Mixture Problems
Learning Objective(s)
· Explain how mixtures are a type of rate.
· Use systems of equations to describe and solve problems involving mixtures.
Mixture problems are excellent candidates for solving with systems of equations methods. These problems arise in many settings, such as when combining solutions in a chemistry lab or adding ingredients to a recipe. Mixtures (and mixture problems) are made whenever different types of items are combined to create a third, “mixed” item.
Learning to think of a mixture as a kind of rate is an important step in learning to solve these types of problems. Any situation in which two or more different variables are combined to determine a third is a type of rate. Speed and time combine to give us distance. Wages and hours worked produce earnings.
In a similar way, lemon juice, sugar, and water mixed together make lemonade. The tartness of the drink will depend on the ratio of the quantities mixed together—that is a rate relationship. A lemonade mixture problem may ask how tartness changes when pure water is added or when different batches of lemonade are combined.
The way to solve most mixture problems is to treat them like other rate problems—identify variables, create equations, and make tables to organize the information and highlight ways to solve the problem.
Let’s work through some mixture problems—that will show how they can be treated as rates. We’ll start with a mixture that contains two types of items with different perunit pricing.
How many pounds of walnuts that cost $0.80 per pound must be mixed with 8 pounds of cashews that costs $1.25 per pound to make a mixture of nuts that costs $1.00 per pound?
The first step here is to figure out the context of the problem and then identify the proper formula that relates all of the information. We have two types of nuts with different perpound prices being combined into a mixture. The perpound price of the mixture is determined by the ratio of the two nuts. Here, our context is total cost—we want a mixture that costs $1.00/pound. We can relate what we know and what we want to find out about total cost using the equation total cost = price • amount.
We’ll create a table to track the different total cost relationships of the two types of nuts and the mixture:
 Total cost ($)  =  Price ($/lb)  •  Amount (lbs) 
Walnuts 
 =  0.80  • 

Cashews 
 =  1.25  •  8 
Total Mix 
 =  1.00  • 

Notice that the columns are labeled with the components of our formula, and that we have filled in the information given in the problem.
The next step in figuring out this problem is to find our unknown quantities. We do know the amount of cashews in the mix (8 lbs), but we do not know the amount of walnuts, so we will call that amount a. The total cost of walnuts, then, will be 0.8a, as walnuts cost $0.80 per pound.
We can also find the total cost of the cashews if we multiply the amount of cashews (8 lbs.) by the price per pound of cashews ($1.25 per pound). We find that the total cost of cashews is $10.
Let’s add all of this information to our table.
 Total cost ($)  =  Price ($/lb)  •  Amount (lbs) 
Walnuts  0.8a  =  0.80  •  a 
Cashews  10  =  1.25  •  8 
Total Mix 
 =  1.00  • 

