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We are tasked with finding the percent concentration of the first saline solution in this problem. Let's break it down step-by-step:

Given:
- 7 ml of the first saline solution (concentration unknown, call it \( x \% \)).
- 8 ml of a 60% saline solution.
- The resulting solution is 15 ml (7 ml + 8 ml) with a 74% saline concentration.

To solve this, we can use the formula for mixtures:

\[
\text{Total amount of solute} = (\text{volume}_1 \times \text{concentration}_1) + (\text{volume}_2 \times \text{concentration}_2)
\]

\[
(7 \times x) + (8 \times 0.60) = 15 \times 0.74
\]

Now, let's solve this equation:

1. Multiply the terms:

\[
7x + 4.8 = 11.1
\]

2. Subtract 4.8 from both sides:

\[
7x = 6.3
\]

3. Divide by 7:

\[
x = \frac{6.3}{7} = 0.9
\]

Thus, the concentration of the first solution is **90%**.

Would you like more details or have any further questions?

Here are 5 related questions to expand on this concept:
1. How would the result change if the second solution had a concentration of 50% instead of 60%?
2. What would happen if equal volumes of the two solutions were mixed?
3. How can you calculate the final concentration if both initial concentrations are unknown?
4. What is the minimum concentration of the second solution to get a mixture concentration of 74%?
5. Can you create a formula for any general mixture of two solutions with known concentrations?

**Tip**: When solving mixture problems, always start by writing an equation that represents the total amount of solute in each component of the mixture.

7 ml of a saline solution was mixed with 8 ml of a 60% saline solution to make a 74% saline solution. Find the percent concentration of the first solution.

Solve a mixture problem where 7 ml of an unknown saline solution is mixed with 8 ml of a 60% saline solution to achieve a 74% concentration. Learn how to find the percent concentration of the first solution using algebraic formulas.

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