How many gallons of a 50​% antifreeze solution must be mixed with 60 gallons of 25​% antifreeze to get a mixture that is 40​% ​antifreeze? Use the​ six-step method.

Let's solve this problem using the six-step method.

### **Step 1: Understand the problem**
You have:
- 60 gallons of a 25% antifreeze solution.
- You want to mix this with some amount \( x \) (in gallons) of a 50% antifreeze solution.
- The goal is to obtain a mixture that is 40% antifreeze.

### **Step 2: Assign a variable**
Let:
- \( x \) = the number of gallons of the 50% antifreeze solution that we need to mix with the 60 gallons of 25% antifreeze.

### **Step 3: Write an equation**
First, calculate the amount of pure antifreeze in each solution:
- Pure antifreeze in the 25% solution: \( 0.25 \times 60 \) gallons.
- Pure antifreeze in the 50% solution: \( 0.50 \times x \) gallons.

The total amount of antifreeze after mixing should be equal to the amount in the final mixture, which will be \( (60 + x) \) gallons at 40% concentration. Therefore, the equation is:
\[
0.25 \times 60 + 0.50 \times x = 0.40 \times (60 + x)
\]

### **Step 4: Solve the equation**
First, calculate the pure antifreeze in the 60 gallons of 25% solution:
\[
0.25 \times 60 = 15 \text{ gallons of pure antifreeze}
\]
So the equation simplifies to:
\[
15 + 0.50x = 0.40 \times (60 + x)
\]
Distribute the 0.40 on the right side:
\[
15 + 0.50x = 24 + 0.40x
\]
Subtract \( 0.40x \) from both sides:
\[
15 + 0.10x = 24
\]
Subtract 15 from both sides:
\[
0.10x = 9
\]
Solve for \( x \):
\[
x = \frac{9}{0.10} = 90 \text{ gallons}
\]

### **Step 5: Check your answer**
Let's verify:
- Pure antifreeze in the 60 gallons of 25% solution = 15 gallons.
- Pure antifreeze in the 90 gallons of 50% solution = 0.50 \times 90 = 45 gallons.
- Total pure antifreeze in the mixture = 15 + 45 = 60 gallons.
- Total mixture volume = 60 + 90 = 150 gallons.
- Concentration of the final mixture = \( \frac{60}{150} = 0.40 \) or 40%.

The solution checks out.

### **Step 6: State the answer**
You need to mix **90 gallons** of a 50% antifreeze solution with 60 gallons of a 25% antifreeze solution to get a mixture that is 40% antifreeze.

Would you like to delve deeper into any step or have any questions?

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Here are 5 related questions you might consider:
1. What if the initial solution was 30% antifreeze instead of 25%? How would that change the amount of 50% solution needed?
2. How much of a 75% antifreeze solution is needed to achieve the same 40% concentration with the 60 gallons of 25% solution?
3. How would you solve for \( x \) if the desired concentration was 35% instead of 40%?
4. If you start with 80 gallons of 30% antifreeze, how much 50% antifreeze solution do you need to add to make a 40% solution?
5. What happens to the concentration if you add pure antifreeze instead of the 50% solution?

**Tip:** Always double-check your work by substituting your solution back into the original context to ensure accuracy.

Learn how to mix antifreeze solutions to achieve a 40% concentration using a step-by-step method. This problem involves calculating the amount of a 50% antifreeze solution needed to mix with 60 gallons of a 25% antifreeze solution. Suitable for high school students and above.

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