Problem 41 - College Algebra and Calculus: An Applied Approach

Question

Acid Mixture Ten gallons of a \(30 \%\) acid solution is obtained by mixing a \(20 \%\) solution with a \(50 \%\) solution. How much of each solution is required to obtain the specified concentration of the final mixture?

Step-by-Step Solution

Answer

In order to create 10 gallons of a 30% acid solution, 6 gallons of the 20% acid solution and 4 gallons of the 50% acid solution are required.

Step 1: Understand the Problem and Define Variables

Let's define \( x \) as the gallons of acid solution with 20% concentration and \( y \) as the gallons of the acid solution with 50% concentration. We are given that the total volume of the mixture is 10 gallons. So we can write this as an equation: \( x + y = 10 \) gallons. This is the equation for the volume of the mixture. We are also given that the final acid concentration is 30%. We infer that 30% of the 10 gallons is pure acid, giving 3 gallons.

Step 2: Set Up Equations

The second equation is derived from considering how much pure acid is in the final mixture. It comes from 20% of \( x \) (the amount of pure acid in the 20% solution), and 50% of \( y \) (the amount of pure acid in the 50% solution). Adding these amounts gives the total amount of pure acid, which is 30% of the final 10-gallon mixture. Hence, the second equation is: \( 0.20x + 0.50y = 3 \) gallons.

Step 3: Solve the System of Equations

We now have a system of two equations, which can be solved simultaneously: \( x + y = 10 \) and \( 0.20x + 0.50y = 3 \). Multiplying the second equation by 10 gives \( 2x + 5y = 30 \). Now, we can solve for \( x \) from the first equation as \( x = 10 - y \), and substitute it into the adjusted second equation. Doing so gives \( y = 4 \). Substituting \( y \) back into the first equation gives \( x = 6 \)

Step 4: Interpret the Results

The solution for the system of equations means that 6 gallons of 20% acid solution and 4 gallons of 50% acid solution are needed to produce 10 gallons of 30% acid solution.

Key Concepts

Mixture Problems

Mixture problems can be found in many real-world scenarios, like chemistry and cooking. Here, the goal is to create a mixture with a specific composition using two or more substances with different properties. In our exercise, we need to mix an acid solution to achieve a desired concentration.

To solve mixture problems step-by-step, you should:

• Identify the components and their concentrations or quantities.
• Set up equations based on the total quantity and the desired mixture's properties.
• Use algebraic methods to find the unknowns, which represent how much of each component you need.

Understanding how mixture problems work not only helps in solving mathematical equations but also in understanding practical situations where such calculations are essential.

Algebra

Algebra is like a language of mathematics that uses symbols and letters to represent numbers and quantities in equations. It's incredibly powerful for solving problems systematically.

In the context of our acid mixture exercise:

• We used variables, like \( x \) and \( y \), to represent unknown quantities.
• We created equations to model the relationships, such as \( x + y = 10 \).
• Algebra allows for manipulation of these equations to isolate and solve for variables.

Getting comfortable with algebra means you'll be able to solve a wide variety of problems by structuring them and breaking down complex problems into manageable parts. It's a fundamental skill in mathematics, offering solutions that apply to many fields.

Linear Equations

Linear equations form the backbone of solving systems with simple relationships between variables. They appear in the form \( ax + by = c \), where \(a\), \(b\), and \(c\) are constants.

When working with linear equations:

• They can be combined to form systems, like the ones seen in our problem: \( x + y = 10 \) and \( 0.20x + 0.50y = 3 \).
• These systems are solved using techniques such as substitution or elimination.
• The solutions to these equations tell us exactly the values of the unknowns that satisfy all the conditions.

Linear equations are used extensively not just in academics, but in fields like economics and engineering, where understanding relationships between variables is crucial. Mastering them is key to advancing in mathematical problem-solving.