Problem 48 - Precalculus: Mathematical for Calculus

Question

A chemist has three acid solutions at various concentrations. The first is \(10 \%\) acid, the second is \(20 \%\) and the third is 40\%. How many milliliters of each should he use to make \(100 \mathrm{mL}\) of \(18 \%\) solution, if he has to use four times as much of the \(10 \%\) solution as the \(40 \%\) solution?

Step-by-Step Solution

Answer

Use 40 mL of 10%, 50 mL of 20%, and 10 mL of 40% solutions.

Step 1: Define Variables

Let \( x \) be the milliliters of the 40% acid solution. Then, according to the problem, the 10% solution should be four times this amount, or \( 4x \). Let \( y \) be the milliliters of the 20% acid solution.

Step 2: Set Up Total Volume Equation

The total volume of the solution is given as 100 mL. Thus, the equation to represent the total volume is: \[ 4x + y + x = 100 \] Simplifying gives: \[ 5x + y = 100 \]

Step 3: Set Up Acid Concentration Equation

The total acid in the final solution is 18% of 100 mL, which is \( 0.18 \times 100 = 18 \) mL. Therefore, the equation for the acid concentration is:\[ 0.10(4x) + 0.20y + 0.40x = 18 \]

Step 4: Simplify the Acid Concentration Equation

Substitute and simplify the equation from Step 3:\[ 0.4x + 0.20y + 0.40x = 18 \] \[ 0.8x + 0.20y = 18 \]

Step 5: Solve the System of Equations

We now solve the system of equations:1. \( 5x + y = 100 \)2. \( 0.8x + 0.20y = 18 \)Multiplying the second equation by 5 to remove decimals:\[ 4x + y = 90 \]Subtract the simplified second equation from the first equation:\( 5x + y - 4x - y = 100 - 90 \)\[ x = 10 \]

Step 6: Find y using x value

Substitute \( x = 10 \) back into the equation \( 5x + y = 100 \):\[ 5(10) + y = 100 \]\[ 50 + y = 100 \]\[ y = 50 \]

Step 7: Calculate the 10% Solution Volume

Using \( x = 10 \), find the amount of 10% solution:\[ 4x = 4(10) = 40 \] mL.

Step 8: Verify the Solution

Check that the solution meets all conditions: 1. Total volume equals 100 mL: \( 40 + 50 + 10 = 100 \).2. Total acid content is correct: \( 0.10(40) + 0.20(50) + 0.40(10) = 4 + 10 + 4 = 18 \) mL.

Key Concepts

Acid Solution Concentration

When mixing acid solutions, the concentration refers to the percentage of acid in each solution. In the exercise, we have three different acid concentrations: 10%, 20%, and 40%. Each percentage tells us how much pure acid is contained in 100 mL of the solution. For instance, a 10% solution means there are 10 milliliters of acid in every 100 milliliters of the solution.
This problem requires us to blend different concentrations to reach a desired overall percentage, in this case, 18%. To do this correctly, we need to consider the volume of acid each solution adds to the mixture. This ensures the final concoction has the correct balance of acids to reach the intended concentration. Combining solutions involves calculating the exact contributions from each concentration. This is usually done by setting up an equation where the sum of all individual acid contributions equals the total acid required in the final mixture. Clear understanding of each solution’s concentration helps in setting up these equations accurately, allowing us to verify that the mixture meets the desired concentration.

Mixture Problems

Mixture problems are a common type of word problem in algebra. They involve combining different substances to achieve a final mixture with specific properties. In this exercise, we are tasked with mixing three distinct acid solutions to create a final mixture with a target concentration of 18% acid. Key aspects to consider when solving mixture problems:

• Identifying the different mixtures or solutions with varying concentrations.
• Understanding the relationships or specific conditions given, such as using four times as much of the 10% solution as the 40% solution.
• Setting up equations that express the total volume and concentration of the mixture.

Using logical steps and algebra, we can solve these equations to find the volume of each solution required. This leads to a balanced mix that fulfills the conditions and requirements specified in the problem.

Variable Definition in Equations

In algebra, defining variables is the first step in solving equations efficiently. It simplifies the problem by giving a symbolic representation to unknown quantities. In this problem, we defined:
- Let \( x \) be the amount of the 40% acid solution. - The 10% solution is four times this amount, denoted as \( 4x \). - Let \( y \) be the amount of the 20% solution. These definitions help translate word problems into mathematical equations that can be solved systematically. Establishing these variables means we can express the total volume and concentration parameters analytically as equations.
The power of variable definition is that it not only simplifies the initial setup but also helps ensure our calculations remain organized. By following these definitions and solving the equations, we find the necessary values of each variable, ensuring the final mixture meets all required conditions.