Question

How many gallons of a 60​% antifreeze solution must be mixed with 80 gallons of 30​% antifreeze to get a mixture that is 50​% ​antifreeze? Use the​ six-step method.

Answer 1

To solve the problem of mixing two solutions of different concentrations to obtain a target concentration, we can use a systematic approach. Let's follow the six-step method:

### Step 1: Identify the variables
Let \( x \) be the number of gallons of the 60% antifreeze solution to be mixed with the 30% antifreeze solution.

### Step 2: Identify the known values
- Volume of 30% solution: \( 80 \) gallons
- Concentration of 30% solution: \( 0.30 \)
- Concentration of 60% solution: \( 0.60 \)
- Desired concentration of the final mixture: \( 0.50 \)

### Step 3: Set up the equation
We can express the total amount of antifreeze in each solution:

1. The amount of antifreeze in the 30% solution:
\[
\text{Antifreeze from 30% solution} = 80 \times 0.30 = 24 \text{ gallons}
\]

2. The amount of antifreeze in the 60% solution:
\[
\text{Antifreeze from 60% solution} = x \times 0.60 = 0.60x \text{ gallons}
\]

3. The total volume of the mixture:
\[
\text{Total volume} = 80 + x \text{ gallons}
\]

4. The total amount of antifreeze in the mixture should equal the desired concentration of 50%, which means:
\[
\text{Antifreeze in mixture} = (80 + x) \times 0.50
\]

### Step 4: Write the equation to represent the situation
Setting the equation based on the amount of antifreeze:
\[
24 + 0.60x = (80 + x) \times 0.50
\]

### Step 5: Solve the equation
Expanding the right-hand side:
\[
24 + 0.60x = 40 + 0.50x
\]

Now, subtract \( 0.50x \) from both sides:
\[
24 + 0.10x = 40
\]

Subtract 24 from both sides:
\[
0.10x = 16
\]

Now, divide by \( 0.10 \):
\[
x = \frac{16}{0.10} = 160
\]

### Step 6: Conclusion
You need to mix **160 gallons** of the 60% antifreeze solution with 80 gallons of 30% antifreeze to achieve a mixture that is 50% antifreeze.