Problem 58 - Algebra for College Students

Question

Solve each of Problems \(47-62\) by setting up. How many gallons of a \(12 \%\)-salt solution must be mixed with 6 gallons of a \(20 \%\)-salt solution to obtain a \(15 \%\)-salt solution?

Step-by-Step Solution

Answer

You need to mix 10 gallons of a 12% salt solution.

Step 1: Identify Unknowns

Let's identify what we need to find. We need to find the number of gallons of a 12% salt solution, which we'll call \( x \), to mix with 6 gallons of a 20% salt solution to get a 15% salt solution.

Step 2: Write the Salt Content Equation

To solve this, we need to consider the total amount of salt in the solutions. The equation is \[ 0.12x + 0.20 imes 6 = 0.15(x + 6) \]. This equation represents the balance needed between the existing salt solutions to achieve a 15% saline solution.

Step 3: Simplify the Equation

First, calculate the salt in 6 gallons of the 20% solution: \( 0.20 imes 6 = 1.2 \). Substitute into the equation: \[ 0.12x + 1.2 = 0.15(x + 6) \].

Step 4: Distribute and Simplify the Right Side

Distribute the 0.15 on the right side of the equation: \( 0.15x + 0.90 \). This gives us the equation \[ 0.12x + 1.2 = 0.15x + 0.90 \].

Step 5: Solve for x

Subtract \( 0.12x \) from both sides, yielding \( 1.2 = 0.03x + 0.90 \). Then subtract 0.90 from both sides to find \( 0.30 = 0.03x \). Divide both sides by 0.03 to solve for \( x \): \( x = 10 \).

Key Concepts

Mixture Problems

Mixture problems involve combining substances with different characteristics to obtain a mixture with a desired characteristic. It's a common type of algebra problem that requires us to think about how different components interact.
In this exercise, we mix liquids with varying percentages of salt to create a precise new mixture. The goal is to find how much of one component we need to add to achieve a specific overall concentration.

• Identify the individual components and their properties: Here, it's the 12% and 20% salt solutions.
• Determine what you need to find: The amount of 12% solution required to achieve a 15% solution.
• Use equations to represent the problem: These help us balance the mixture's properties effectively.

The key to mixture problems is setting up the right equation that connects the quantities and concentration levels of each part involved.

Percentages in Algebra

Percentages are a special way of representing fractions or parts of a whole. In algebra, percentages are crucial when working with problems involving proportions or concentrations, like in mixture problems.
Here, we express the concentrations of solutions in terms of percentages (e.g., 12%, 20%). To work with these mathematically, convert percentages into decimals (12% becomes 0.12) when formulating equations.
Understanding percentages allows us to grasp how much of one substance contributes to the whole, which is vital in:

• Setting up equations that represent real-world scenarios in mathematical terms.
• Determining how changes in one component's concentration affect the overall mixture.

Mastering percentages in algebra helps visualize and solve problems where measurements of concentration or proportion are involved.

Solving Linear Equations

Solving linear equations is a foundational algebraic skill. Linear equations involve unknown variables and constants, and our goal is to determine the value of these variables.
In this exercise, the unknown is the amount of the 12% solution, represented by the variable \( x \). We formulated an equation based on the salt content equations of all mixtures.
Steps to solve the linear equation include:

• Combining like terms on each side (simplifying the equation).
• Using arithmetic operations to isolate the variable (e.g., subtraction, division).
• Checking the solution by plugging it back into the original context to ensure accuracy.

Understanding how to solve these equations effectively allows us to find unknown quantities in various contexts, from simple mix problem scenarios to more complex applications.