We can solve this problem using a system of linear equations. Let:

• $x$ be the number of gallons of 30% alcohol.
• $y$ be the number of gallons of 10% alcohol.

We are given two important pieces of information:

1. The total amount of alcohol solution after mixing is 20 gallons.
2. The final mixture should be 14% alcohol.

Step 1: Total amount of solution

We know the total volume of the mixture is 20 gallons: $x + y = 20$

Step 2: Total amount of alcohol in the solution

The amount of alcohol in each solution is calculated as:

• $0.30x$ gallons of pure alcohol from the 30% solution,
• $0.10y$ gallons of pure alcohol from the 10% solution.

The total amount of pure alcohol in the final 20-gallon mixture must be: $0.14 \times 20 = 2.8 \, \text{gallons}.$

So, the equation for the amount of alcohol is: $0.30x + 0.10y = 2.8$

Step 3: Solving the system of equations

We now have two equations:

1. $x + y = 20$
2. $0.30x + 0.10y = 2.8$

We can solve these equations simultaneously.

From the first equation, solve for $y$:

$y = 20 - x$

Substitute this into the second equation:

$0.30x + 0.10(20 - x) = 2.8$ $0.30x + 2 - 0.10x = 2.8$ $0.20x + 2 = 2.8$ $0.20x = 0.8$ $x = \frac{0.8}{0.20} = 4$

Now, substitute $x = 4$ into the equation $y = 20 - x$:

$y = 20 - 4 = 16$

Conclusion

You should mix 4 gallons of 30% alcohol with 16 gallons of 10% alcohol to obtain 20 gallons of a 14% alcohol solution.

Would you like a breakdown of each step or have any questions?

Here are 5 related questions for practice:

1. How many gallons of 40% alcohol and 20% alcohol are needed to make 30 gallons of 28% alcohol?
2. If you mix 10 gallons of 50% alcohol with water, what will be the final concentration of the solution?
3. How much 15% salt solution should be mixed with 25% salt solution to get 50 gallons of 20% salt solution?
4. What amount of 12% acid and 8% acid is needed to create 100 liters of 10% acid solution?
5. How can you adjust the concentration of a solution if you only have one type of alcohol?

Tip: Always check your final answer by substituting the values into the original conditions!