To solve the problem of how many gallons of 25% alcohol should be mixed with 5% alcohol to obtain 20 gallons of 24% alcohol, we can use a system of equations. Here's how:

1. Let $x$ be the gallons of 25% alcohol.
2. Let $y$ be the gallons of 5% alcohol.

We have two conditions:

• Volume equation: $x + y = 20$ (since the total mixture is 20 gallons).
• Alcohol concentration equation: $0.25x + 0.05y = 0.24(20)$ (total alcohol content).

Step 1: Simplify the alcohol equation

$0.25x + 0.05y = 4.8$

Step 2: Solve the system of equations

From the volume equation: $y = 20 - x$

Substitute $y = 20 - x$ into the alcohol equation: $0.25x + 0.05(20 - x) = 4.8$

Expand and simplify: $0.25x + 1 - 0.05x = 4.8$ $0.2x + 1 = 4.8$ $0.2x = 3.8$ $x = \frac{3.8}{0.2} = 19$

Step 3: Solve for $y$

$y = 20 - x = 20 - 19 = 1$

Final Answer:

• $x = 19$ gallons of 25% alcohol.
• $y = 1$ gallon of 5% alcohol.

Would you like further explanation or help with related questions?

Here are 5 related questions for practice:

1. How would the solution change if the final concentration were 30% instead of 24%?
2. What is the total amount of pure alcohol in the final solution?
3. What equation represents the relationship between the concentrations in mixtures?
4. Can you generalize the formula for mixing two solutions with different concentrations?
5. What happens if no solution meets the required final concentration?

Tip: Always double-check that the total mixture's alcohol content matches the expected concentration when solving mixture problems!