Problem 48 - Algebra and Trigonometry

Question

Mixtures 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

40 mL of 10\%, 50 mL of 20\%, and 10 mL of 40\% solutions are needed.

Step 1: Define variables

Let \( x \) be the volume of the 10\(\%\) solution in milliliters, \( y \) be the volume of the 20\(\%\) solution, and \( z \) be the volume of the 40\(\%\) solution.

Step 2: Establish equations based on total volume

Since the total volume of the solution should be 100 \(\text{mL} \), we write the equation: \[ x + y + z = 100 \].

Step 3: Establish equations based on acid concentration

The total acid from each must add up to 18\(\%\) of 100 \(\text{mL} \), or 18 \(\text{mL of pure acid} \). Thus, the equation based on concentration is: \[ 0.1x + 0.2y + 0.4z = 18 \].

Step 4: Use condition for solution volumes

According to the problem, the chemist uses four times as much of the 10\(\%\) solution as the 40\(\%\) solution, giving another equation: \[ x = 4z \].

Step 5: Substitute and solve equations

Substitute \( x = 4z \) in the first two equations. \[ 4z + y + z = 100 \Rightarrow y + 5z = 100 \] \[ 0.1(4z) + 0.2y + 0.4z = 18 \Rightarrow 0.4z + 0.2y + 0.4z = 18 \Rightarrow 0.8z + 0.2y = 18 \].

Step 6: Solve system of equations

From \( y + 5z = 100 \) we get \( y = 100 - 5z \). Substitute in the second equation: \[ 0.8z + 0.2(100 - 5z) = 18 \] \[ 0.8z + 20 - z = 18 \] \[ -0.2z + 20 = 18 \] \[ -0.2z = -2 \] \[ z = 10 \].

Step 7: Find x and y based on z

Substitute \( z = 10 \) back to find \( x \) and \( y \). \( x = 4z = 40 \) and \( y = 100 - 5z = 100 - 50 = 50 \). Thus, \( x = 40 \), \( y = 50 \), \( z = 10 \).

Key Concepts

Concentration Equations

When dealing with mixtures, concentration equations help us understand how a certain percentage of an ingredient (in this case, acid) is distributed in a solution. If you have a 10\(\%\) acid solution, it means that in every milliliter of solution, 10\(\%\) of it is pure acid. Similarly, for 20\(\%\) and 40\(\%\), you find that proportion of acid in the solutions. In our problem, we want a total of 18 mL of pure acid in a 100 mL solution. This is expressed mathematically as:- \(0.1x + 0.2y + 0.4z = 18\), where \(x\), \(y\), and \(z\) represent the milliliters of 10\(\%\), 20\(\%\), and 40\(\%\) acid solutions respectively. By setting up an equation like this, we lay the foundation for finding how much of each solution is needed.

System of Equations

A system of equations is a set of two or more equations that share variables. Solving a system means finding the values for these variables that satisfy all the equations in the system.In this problem:- We have three equations involving three unknowns. These represent different conditions: - Total volume: \(x + y + z = 100\) - Acid amount based on concentration: \(0.1x + 0.2y + 0.4z = 18\) - Relationship between 10\(\%\) and 40\(\%\) solutions: \(x = 4z\)By solving this system, we can determine exactly how many milliliters of each acid solution are needed to create the desired mixture.

Variables Substitution

Variables substitution is a powerful technique used to simplify solving systems of equations.In our exercise, we have:- \(x = 4z\) from the given condition about the ratios of solutions. This allows us to replace \(x\) with \(4z\) in our other equations. For instance:- The total volume equation becomes \(4z + y + z = 100\), simplifying to \(y + 5z = 100\).By substituting, we reduce the number of variables, making it easier to solve for the remaining ones.

Acid Concentration Calculation

Calculating the specific concentrations of acids in a final solution is the goal of our initial problem. Let's see how we apply our solved variables.We know:- \(x = 40\)- \(y = 50\)- \(z = 10\)Checking our conditions:- The first equation ensures the volumes sum up correctly: \(x + y + z = 40 + 50 + 10 = 100\).- The second checks acid concentration: \(0.1(40) + 0.2(50) + 0.4(10) = 4 + 10 + 4 = 18\).Thus, these calculations confirm that our solutions were placed in the correct amounts to achieve the desired concentration.