Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 58

How many liters of a $10\%$ alcohol solution must be mixed with $40$ L of a $50\%$ solution to get a $40\%$ solution?

Step 1: Read the problem, we are asked to find the amount of the $10\%$ alcohol solution.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x =$ amount of the $10\%$ alocohol solution.


$\begin{array}{|c|c|c|c|} \hline & \text{Liters of solution} & \text{Percent Concentration} & \text{Liters of Pure Alcohol} \\ \hline 10\% \rm{alcohol} & x & 0.10 & 0.10x \\ \hline 50\% \rm{alcohol} & 40 & 0.50 & 0.50(40) \\ \hline \text{Resulting mixture ($40\%$ alcohol)} & x + 40 & 0.40 & 0.40(x + 40) \\ \hline \end{array}$

The sum of the quantities of each solution is equal to the quantity of the resulting solution

Step 3: Write an equation from the last column of the table
$0.10x + 0.50(40) = 0.40(x + 40)$

Step 4: Solve

$\begin{equation} \begin{aligned} 0.10x + 20 &= 0.40x + 16 \\ \\ 0.10x - 0.40x &= 16 - 20\\ \\ -0.30x &= -4\\ \\ x &= \frac{40}{3} \end{aligned} \end{equation}$


Step 5: State the answer
In other words, $\displaystyle \frac{40}{3} L$ of $10\%$ alcohol solution.