Problem 59 - Algebra for College Students

Question

Solve each of Problems \(47-62\) by setting up. Suppose that you have a supply of a \(30 \%\) solution of alcohol and a \(70 \%\) solution of alcohol. How many quarts of each should be mixed to produce 20 quarts that is \(40 \%\) alcohol?

Step-by-Step Solution

Answer

Mix 15 quarts of 30% alcohol solution with 5 quarts of 70% alcohol solution.

Step 1: Define Variables

Let \( x \) be the amount of the \( 30\% \) alcohol solution and \( y \) be the amount of the \( 70\% \) alcohol solution. We know that the total volume of the solution is 20 quarts, so we can write the equation: \( x + y = 20 \).

Step 2: Set Up Alcohol Concentration Equation

The total amount of pure alcohol in the mixture should be \( 0.40 \times 20 = 8 \) quarts. The amount of pure alcohol from the \( 30\% \) solution is \( 0.30x \), and from the \( 70\% \) solution is \( 0.70y \). So we write: \( 0.30x + 0.70y = 8 \).

Step 3: Solve the System of Equations

We now have the system of equations: 1. \( x + y = 20 \) 2. \( 0.30x + 0.70y = 8 \)We can solve the first equation for \( y \): \( y = 20 - x \)Substitute \( y = 20 - x \) into the second equation:\( 0.30x + 0.70(20 - x) = 8 \)

Step 4: Simplify and Solve for x

Simplify \( 0.30x + 0.70(20 - x) = 8 \): \( 0.30x + 14 - 0.70x = 8 \)Combine like terms: \( -0.40x + 14 = 8 \) Subtract 14 from both sides: \( -0.40x = -6 \) Divide both sides by \( -0.40 \): \( x = 15 \).

Step 5: Find y Using x

Now that we know \( x = 15 \), substitute back into the equation \( y = 20 - x \): \( y = 20 - 15 = 5 \).

Key Concepts

Systems of Equations

A system of equations is a set of equations with multiple variables that are solved simultaneously. In algebra, solving these equations aims to find the values of the variables that satisfy all the equations in the system.
In mixture problems like the one given, systems of equations help you determine how much of each component to mix to achieve a desired outcome.

To understand how systems of equations work in this context, consider the two equations we formed:

• The first equation represents the total amount of the solution: \( x + y = 20 \) quarts.
• The second equation accounts for the concentration of alcohol within the mixture: \( 0.30x + 0.70y = 8 \) quarts of pure alcohol.

By solving the system, you're simultaneously finding the solution quantities \( x \) and \( y \) that make both scenarios true. The first step often involves substituting one equation into the other, allowing you to solve for a single variable more efficiently.

Alcohol Concentration

Alcohol concentration in a solution is the percentage of alcohol relative to the total volume. In mixture problems, this determines how various solutions can be combined to reach a specific concentration target.
One must calculate the amount of pure alcohol and how it mixes together to achieve the desired concentration.

In the problem, you're mixing two solutions:

• A 30% alcohol solution contributes \( 0.30x \) quarts of pure alcohol.
• A 70% alcohol solution contributes \( 0.70y \) quarts of pure alcohol.

When mixed together, these result in an overall volume with a 40% alcohol concentration, equating to 8 quarts of pure alcohol in the final mix (as calculated by \( 0.40 \times 20 \)).
This part of the problem requires careful calculation to ensure you achieve the desired strength efficiently. By setting up the corresponding alcohol concentration equation, you're using algebra to model and solve for the necessary solution quantities.

Solution Types

In algebra, mixture problems typically deal with different 'solutions' – not just the answers to the equation, but also the combinations of ingredients or components.
In this specific problem, the term "solution" refers to the liquid mixtures, each with a distinct alcohol concentration. These are often characterized by their specific percentage of a component, like alcohol, that distinguishes them from pure ingredients.

Different solution types in this context include:

• The 30% alcohol solution with the variable \( x \), which is typically more diluted.
• The 70% alcohol solution with the variable \( y \), which is more concentrated.

Selecting how much of each solution to mix depends on the desired characteristics of the final product. In this case, you want a final mix that results in 20 quarts with a 40% alcohol concentration. Such mixture problems illustrate how varying types and proportions contribute to the compound solution's final properties.