How much water should be added to 1 gallo of pure antifreeze to obtain a solution that is 95 percent antifreeze

To solve this problem, we need to determine how much water should be added to 1 gallon of pure antifreeze to dilute it to a solution that is 95% antifreeze.

Let:
- \( x \) be the amount of water (in gallons) to add.

We know:
- The total volume of the final solution will be \( 1 + x \) gallons.
- We want the solution to be 95% antifreeze, which means 5% of the solution is water.

The amount of antifreeze in the final solution is 1 gallon (since we started with pure antifreeze). To satisfy the condition that 95% of the final solution is antifreeze, we can set up the equation:

\[
\frac{1}{1 + x} = 0.95
\]

### Solving the equation:
1. Multiply both sides by \( 1 + x \):

\[
1 = 0.95(1 + x)
\]

2. Expand the right-hand side:

\[
1 = 0.95 + 0.95x
\]

3. Subtract 0.95 from both sides:

\[
0.05 = 0.95x
\]

4. Solve for \( x \):

\[
x = \frac{0.05}{0.95} \approx 0.0526 \text{ gallons}
\]

Thus, you need to add approximately **0.0526 gallons** of water, or about **6.73 ounces** (since 1 gallon = 128 ounces).

Would you like more details or have any questions?

Here are 5 related questions:
1. How much water would be needed for an 80% antifreeze solution?
2. What if you had 2 gallons of antifreeze and wanted a 95% solution?
3. How would you adjust the solution to make it 50% antifreeze?
4. How does the concentration change if more water is added?
5. What is the ratio of antifreeze to water in a 95% solution?

**Tip:** Converting between different units (gallons, liters, ounces) can help ensure precision in real-world problems.

How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 95 percent antifreeze?

Learn how to determine the amount of water needed to dilute 1 gallon of pure antifreeze to a solution that is 95% antifreeze. This problem involves using basic algebraic concepts and concentration formulas, making it suitable for students in Grades 9-12.

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