How many ounces of pure water must be added to 50 oz of a 15% saline solution to make a saline solution that is 10% salt

Let the number of ounces of pure water to be added be \( x \).

We are starting with 50 oz of a 15% saline solution, meaning it contains:
\[
0.15 \times 50 = 7.5 \, \text{oz of salt}.
\]
After adding \( x \) ounces of water, the total volume of the solution becomes \( 50 + x \) ounces. We want the final solution to be 10% salt, so the amount of salt should be:
\[
0.10 \times (50 + x) \, \text{oz of salt}.
\]
Since the amount of salt remains constant at 7.5 oz, we set up the equation:
\[
7.5 = 0.10 \times (50 + x).
\]
Solving for \( x \):
\[
7.5 = 5 + 0.10x,
\]
\[
7.5 - 5 = 0.10x,
\]
\[
2.5 = 0.10x,
\]
\[
x = \frac{2.5}{0.10} = 25.
\]

Thus, you need to add **25 oz** of pure water.

Would you like more details or have any further questions?

Here are 5 related questions to expand this topic:
1. How would you adjust the problem if the initial solution was 20% saline instead of 15%?
2. What if you wanted a final solution of 5% salt instead of 10%?
3. How can you calculate the final volume and concentration if 30 oz of water were added instead of 25 oz?
4. How do you approach this problem if a different solute was involved (e.g., sugar in water)?
5. How does the ratio of water and saline affect the concentration in large volumes?

**Tip:** When dealing with percentage problems, always think of percentages as fractions or decimals to simplify calculations.

How many ounces of pure water must be added to 50 oz of a 15% saline solution to make a saline solution that is 10% salt?

Discover how to calculate the amount of pure water needed to dilute a 50 oz 15% saline solution to a 10% concentration. This step-by-step solution uses key algebraic and percentage concepts. Suitable for students in Grades 8-10.

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