Problem 59 - Algebra for College Students

Question

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% solution and 5 quarts of 70% solution.

Step 1: Define Variables

Let \( x \) represent the number of quarts of the 30% alcohol solution, and \( y \) represent the number of quarts of the 70% alcohol solution. We know that the total mixture should be 20 quarts and have 40% alcohol.

Step 2: Set Up Equations

We need two equations: one for the total volume and one for the alcohol content. \( x + y = 20 \) represents the total volume of the mixture. \( 0.3x + 0.7y = 0.4 \times 20 \) represents the alcohol content.

Step 3: Simplify Alcohol Content Equation

Simplify the alcohol content equation: \( 0.3x + 0.7y = 8 \) since \( 0.4 \times 20 = 8 \).

Step 4: Solve the System of Equations

We have the system of equations: \( x + y = 20 \) and \( 0.3x + 0.7y = 8 \). Use substitution or elimination to solve. First, solve for one variable in terms of the other, e.g., \( y = 20 - x \).

Step 5: Substitute and Solve for a Single Variable

Substitute \( y = 20 - x \) into \( 0.3x + 0.7y = 8 \), giving \( 0.3x + 0.7(20-x) = 8 \). Simplify: \( 0.3x + 14 - 0.7x = 8 \).

Step 6: Simplify and Solve for x

Combine the \( x \) terms: \(-0.4x + 14 = 8 \). Solve for \( x \): \(-0.4x = 8 - 14 \). Thus, \(-0.4x = -6 \), and \( x = 15 \).

Step 7: Solve for y

Substitute \( x = 15 \) back into \( y = 20 - x \) to find \( y = 20 - 15 \). Therefore, \( y = 5 \).

Step 8: Verify the Solution

Check that both equations hold true: \( x + y = 15 + 5 = 20 \) and \( 0.3(15) + 0.7(5) = 4.5 + 3.5 = 8 \). Both equations satisfy the condition.

Key Concepts

Understanding Solution Mixing Problems

Mixing problems involve combining different substances to create a mixture with specific characteristics. Here, the problem involves mixing two different alcohol solutions. One is a 30% alcohol solution, and the other is a 70% alcohol solution. The goal is to create 20 quarts of a new mixture that is 40% alcohol. To solve this, we need to figure out how many quarts of each of the original solutions to mix together.
For such problems, you usually need a mathematical approach to establish a relationship between the quantities of each component needed in the final mixture. This often involves setting up equations based on the total quantity and the concentration needed.

System of Linear Equations Explained

A system of linear equations is a collection of two or more equations with the same set of variables. In the alcohol mixing problem, we have two variables: the quantity of the 30% alcohol solution (\(x\)) and the quantity of the 70% alcohol solution (\(y\)).

The equations representing the problem are:

• \(x + y = 20\) represents the total amount of mixture (in quarts) needed.
• \(0.3x + 0.7y = 8\), which comes from the requirement that the final mixture is 40% alcohol, totaling 8 quarts of pure alcohol.

Solving these equations together (as a system) will give us the values of \(x\) and \(y\) that satisfy both conditions.

Examining Alcohol Solutions

In mixing problems, alcohol solutions are typically expressed in terms of percentages. These percentages tell us how much of the solution is made up of alcohol and how much is water or other substances. For example:

• A 30% alcohol solution means that 30% of the solution's volume is pure alcohol, and the rest is something else (usually water).
• A 70% alcohol solution similarly has 70% of its volume as alcohol.

Understanding these percentages is crucial because they guide how we set up equations. When mixing solutions, each type adds to the total volume of alcohol, so the math involved must ensure the final solution percentage meets the desired requirements.

The Substitution Method for Solving Equations

The substitution method is one of the ways to solve a system of linear equations like the one in our problem. The idea is to solve one of the equations for one variable and then "substitute" that expression into the other equation.

Here’s how it works in our example:

• First, we solve the simpler equation, \(x + y = 20\), to express \(y\) in terms of \(x\): \(y = 20 - x\).
• Then, substitute \(y\) in the alcohol content equation: \(0.3x + 0.7(20-x) = 8\).
• This substitution turns the problem into a single equation with one variable, which we can solve easily.

The substitution method is helpful because it reduces the complexity by eliminating one variable, simplifying the calculation process.