or a chemist creating a standard buffer solution uses these exact principles. By setting up a system of equations, one can accurately calculate the required volumes to ensure the final product is both safe and effective for its intended application.