10.3.a – Mixture Problems | Developmental Math Emporium

Learning Outcomes

Solve mixture problems

Write a system of linear equations representing a mixture problem, solve the system and interpret the results

One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. There are many other disciplines that use solutions as well.

The concentration or strength of a liquid solution is often described as a percentage. This number comes from the ratio of how much mass is in a specific volume of liquid. For example if you have [latex]50[/latex] grams of salt in a [latex]100mL[/latex] of water you have a [latex]50\%[/latex] salt solution based on the following ratio:

[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]

Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.

We will use the following table to help us solve mixture problems:

	Amount	Concentration (%)	Total
Solution 1
Solution 2
Final Solution

To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is [latex]120mL[/latex] of a [latex]9\%[/latex] solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.

First, find the total mass of solids for each solution by multiplying the volume by the concentration.

	Amount	Concentration (%)	Total Mass
Solution 1	[latex]120 mL[/latex]	0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]	[latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex]
Solution 2	[latex]75 mL[/latex]	0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]	[latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex]
Final Solution

Next we add the new volumes and new masses.

	Amount	Concentration (%)	Total Mass
Solution 1	[latex]120 mL[/latex]	0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]	[latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex]
Solution 2	[latex]75 mL[/latex]	0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]	[latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex]
Final Solution	[latex]195 mL[/latex]	[latex]\frac{28.05\text{ grams }}{ 195 \text{ mL }}=0.14=14\text{ % }[/latex]	[latex]10.8\text{ grams }+17.25\text{ grams }=28.05\text{ grams }[/latex]

Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.

Example

A chemist has [latex]70 mL[/latex] of a [latex]50\%[/latex] methane solution. How much of an [latex]80\%[/latex] solution must she add so the final solution is [latex]60\%[/latex] methane?

Show Solution

Let’s use the problem solving process outlined previously to help us work through a solution to the problem.

Read and Understand: We are looking for a new amount – in this case a volume – based on the words “how much”. We know two starting concentrations and the final concentration, as well as one volume.

Define and Translate: Solution 1 is the [latex]70 mL[/latex] of [latex]50\%[/latex] methane and solution 2 is the unknown amount with [latex]80\%[/latex] methane. We can call our unknown amount [latex]x[/latex].

Write and Solve: Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.

	Amount	Concentration (%)
Solution 1	[latex]70[/latex]	[latex]0.5[/latex]
Solution 2	[latex]x[/latex]	[latex]0.8[/latex]
Final Solution		[latex]0.6[/latex]

Multiply amount by concentration to get total, be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]Add the entries in the amount column to get final amount. The concentration for this amount is [latex]0.6[/latex] because we want the final solution to be [latex]60\%[/latex] methane.

	Amount	Concentration (%)	Total Mass
Solution 1	70	0.5	35
Solution 2	x	0.8	0.8x
Final Solution	70+x	0.6	[latex]42+0.6x[/latex]

Add the total mass for solution 1 and solution 2 to get the total mass for the [latex]60\%[/latex] solution. This is our equation for finding the unknown volume.

[latex]35+0.8x=42+0.6x[/latex]

[latex]\begin{array}{c}35+0.8x=42+0.6x\\\underline{-0.6x}\,\,\,\,\,\,\,\underline{-0.6x}\\35+0.2x=42\\\end{array}[/latex]

Subtract [latex]35[/latex] from both sides

[latex]\begin{array}{c}35+0.2x=42\\\underline{-35}\,\,\,\,\,\,\,\underline{-35}\\0.2x=7\end{array}[/latex]

Divide both sides by [latex]0.2[/latex]

[latex]\begin{array}{c}0.2x=7\\\frac{0.2x}{0.2}=\frac{7}{0.2}\end{array}[/latex]
[latex]x=35[/latex]

Answer

[latex]35mL[/latex] of the [latex]80\%[/latex] solution must be added to the original [latex]70mL[/latex] of the [latex]50\%[/latex] solution to gain a solution with a concentration of [latex]60\%[/latex].

The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.

Example

A farmer has two types of milk, one that is [latex]24\%[/latex] butterfat and another which is [latex]18\%[/latex] butterfat. How much of each should he use to end up with [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat?

Show Solution

Read and Understand: We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is [latex]42[/latex] gallons. There are two unknowns in this problem.

Define and Translate: We will call the unknown volume of the [latex]24\%[/latex] solution [latex]x[/latex], and the unknown volume of the [latex]18\%[/latex] solution [latex]y[/latex].

Write and Solve: Fill in the table with the information we know.

	Amount	Concentration (%)
Solution 1	[latex]x[/latex]	[latex]0.24[/latex]
Solution 2	[latex]y[/latex]	[latex]0.18[/latex]
Final Solution	[latex]42[/latex]	[latex]0.2[/latex]

Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from

	Amount	Concentration (%)	Total Mass
Solution 1	[latex]x[/latex]	[latex]0.24[/latex]	[latex]0.24x[/latex]
Solution 2	[latex]y[/latex]	[latex]0.18[/latex]	[latex]0.18y[/latex]
Final Solution	[latex]x+y=42[/latex]	[latex]0.2[/latex]	[latex]8.4[/latex]

When you sum the amount column you get one equation: [latex]x+ y = 42[/latex]
When you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[/latex]

Use elimination to find a value for [latex]x[/latex] and [latex]y[/latex].

Multiply the first equation by [latex]-0.18[/latex]

[latex]-0.18(x+y) = (42)(-0.18)[/latex]

[latex]-0.18x - 0.18y = -7.56[/latex]

Now our system of equations looks like this:

[latex]-0.18x - 0.18y = -7.56[/latex]

[latex]0.24x + 0.18y = 8.4[/latex]

Adding the two equations together to eliminate the y terms gives this equation:

[latex]0.06x = 8.4[/latex]

Divide by [latex]0.06[/latex] on each side:

[latex]x = 14[/latex]

Now substitute the value for [latex]x[/latex] into one of the equations in order to solve for [latex]y[/latex].

[latex](14) + y = 42[/latex]

[latex]y = 28[/latex]

Answer

This can be interpreted as [latex]14[/latex] gallons of [latex]24\%[/latex] butterfat milk added to [latex]28[/latex] gallons of [latex]18\%[/latex] butterfat milk will give [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat milk.

In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.

Module 10: Linear Systems