Solve mixture problems with proportions quick check

Answer 1

Mixture problems typically involve combining different substances or solutions that have different proportions of a certain quantity (like price, concentration, etc.). Here's a quick guide to help you solve these types of problems:

Steps to Solve Mixture Problems

1. Identify the Components: Clearly define the substances or mixtures involved.

2. Define Variables: Let variables represent the quantities of each component in the mixture.

3. Set Up Equations: Based on the proportions or values given in the problem, set up equations. Typically, there are two main types of equations:

   • Total Amount Equation: This represents the total quantities of the components.
   • Total Value/Concentration Equation: This represents the combined value/concentration of the mixture.

4. Solve the Equations: Use algebraic methods to solve the equations for the unknowns.

5. Check the Solution: Verify that the solution meets all conditions laid out in the problem.

Quick Example

Problem: You have a 10% salt solution and a 30% salt solution. How much of each solution do you need to mix to obtain 200 mL of a 20% salt solution?

Solution Steps:

1. Define Variables:

   • Let \( x \) = the volume of the 10% solution.
   • Let \( y \) = the volume of the 30% solution.

2. Set Up the Equations:

   • Total volume equation: \[ x + y = 200 \quad \text{(1)} \]
   • Salt concentration equation (for salt content): \[ 0.10x + 0.30y = 0.20(200) \quad \text{(2)} \] Simplifying (2): \[ 0.10x + 0.30y = 40 \]

3. Solve the Equations:

   • From equation (1), express \( y \): \[ y = 200 - x \]
   • Substitute \( y \) in equation (2): \[ 0.10x + 0.30(200 - x) = 40 \] \[ 0.10x + 60 - 0.30x = 40 \] \[ -0.20x + 60 = 40 \] \[ -0.20x = -20 \] \[ x = 100 \]
   • Substitute back to find \( y \): \[ y = 200 - 100 = 100 \]

4. Conclusion: You need 100 mL of the 10% solution and 100 mL of the 30% solution to make 200 mL of a 20% salt solution.

Practice Problems

Now you can try solving these problems using the steps outlined:

1. How many liters of a 5% acid solution must be mixed with 7 liters of a 15% acid solution to create a 10% acid solution?

2. If you mix 3 liters of a 25% sugar solution with how many liters of a 70% sugar solution to get a 40% sugar solution?

If you have any specific problems you'd like to go through, feel free to share!