Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 60

How much water must be added to $6$ gal of a $4\%$ insecticide solution to reduce concentration to $3\%$?

Step 1: Read the problem, we are asked to find the amount of water in the solution.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x = $ amount of water added in the solution.
Since we are concern with the percent of concentration of the insecticide in the solution, then we can say
that the water is $0\%$ insecticide. So,


$
\begin{array}{|c|c|c|c|c|c|}
\hline
& \text{Liters of solution} & \cdot & \text{Percent concentration} & = & \text{Liters of pure insecticide} \\
\hline
\text{Water} & x & \cdot & 0 & = & 0 \\
\hline
\text{$4\%$ insecticide} & 6 & \cdot & 0.04 & = & 0.04(6) \\
\hline
\text{$3\%$ insecticide} & x + 6 & \cdot & 0.03 & = & 0.03(x + 6) \\
\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
$0 + 0.04(6) = 0.03(x + 6)$

Step 4: Solve

$
\begin{equation}
\begin{aligned}
0.24 &= 0.03x + 0.18 \\
\\
-0.03x &= 0.18 - 0.24\\
\\
-0.03x &= -0.06\\
\\
x &= 2
\end{aligned}
\end{equation}
$


Step 5: State the answer
In other words, $2$ gal of water must be added in the solution.