Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 56

How many liters of a $14\%$ alcohol solution must be mixed with $20$ L of a $50\%$ solution get a $30\%$ solution?

Step 1: Read the problem, we are asked to find the amount of the $14\%$ alcohol solution.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x = $ amount of the $14\%$ alocohol solution.


$
\begin{array}{|c|c|c|c|}
\hline
& \text{Liters of solution} & \cdot & \text{Percent Concentration} & = & \text{Liters of Pure Alcohol} \\
\hline
14\% & x & \cdot & 0.014 & = & 0.014x \\
\hline
50\% & 20 & \cdot & 0.50 & = & 0.50(20) \\
\hline
30\% & x + 20 & \cdot & 0.30 & = & 0.30(x + 20) \\
\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
$0.14x + 0.5(20) = 0.30 (x + 20)$

Step 4: Solve

$
\begin{equation}
\begin{aligned}
0.14x - 0.30x &= 6- 10 \\
\\
-0.16x &= -4\\
\\
x &= 25
\end{aligned}
\end{equation}
$


Step 5: State the answer
In other words, $25$ liters of $14\%$ alcohol solution must be mixed.