To solve this problem, we can use a system of equations based on the concentrations and volumes of the indicator solutions.
Let \( x \) be the number of gallons of the 12% solution, and \( y \) be the number of gallons of the 20% solution. We know that:
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The total volume of the mixed solution should be 10 gallons: \[ x + y = 10 \]
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The desired concentration for the final solution is 14%. The equation for the amount of indicator in the solutions can be written as: \[ 0.12x + 0.20y = 0.14 \times 10 \] Simplifying the right-hand side gives: \[ 0.12x + 0.20y = 1.4 \]
Now we have the following system of equations:
\[ \begin{align*}
- & \quad x + y = 10 \
- & \quad 0.12x + 0.20y = 1.4 \end{align*} \]
From equation (1), we can express \( y \) in terms of \( x \): \[ y = 10 - x \]
Now, substitute \( y \) into equation (2): \[ 0.12x + 0.20(10 - x) = 1.4 \]
Expand and simplify: \[ 0.12x + 2 - 0.20x = 1.4 \] \[ -0.08x + 2 = 1.4 \]
Now, isolate \( x \): \[ -0.08x = 1.4 - 2 \] \[ -0.08x = -0.6 \] \[ x = \frac{-0.6}{-0.08} = 7.5 \]
Now that we have \( x \), substitute back to find \( y \): \[ y = 10 - x = 10 - 7.5 = 2.5 \]
Thus, the solution is:
- \( x = 7.5 \) gallons of the 12% solution
- \( y = 2.5 \) gallons of the 20% solution
Final Answer: Mix 7.5 gallons of a 12% indicator solution with 2.5 gallons of a 20% indicator solution.