3.) How much of a \( 30 \% \) acid solution must be mixed with 70 gallons of a \( 12 \% \) acid solution to obtain a solution that is \( 20 \% \) acid? \( \begin{array}{llll}\text { (A) } 5.6 \mathrm{gal} & \text { (B) } 49 \mathrm{gal} & \text { [C] } 56 \mathrm{gal} & \text { (D) none of these }\end{array} \)

To solve the problem, we can set up an equation based on the amount of acid in each solution. Let \( x \) be the number of gallons of the \( 30\% \) acid solution that we need to mix with \( 70 \) gallons of the \( 12\% \) acid solution. 1. **Calculate the amount of acid in each solution:** - The amount of acid in the \( 30\% \) solution is \( 0.30x \). - The amount of acid in the \( 12\% \) solution is \( 0.12 \times 70 = 8.4 \) gallons. 2. **Set up the equation for the final solution:** The total volume of the new solution will be \( x + 70 \) gallons, and we want this solution to be \( 20\% \) acid. Therefore, the amount of acid in the final solution should be: \[ 0.20(x + 70) \] 3. **Set up the equation:** The total amount of acid from both solutions should equal the amount of acid in the final solution: \[ 0.30x + 8.4 = 0.20(x + 70) \] 4. **Distribute and simplify:** \[ 0.30x + 8.4 = 0.20x + 14 \] 5. **Rearranging the equation:** Subtract \( 0.20x \) from both sides: \[ 0.30x - 0.20x + 8.4 = 14 \] \[ 0.10x + 8.4 = 14 \] 6. **Isolate \( x \):** Subtract \( 8.4 \) from both sides: \[ 0.10x = 14 - 8.4 \] \[ 0.10x = 5.6 \] 7. **Solve for \( x \):** \[ x = \frac{5.6}{0.10} = 56 \] Thus, the amount of the \( 30\% \) acid solution that must be mixed is \( \boxed{56} \) gallons.