Problem 45 - Elementary Algebra

Question

A \(10 \%\) salt solution is to be mixed with a \(20 \%\) salt solution to produce 20 gallons of a \(17.5 \%\) salt solution. How many gallons of the \(10 \%\) solution and how many gallons of the \(20 \%\) solution will be needed?

Step-by-Step Solution

Answer

Mix 5 gallons of the 10% solution with 15 gallons of the 20% solution.

Step 1: Define Variables

Let \(x\) be the gallons of the \(10\%\) salt solution and \(y\) be the gallons of the \(20\%\) salt solution.

Step 2: Set Up Equations

We know the total volume of the new mixture is 20 gallons, so:\[x + y = 20\]The equation for the total salt content is:\[0.10x + 0.20y = 0.175 \times 20\]

Step 3: Simplify Salt Content Equation

Calculate the total amount of salt required for the \(17.5\%\) solution:\[0.10x + 0.20y = 3.5\] (because \(0.175 \times 20 = 3.5\)).

Step 4: Solve System of Equations

Substitute \(y = 20 - x\) from the first equation into the second:\[0.10x + 0.20(20 - x) = 3.5\]Simplifying results in:\[0.10x + 4 - 0.20x = 3.5\]

Step 5: Solve for x

Combine like terms in the equation:\[-0.10x = -0.5\]Solve for \(x\):\[x = 5\]

Step 6: Solve for y

Use the value of \(x\) in the equation \(y = 20 - x\):\[y = 20 - 5\]\[y = 15\]

Step 7: Verify Solution

Check that the solution adds up to 20 gallons and checks the salt content:\[0.10(5) + 0.20(15) = 0.5 + 3 = 3.5\]Both total volume and salt content match the requirements.

Key Concepts

System of Equations

In the context of mixture problems, systems of equations allow us to find unknown quantities by setting up multiple expressions that delineate the relationships between those quantities. For instance, when mixing salt solutions of different concentrations, we use equations to express both volume and salt content relationships.

Here's how it works for our exercise:

• We defined two variables: x for the gallons of the 10% solution and y for the gallons of the 20% solution.
• The first equation comes from the total volume requirement, giving us x + y = 20.
• The second equation emerges from the desired concentration. The salt content equation is based on the concentration percentage of each solution and the total salt percentage in the final mixture: 0.10x + 0.20y = 3.5, where 3.5 gallons of salt is required in total.

These equations form a system that can be solved using substitution or elimination to find the volumes of each solution needed.

Percent Concentration

Percent concentration is a measure that indicates the amount of solute (in this case, salt) present in a solution relative to the total amount of solution. This metric is crucial in preparing solutions in specified concentrations.

• The 10% salt solution means 10 parts of salt in every 100 parts of the solution.
• The 20% solution has 20 parts of salt per 100 parts of solution.

In the exercise, we're looking to achieve a 17.5% concentration in a 20-gallon mixture. To find how much each solution contributes to the total salt content, we multiply the concentration percentage by the volume of each solution, aligning this with our final target concentration of 17.5%. This cumulative percentage of salt determines the balance needed between the two solutions.

Solution Mixtures

Solution mixtures involve combining two or more solutions to achieve a specific characteristic, like concentration or volume. These mixtures are common in chemistry when creating dilutions or mixing substances.

• In our problem, the goal is to produce a mixture that is precisely 17.5% salt by combining two solutions with known percentages.
• The desired mixture volume is 20 gallons, which must include the precise percentage of salt.

To achieve the specific concentration, we must consider both the amount and concentration of each component. By finding the right balance (as solved in the system of equations), we confirm that the resulting mixture meets both the specified volume and concentration of salt. This balancing act of combining different solutions demonstrates the practical application of algebraic equations in real-world scenarios.