Problem 54 - Intermediate Algebra

Question

Suppose that one solution contains \(50 \%\) alcohol and another solution contains \(80 \%\) alcohol. How many liters of each solution should be mixed to make \(10.5\) liters of a \(70 \%\)-alcohol solution?

Step-by-Step Solution

Answer

Mix 3.5 liters of 50% alcohol with 7 liters of 80% alcohol.

Step 1: Define Variables

Let the volume of the 50% alcohol solution be \(x\) liters and that of the 80% alcohol solution be \(y\) liters. Then we have two unknowns to find: \(x\) and \(y\).

Step 2: Set Up the Volume Equation

We know that the total volume of the mixture is 10.5 liters. Therefore, \(x + y = 10.5\).

Step 3: Set Up the Alcohol Content Equation

We need the mixture to have 70% alcohol by volume. So, the equation for the alcohol content will be \(0.50x + 0.80y = 0.70 \times 10.5\).

Step 4: Simplify the Alcohol Content Equation

Calculate \(0.70 \times 10.5 = 7.35\). Thus, the equation becomes \(0.50x + 0.80y = 7.35\).

Step 5: Solve the System of Equations

We have the system:1. \(x + y = 10.5\)2. \(0.50x + 0.80y = 7.35\).First, solve for \(x\) in terms of \(y\) from the first equation: \(x = 10.5 - y\). Substitute into the second equation: \[0.50(10.5 - y) + 0.80y = 7.35\].

Step 6: Simplify and Solve for y

Expand and simplify the equation: \[5.25 - 0.50y + 0.80y = 7.35\]\[5.25 + 0.30y = 7.35\]Subtract 5.25 from both sides to isolate \(y\): \[0.30y = 2.10\]Divide both sides by 0.30: \[y = 7\] liters.

Step 7: Find x Using y

Substitute \(y = 7\) into the equation \(x = 10.5 - y\): \[x = 10.5 - 7 = 3.5\] liters.

Key Concepts

Algebraic Equations

Algebraic equations play a crucial role in solving mixture problems. Mixture problems involve combining substances with different properties to create a mixture with a desired attribute, like in our exercise where different alcohol solutions are mixed. In algebra, equations are mathematical statements asserting the equality of two expressions given unknown variables.

To solve these problems, we assign variables to unknown quantities, such as the volume of each solution in our example. By doing this, we set the stage to create equations based on given conditions.

• The assignment of variables helps in systematically approaching the problem.
• Establishing relationships between the variables and the desired outcome is crucial.
• Equations are used to model the given conditions and constraints.

In our exercise, we defined two variables:

• \(x\), representing the volume of the 50% alcohol solution.
• \(y\), representing the volume of the 80% alcohol solution.

The next step involved creating the equations based on the problem requirements. Efficiently setting up these equations is a foundational skill in algebra and necessary to solve mixture problems effectively.

Solution Concentration

Solution concentration refers to the amount of solute present in a solution relative to the total amount of solution. In chemistry and related fields, concentration is usually expressed in percentage terms or other forms like molarity.

In our specific problem, the solution concentration indicates the percentage of alcohol in each solution involved.

• The 50% solution has 50 milliliters of alcohol per 100 milliliters of solution.
• The 80% solution has 80 milliliters of alcohol per 100 milliliters of solution.

Understanding concentration is vital to set up equations that account for the alcohol content in the mixed solution. When discussing concentration in algebraic terms, we use the equation:\[ C_1V_1 + C_2V_2 = C_M V_M \]Where

• \(C_1\) and \(C_2\) are the concentrations of the initial solutions.
• \(V_1\) and \(V_2\) are their respective volumes.
• \(C_M\) and \(V_M\) represent the concentration and volume of the mixture.

Using this formula helps ensure the new solution has the desired concentration, like achieving 70% in the combined solution.

Systems of Equations

Systems of equations are sets of two or more equations with multiple variables. These systems are crucial in solving problems involving more than one unknown. They allow us to model and solve problems where it's necessary to find multiple values that simultaneously satisfy all given equations.

In our example, we set up a system of equations to find out how much of each alcohol solution is needed. The system consisted of:

• The volume equation: \(x + y = 10.5\)
• The alcohol content equation: \(0.50x + 0.80y = 7.35\)

Dealing with systems of equations involves:

• Expressing one variable in terms of another, using methods like substitution or elimination.
• Substituting this expression into another equation to find one variable's value.
• Using the found value to solve for the other variable.

In our solution, we used substitution to find \(y\) first, and then plugged \(y\) back into the other equation to find \(x\). This step-by-step approach is effective for solving systems of equations, especially when both equations are linear. Mastering these techniques equips students with powerful tools for tackling complex algebraic problems and mixture equations alike.