Now that we have assigned a variable to the amount of walnuts, we can use these existing relationships (and a bit of logic) to figure out how this relates to the total mix. The total amount of nuts in the mixture will be the number of pounds of walnuts (a) plus the number of pounds of cashews (8), or a + 8. And since we know that the price of the mixture will be $1.00 per pound, we can figure out the total cost of the mix by multiplying the amount by the price: 1.00(a + 8) = a + 8. (We luck out a bit in this problem since the price is $1.00 per pound; because we are multiplying by 1, the total cost and the amount are both represented as a + 8.
 Total cost ($)  =  Price ($/lb)  •  Amount (lbs) 
Walnuts  0.8a  =  0.80  •  a 
Cashews  10  =  1.25  •  8 
Total Mix  a + 8  =  1.00  •  a + 8 
We have completed this table. But are we any closer to finding out how many pounds of walnuts we need for this mixture? We are very close, in fact. The key to the problem lies in the total cost column. Just as the total amount of nuts in the mixture could be determined by the amount of walnuts + the number of cashews, the total cost of the mix (a + 8) must be the sum of the total cost of the walnuts (0.8a) and the total cost of the cashews (10) within the mix. We can set these quantities equal to each other and then solve for a.
0.8a + 10  =  a + 8 
0.8a + 10 − 0.8a  =  a + 8 − 0.8a 
10  =  0.2a + 8 
10 − 8  =  0.2a + 8 − 8 
2  =  0.2a 
_{}  =  _{} 
10  =  a 
a = 10, so there must be 10 pounds of walnuts in the mixture.
Since this is our first mixture problem and we aren’t sure we did it right, let’s check the answer. We calculated that 10 pounds of walnuts (the variable a) plus 8 pounds of cashews would give us a mixture that costs $1.00/pound. Let’s see if it will. Ten pounds of walnuts at $0.80/pound cost $8. Eight pounds of cashews cost $10. Combined, that comes to 18 pounds of the mixture for $18. That is indeed a dollar a pound, which is just what we were looking for. Whew!
Okay then. We figured out the quantity of walnuts in this mixture by creating a table, organizing our existing information, and then assigning a variable, a, to represent a missing quantity (walnuts).
Then we recognized an equivalent relationship in the table: the total cost of the mix must equal the combined costs of the individual quantities that make up the mix. Identifying this fact led us to the equation 0.8a + 10 = a + 8, which helped us solve for a.
That was complicated, but notice that we did not need to use a system of equations to solve this problem. We were able to figure out all of the relationships in terms of a, and our equation of 0.8a + 10 = a + 8 only involves one variable. (We only need to use a system of equations if there are two variables to solve for.)
Create and fill in a table for the following problem, and identify an equation to solve:
How many pounds of Kenyan coffee beans that cost $5.00 per pound must be mixed with 8 pounds of Ethiopian coffee beans that costs $8.00 per pound to make a blend that costs $6.00 per pound?
Liquid Mixtures
Combining liquids of different salinity or acidity is another kind of mixture problem. These liquid mixture problems have many applications in the sciences, where finding a solution with a specific concentration of a chemical is often important to experiments.
Dealing with concepts like acidity in liquid mixture problems may be confusing at first because it is hard to visualize how “acidic” something is. So here’s a quick primer on how to think about this type of a problem: the percentage of acidity tells us how much pure acid is in the solution. For example, if a solution is 10% acid, one liter of the solution would have 0.1 liters of pure acid. A typical way to set up a formula for these types of problems is Amount of acid = Percent acidity • Amount of solution.
You need 20 liters of 20% acid solution. You have jugs of 10% solution and 25% solution. How many liters of each should you combine to get the needed solution?
We can think about this problem in the same way as we thought about the dry mixture problem. Our strategy will be to identify the quantities we know and then use variables (and quantities based on those variables) to account for the rest of the relationships. We will construct our table with the Amount of acid = Percent acidity • Amount of solution formula in mind.
Let’s start by creating a table and putting in the information that is provided in the problem.
 Amount of acid (liters)  =  Acidity (%)  •  Amount of solution (liters) 
Solution 1 
 =  10%  • 

