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.