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.