Problem 57 - Intermediate Algebra

Question

Antifreeze. How many pints of a \(10 \%\) antifreeze solution and how many pints of a 40 \% antifreeze solution must be mixed to obtain 24 pints of a \(30 \%\) solution?

Step-by-Step Solution

Answer

Mix 8 pints of 10% antifreeze solution and 16 pints of 40% solution.

Step 1: Set Up Variables

Let \( x \) represent the number of pints of the 10% antifreeze solution, and \( y \) represent the number of pints of the 40% antifreeze solution. We want a total of 24 pints, so we can express this with the equation: \( x + y = 24 \).

Step 2: Create Equation for Concentration

We know that the 10% solution contributes \( 0.1x \) pints of antifreeze, and the 40% solution contributes \( 0.4y \) pints of antifreeze. The total solution should be 30% antifreeze, so the equation for concentration is: \( 0.1x + 0.4y = 0.3 \times 24 \).

Step 3: Simplify Concentration Equation

First, calculate the total amount of antifreeze needed: \( 0.3 \times 24 = 7.2 \) pints. Now, the concentration equation simplifies to: \( 0.1x + 0.4y = 7.2 \).

Step 4: Solve the System of Equations

We now solve the system of equations: \[\begin{align*} x + y & = 24, \ 0.1x + 0.4y & = 7.2. \end{align*}\] Solve the first equation for \( x \): \( x = 24 - y \). Substitute \( x \) in the second equation: \( 0.1(24 - y) + 0.4y = 7.2 \).

Step 5: Solve for y

Substitute \( x = 24 - y \) into the second equation and simplify: \( 2.4 - 0.1y + 0.4y = 7.2 \). Combine like terms: \( 2.4 + 0.3y = 7.2 \). Subtract 2.4 from both sides: \( 0.3y = 4.8 \). Divide both sides by 0.3: \( y = 16 \).

Step 6: Solve for x

Now that we know \( y = 16 \), substitute back into \( x = 24 - y \): \( x = 24 - 16 \). Therefore, \( x = 8 \).

Step 7: Check Your Solution

Verify your results by checking the total volume and concentration: \( x + y = 8 + 16 = 24 \) pints, and \( 0.1 \times 8 + 0.4 \times 16 = 0.8 + 6.4 = 7.2 \) pints of antifreeze, which matches the required concentration.

Key Concepts

Mixture Problems

Mixture problems are common in algebra and involve combining different substances to achieve a new mixture with a specific property or concentration. In this context, these problems typically ask how much of each substance should be mixed to create a solution of a desired concentration. To solve them, you must:

• Identify the different components to be mixed.
• Determine the desired total amount and concentration of the mixture.
• Set up equations based on the relationships between the components.

In our antifreeze example, you have two solutions of different concentrations (10% and 40%) and want to create a 30% solution. The key to solving mixture problems is setting up an equation for the total amount and another for the concentration.

System of Equations

A system of equations is a set of two or more equations that are solved together, as they share variables. This approach is useful in mixture problems when you have two conditions to meet, such as total volume and concentration.
For the antifreeze problem, we set up the following system of equations:

• The total volume of the mixture: \[ x + y = 24 \]where \( x \) is the volume of the 10% solution and \( y \) is the volume of the 40% solution.
• The concentration condition:\[ 0.1x + 0.4y = 7.2 \]This equation represents the total amount of antifreeze, ensuring it equals 30% of the desired 24 pints.

Solving these equations involves substituting one equation into the other to find the values of \( x \) and \( y \). Using systems of equations is a powerful technique for solving problems with multiple unknowns.

Percentage Concentration

Percentage concentration is a way of expressing the amount of a substance in a mixture as a percentage of the total mixture. It is calculated by taking the amount of the substance, dividing it by the total amount of the mixture, and then multiplying by 100.
In the case of the antifreeze problem, the percentage concentration helps us understand how much pure antifreeze each solution contributes:

• The 10% solution contributes \( 0.1 \times x \) pints of pure antifreeze.
• The 40% solution contributes \( 0.4 \times y \) pints of pure antifreeze.

The goal is to combine these solutions to match the desired 30% concentration for the entire 24-pint mixture. Understanding percentage concentration is crucial for mixing substances in desired proportions, whether it be in mathematical problems or real-life applications.