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Chapter 4: Problem 44

A \(10 \%\) acid solution is to be mixed with a \(50 \%\) acid solution in order to get 120 ounces of a \(20 \%\) acid solution. How many ounces of the \(10 \%\) solution and \(50 \%\) solution should be mixed?

Short Answer

Expert verified
Mix 90 ounces of the 10% solution with 30 ounces of the 50% solution.

Step by step solution

01

Define the Variables

Let \( x \) be the number of ounces of the \(10\%\) solution, and \( y \) be the number of ounces of the \(50\%\) solution. We are tasked to find the values of \( x \) and \( y \).
02

Set Up the System of Equations

We know that the total solution should be 120 ounces, which gives us the first equation: \( x + y = 120 \). Additionally, the concentration equation for the acid content is \( 0.10x + 0.50y = 0.20 \times 120 \) since the resulting solution should have a \(20\%\) concentration.
03

Simplify the Second Equation

Calculate \(0.20 \times 120\), which equals 24. Thus, the second equation becomes \( 0.10x + 0.50y = 24 \).
04

Solve the System of Equations

Start by solving the first equation for one of the variables. Let \( y = 120 - x \), then substitute this into the second equation: \( 0.10x + 0.50(120 - x) = 24 \).
05

Substitute and Simplify

Expand the substituted equation to \(0.10x + 60 - 0.50x = 24\). Further simplify to \( -0.40x + 60 = 24 \).
06

Isolate \( x \)

Subtract 60 from both sides: \(-0.40x = -36\). Then divide both sides by \(-0.40\) to find \( x = 90 \).
07

Find \( y \)

Use the equation \( y = 120 - x \), substituting \( x = 90 \) to find \( y = 30 \).
08

Verify the Solution

Check that \( 0.10 \times 90 + 0.50 \times 30 = 24 \). Calculate \(9 + 15 = 24\), which verifies the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Equations
In algebra, a system of equations is a collection of two or more equations with a set of variables. To find the values of these variables, the equations are solved simultaneously, meaning each equation is considered at the same time. By doing so, you can find a common solution that satisfies all equations in the system.
  • Basic Concept: A system of equations looks like this:
    • The first equation could be: \( ax + by = c \)
    • The second equation could be: \( dx + ey = f \)
  • Goal: Find the values of \(x\) and \(y\) that make both equations true at the same time.
  • Methods: The two most common methods to solve these systems are:
    • Substitution: Solve one equation for one variable and substitute this expression into the other equation.
    • Elimination: Add or subtract equations to eliminate one variable, making it possible to solve for the other.
In the acid mixture problem, we use the substitution method. First, we solve for \(y\) in terms of \(x\) from \( x + y = 120 \). Then, substitute \(y\) in the second equation to find \(x\). This approach is clear and helps us reach the solution efficiently.
Mixture Problems
Mixture problems are a type of word problem commonly found in algebra. They involve mixing two or more substances to achieve a desired composition or concentration. The trick is to understand and use the concept of balancing the quantities and properties involved, like concentration in our problem.
  • Identifying Parts: Recognize what you're mixing. For example, in the acid solution problem, you have a \(10\%\) solution and a \(50\%\) solution being mixed.
  • Total Quantity: Decide the total amount for the final mixture. Here, it is 120 ounces of a \(20\%\) solution.
  • Form Equations: Mixture problems typically involve forming two types of equations:
    • Total quantity equation, like \(x + y = 120\).
    • Concentration equation, like \(0.10x + 0.50y = 0.20 \times 120\).
  • Solving the Problem: Use algebraic methods to solve the system, determining the amount of each substance needed.
Mixture problems often seem complex at first glance, but breaking them down into parts can make them easier to handle. It's all about setting up the right equations and solving them methodically.
Percentage Concentration
Percentage concentration is a way of expressing how much of one component is present in a solution. It is often used in chemistry and involves calculations based on the ratio of the solute (substance being dissolved) to the total solution.
  • Basic Formula:
    • The concentration of a solution is calculated as \( \left( \frac{\text{Amount of Solute}}{\text{Total Solution}} \right) \times 100\% \).
  • Understanding Percentage: For example, a \(10\%\) acid solution means that 10% of the total weight or volume is acid.
  • Application in Problems: In mixture problems, knowing the percentage helps set up equations. It helps you determine how much of each component is needed.
    • For a \(20\%\) final solution, you need a balance such that the mixture of \(10\%\) and \(50\%\) solutions achieves this specific concentration.
Understanding percentage concentration helps in not only calculating the right mixture but also in ensuring that solutions meet required standards, especially in practical scenarios like manufacturing or cooking. In algebra problems, it helps in structuring accurate equations and solving for unknowns.

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