Math Problem Statement

How much 

8

%

 saline solution should Kent mix with 

60

 cubic centimeters (cc) of a 

16

%

 saline solution to produce an 

11

%

 saline solution?

Solution

To solve this problem, we can use the concept of a weighted average of concentrations.

Let's denote the volume of the 8% saline solution that Kent needs to mix as xx cc.

Step 1: Set up the equation

The total amount of salt in the final mixture must equal the sum of the amounts of salt in the two initial solutions.

The amount of salt in the 60 cc of 16% saline solution is: \text{Salt from 16% solution} = 0.16 \times 60 \text{ cc}

The amount of salt in the xx cc of 8% saline solution is: \text{Salt from 8% solution} = 0.08 \times x \text{ cc}

The total volume of the final solution will be x+60x + 60 cc, and the concentration of this solution is 11%. So, the amount of salt in the final solution is: Salt in final solution=0.11×(x+60) cc\text{Salt in final solution} = 0.11 \times (x + 60) \text{ cc}

Step 2: Form the equation

Since the amount of salt in the final solution is the sum of the salt from both initial solutions, we set up the equation: 0.16×60+0.08×x=0.11×(x+60)0.16 \times 60 + 0.08 \times x = 0.11 \times (x + 60)

Step 3: Solve the equation

First, calculate the constant term: 0.16×60=9.60.16 \times 60 = 9.6

Substitute this into the equation: 9.6+0.08x=0.11(x+60)9.6 + 0.08x = 0.11(x + 60)

Expand and simplify: 9.6+0.08x=0.11x+6.69.6 + 0.08x = 0.11x + 6.6

Subtract 0.08x0.08x from both sides: 9.6=0.03x+6.69.6 = 0.03x + 6.6

Subtract 6.6 from both sides: 3=0.03x3 = 0.03x

Finally, divide by 0.03: x=30.03=100 ccx = \frac{3}{0.03} = 100 \text{ cc}

Conclusion

Kent should mix 100 cubic centimeters (cc) of 8% saline solution with 60 cc of 16% saline solution to produce an 11% saline solution.


Would you like more details on this calculation or have any other questions?

Related Questions:

  1. How would the answer change if the final concentration needed was 10% instead of 11%?
  2. What would be the concentration if Kent mixed equal parts of 8% and 16% solutions?
  3. How much 8% saline would be needed if the initial solution was 20% instead of 16%?
  4. How does the volume of the final solution affect the required concentrations?
  5. What is the mathematical basis for the equation used to find the solution?

Tip:

When mixing solutions with different concentrations, always ensure that the total amount of the substance (e.g., salt) is conserved across the mixture.

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Math Problem Analysis

Mathematical Concepts

Weighted average
Concentration calculations
Linear equations

Formulas

-

Theorems

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Suitable Grade Level

Grades 9-12