Problem 24
Question
Solve. For each exercise, a table is given for you to complete and use to write an equation that models the situation. How many cubic centimeters \((\mathrm{cc})\) of a \(25 \%\) antibiotic solution should be added to 10 cubic centimeters of a \(60 \%\) antibiotic solution in order to get a \(30 \%\) antibiotic solution?
Step-by-Step Solution
Verified Answer
Add 60 cc of the 25% solution.
1Step 1: Define Variables
Let's define the variables for the problem:- Let \( x \) be the number of cubic centimeters of the 25% solution added.- We already have 10 cubic centimeters of the 60% solution.- We want a 30% solution of the combined volumes.
2Step 2: Set Up the Equation
The total amount of antibiotic in the mixed solution can be expressed. For the 25% solution, the antibiotic amount is \( 0.25x \), and for the 60% solution, it is \( 0.6 \times 10 = 6 \). We want the final mixture to have a 30% concentration, so the equation is:\[0.25x + 6 = 0.30(x + 10).\]
3Step 3: Simplify the Equation
Distribute the 30% concentration across the combined solution volume:\[0.25x + 6 = 0.3x + 3.\]
4Step 4: Solve for x
Rearrange the equation to solve for \( x \):1. Subtract \( 0.25x \) from both sides: \[ 6 = 0.05x + 3. \]2. Subtract 3 from both sides: \[ 3 = 0.05x. \]3. Divide both sides by 0.05: \[ x = \frac{3}{0.05} = 60. \]
5Step 5: Conclusion
We find that 60 cubic centimeters of the 25% antibiotic solution should be added to achieve the desired 30% concentration for the solution.
Key Concepts
Mixture ProblemsLinear EquationsPercentages
Mixture Problems
Mixture problems in algebra involve combining different substances with varying concentrations to achieve a desired mixture. These types of problems are common across various disciplines, including chemistry and pharmacy. They generally require a good understanding of proportions and percentages and the ability to manipulate algebraic expressions.
In our exercise, we are tasked with finding how much of a weaker solution needs to be added to a stronger solution to achieve a specific concentration. Here, the key is to set up the relationship between different parts of the solution, usually requiring an equation that represents the total concentration of the resulting mixture.
In our exercise, we are tasked with finding how much of a weaker solution needs to be added to a stronger solution to achieve a specific concentration. Here, the key is to set up the relationship between different parts of the solution, usually requiring an equation that represents the total concentration of the resulting mixture.
- The known variables are the amounts and concentrations of the original solutions.
- The unknown variable is the amount of the weaker solution to be added, represented in terms of the desired concentration.
Linear Equations
Linear equations are algebraic expressions that involve variables raised to the first power and have the general form \( ax + b = c \). They are crucial in solving mixture problems because they allow you to find unknown quantities through balancing equations.
In our exercise, the different concentrations and volumes of the solutions are combined to form a linear equation. The steps to solve this equation usually involve:
In our exercise, the different concentrations and volumes of the solutions are combined to form a linear equation. The steps to solve this equation usually involve:
- Identifying terms on both sides of the equation that represent the different components of the solutions.
- Simplifying the equation into a standard linear form.
- Finally, solving for the unknown variable using basic algebraic techniques such as addition, subtraction, multiplication, and division.
Percentages
Percentages are a way to express parts of a whole, commonly used in mixture problems to describe concentrations. The concentration of solutions is usually given in percentages, representing the part of the solution that contains the active substance.
In any mixture problem, understanding how to convert percentages into their decimal form is crucial. For instance, 25% becomes 0.25, 60% becomes 0.6, etc. This conversion allows you to set up the equation where the total active ingredient is balanced against the desired concentration of the entire mixture.
In any mixture problem, understanding how to convert percentages into their decimal form is crucial. For instance, 25% becomes 0.25, 60% becomes 0.6, etc. This conversion allows you to set up the equation where the total active ingredient is balanced against the desired concentration of the entire mixture.
- Each part of the solution contributes a specific percentage of the active ingredient.
- The aim is to adjust these contributions to reach a target percentage of the mixture.
Other exercises in this chapter
Problem 24
Solve each inequality. Graph the solution set. $$ -5 x
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Solve each equation. See Examples 3 through \(5 .\) $$ \frac{3(y+3)}{5}=2 y+6 $$
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Solve each equation. Don't forget to first simplify each side of the equation, if possible. Check each solution. See Examples 5 through 7 . $$ 2 x+7=x-10 $$
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Solve each equation. Check each solution. See Examples 7 and 8 . \(\frac{b}{4}-1=-7\)
View solution