Problem 68 - Algebra and Trigonometry

Question

Mixture Problem A biologist has two brine solutions, one containing 5\(\%\) salt and another containing 20\(\%\) salt. How many milliters of each solution should she mix to obtain 1 L of a solution that contains 14\(\%\) salt?

Step-by-Step Solution

Answer

Mix 400 mL of 5\(\%\) solution with 600 mL of 20\(\%\) solution.

Step 1: Define the Variables

Let \(x\) represent the amount of 5\(\%\) salt solution in milliliters, and \(y\) represent the amount of 20\(\%\) salt solution in milliliters. We need to find the values of \(x\) and \(y\).

Step 2: Set Up the Equation for Total Volume

The total volume of the solution should be 1000 milliliters (since 1 L = 1000 mL). Therefore, we have the equation: \[x + y = 1000.\]

Step 3: Set Up the Equation for Salt Concentration

To achieve a solution with 14\(\%\) salt, the combined contribution of salt from both solutions should be 14\(\%\) of the 1000 mL, which is 140 mL of salt. Thus, we have another equation: \[0.05x + 0.20y = 140.\]

Step 4: Solve the System of Equations

We have the system of equations: \[x + y = 1000\] and \[0.05x + 0.20y = 140.\]Solve the first equation for \(y\): \[y = 1000 - x\].Substitute \(y\) in the second equation: \[0.05x + 0.20(1000 - x) = 140.\]

Step 5: Simplify and Solve for x

Calculate \(0.20(1000 - x)\) which equals \(200 - 0.20x\).The equation becomes \[0.05x + 200 - 0.20x = 140.\]Simplify to \[-0.15x + 200 = 140.\]Rearrange to find \(-0.15x = -60\) giving \(x = 400\).

Step 6: Determine the Value of y

Substitute \(x = 400\) back into the equation \(y = 1000 - x\): \[y = 1000 - 400 = 600.\]

Step 7: Verify the Solution

Plug \(x = 400\) and \(y = 600\) back into the concentration equation: \[0.05(400) + 0.20(600) = 20 + 120 = 140\].The solution satisfies both the volume and salt concentration requirements.

Key Concepts

Algebraic Equations

In a mixture problem, formulating algebraic equations is vital as it helps in defining and solving the problem systematically. To tackle these problems, one must first identify knowns and unknowns by defining variables.

• In our biologist scenario, let's assign the variable \( x \) to the volume of the 5\(\%\) salt solution in milliliters and \( y \) to the 20\(\%\) salt solution. These variables will represent the unknown quantities we aim to determine.
• The next step is to translate the problem's conditions into algebraic equations. We express the total mixture volume requirement and its salt concentration as two separate equations.
• This structured approach allows for the logical construction of a system of equations that models the situation accurately.

Understanding these equations helps connect the word problem to mathematical expressions, providing a clear route to resolution.

Percent Concentration

Percent concentration plays a pivotal role in mixture problems as it conveys the amount of solute (in this case, salt) relative to the solution's total volume.

• For the problem at hand, each solution has a specific percent concentration: the first solution contains 5\(\%\) salt and the second contains 20\(\%\), which affects how we calculate their contributions to the final mixture.
• If you mix these solutions, the aim is to get a final mixture with a 14\(\%\) salt concentration. This requires precise calculations, as the percent concentration of a solution refers to the ratio of solute volume to total solution volume.
• Using the formula for the concentration of a solution adds another equation reflecting the mixture's solute requirement. For example, 140 mL of salt must be obtained from the mixture to achieve a 14\(\%\) concentration within 1000 mL.

A solid grasp of percent concentration can guide you through the arithmetic needed to find the desired mix ratios.

System of Equations

Solving systems of equations is a powerful approach in mixture problems, especially when dealing with multiple requirements or constraints. In the example, we have two main requirements: total volume and desired salt concentration.

• The first equation comes from the condition that the total solution must equal 1000 mL, thus \( x + y = 1000 \).
• The second equation hinges on the salt concentration condition: \(0.05x + 0.20y = 140 \), representing the total mL of salt.
• By solving this system of linear equations, we find values for \( x \) and \( y \) that satisfy both conditions simultaneously. This is often achieved through substitution or elimination methods.

Techniques like these allow for precise solutions, ensuring the mixture requirements are perfectly balanced. Understanding and applying systems of equations are crucial for effectively solving mixture problems and similar algebraic challenges.