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:
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The amount of antifreeze in the 30% solution: \[ \text{Antifreeze from 30% solution} = 80 \times 0.30 = 24 \text{ gallons} \]
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The amount of antifreeze in the 60% solution: \[ \text{Antifreeze from 60% solution} = x \times 0.60 = 0.60x \text{ gallons} \]
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The total volume of the mixture: \[ \text{Total volume} = 80 + x \text{ gallons} \]
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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.