Let the concentration of the first acid solution be $x \%$. The total amount of acid in the mixture can be expressed as:

• Acid from the first solution: $2 \times \frac{x}{100}$ gallons
• Acid from the second solution (85% acid): $4 \times \frac{85}{100} = 3.4$ gallons

The total volume of the mixture is $2 + 4 = 6$ gallons, and the final concentration is 60%, meaning that the total acid content in the mixture is:

$6 \times \frac{60}{100} = 3.6 \text{ gallons of acid.}$

Now, we can set up the equation for the total amount of acid:

$2 \times \frac{x}{100} + 3.4 = 3.6$

Solve for $x$:

$2 \times \frac{x}{100} = 3.6 - 3.4$ $2 \times \frac{x}{100} = 0.2$ $\frac{x}{100} = \frac{0.2}{2}$ $\frac{x}{100} = 0.1$ $x = 10$

Thus, the concentration of the first solution is $10\%$.

Would you like further details or have any questions?

Here are five related questions:

1. How do we calculate the concentration of a solution given the amount of solute and solvent?
2. What is the general formula for mixing solutions of different concentrations?
3. How would the solution change if the volume of the first solution were different?
4. How can we derive the concentration for more than two solutions?
5. What happens to the final concentration if a diluting substance is added?

Tip: When mixing solutions, always ensure that the units for volume and concentration are consistent throughout the calculations.