Solution 2 
 =  25%  • 

Mixture 
 =  20%  •  20 
Although the amount of acid in the mixture is not explicitly provided for us in the problem, we can quickly figure it out by multiplying the amount of needed solution (20 liters) times the acidity (20%, or 0.2) to arrive at 4 liters of acid.
Now we need to figure out the remaining relationships. Let’s use x to represent the amount of 10% solution. This also means we can use 20 − x to represent the amount of 25% solution, since we know that the total amount of the solution will be 20 liters.
 Amount of acid (liters)  =  Acidity (%)  •  Amount of solution (liters) 
Solution 1 
 =  10%  •  x 
Solution 2 
 =  25%  •  20 − x 
Mixture  4  =  20%  •  20 
Having identified the total amounts of each solution, we can multiply each by their acidity to obtain the amount of acid in each solution.
 Amount of acid (liters)  =  Acidity (%)  •  Amount of solution (liters) 
Solution 1  0.1x  =  10%  •  x 
Solution 2  0.25(20 − x)  =  25%  •  20 − x 
Mixture  4  =  20%  •  20 
Our table is now complete. But where is the relationship that will help us unlock this problem?
Let’s look down each column and see if there are any relationships we can use. The column titled “Amount of solution (liters)” does not help because x + (20 − x) = 20 resolves to 20 = 20, so it does not help us figure out the value of x. Nor does looking at “Acidity (%)” help us, because x is not present in that column.
Look at the column titled “Amount of acid (liters).” We know that the amount of acid in the mixture must be equivalent to the sum of the amounts of acid in each individual solution, so if we set these quantities equal to each other, we can find a value for our x variable.
0.1x + 0.25(20 − x)  =  4 
0.1x + 5 − 0.25x  =  4 
5 − 0.15x  =  4 
5 − 5 − 0.15x  =  4 − 5 
0.15x  =  1 
_{}  =  _{} 
x  =  _{} 
Solving the equation for x, the amount of Solution 1, we find that x = _{}. We can now use this information to find the amount of Solution 2, which we called 20 − x: _{}, or _{} liters.
As with the earlier nuts problem, we see that the key to setting up a mixture problem is to identify the context for the problem, figure out a formula that can be used to represent the relationships, and then use known amounts and variables to fill in the table. The key to solving the problem is then finding equivalent relationships that allow us to solve for the variable.
A Second Variable
It is worth mentioning that there is no single right way to solve any of these problems. The methods described above do work, but you may have noticed that both solutions only used one variable. Some people may find it easier to think of these problems in terms of two unknown quantities and treat the resulting equations as a system of two linear equations.
For example, let’s revisit the dry mixture question presented earlier. Our final table looked like this:
 Total cost ($)  =  Price ($/lb)  •  Amount (lbs) 
Walnuts  0.8a  =  0.80  •  a 
Cashews  10  =  1.25  •  8 
Total Mix  a + 8  =  1.00  •  a + 8 
It could just as easily have been written using two variables. We could have kept a as the amount of walnuts, but used b for the amount of the mixture. That would have given us this table:
 Total cost ($)  =  Price ($/lb)  •  Amount (lbs) 
Walnuts  0.8a  =  0.80  •  a 
Cashews  10  =  1.25  •  8 
Total Mix  b  =  1.00  •  b 
Using a second variable changes how we solve the problem, but it does not change the fundamental relationships within the problem. Our system of equations would look like this:
a + 8 = b 
0.8a + 10 = b 
We could then use the substitution method to obtain the following equation:
0.8a + 10 = a + 8 
Notice that this is the exact same equation we solved in the solution the first time we did the problem.
We could also have solved the acidity problem using two variables. Our final table looked like this:
 Amount of acid (liters)  =  Acidity (%)  •  Amount of solution (liters) 
Solution 1  0.1x  =  10%  •  x 
Solution 2  0.25(20 − x)  =  25%  •  20 − x 
Mixture  4  =  20%  •  20 
Instead of using 20 − x to represent the amount of Solution 2, we could have represented that quantity with the variable y.
 Amount of acid (liters)  =  Acidity (%)  •  Amount of solution (liters) 
Solution 1  0.1x  =  10%  •  x 
Solution 2  0.25y  =  25%  •  y 
Mixture  4  =  20%  •  20 
Some people prefer this method because it uses different variables for the different solutions, and that helps them keep all the players straight. Our system would now look like this:
x + y = 20 
0.1x + 0.25y = 4 
Rearranging the first equation would result in either y = 20 − x or x = 20 − y. Either of these equations could be used to solve the system using the substitution method. Finally, notice that using y = 20 − x recreates the equation that we developed when we were setting up the system using only one variable: 0.1x +0.25(20 − x) = 4.
The important lesson here is that there is more than one way to solve a rate problem. You can follow whichever method makes the most sense to you.
Oren created the following table to figure out a mixture problem. If this table is a correct representation of the problem, which of the following problems was he solving?
Total cost ($) = Price ($/lb) • Amount of candy (lbs) Jellybeans 0.5(p + 2) = p + 2 • 0.5 Gumdrops 1.5p = p • 1.5 “Beanbag” 10.00 = • 2
A) A candy maker sells jellybeans for $2 less per pound than gumdrops. One of his most popular items is called a “beanbag,” which contains 1.5 lb of jellybeans and 0.5 lbs of gumdrops. It sells for $10. What is the price of gumdrops?
B) A candy maker sells jellybeans for $2 more per pound than gumdrops. One of his most popular items is called a “beanbag,” which contains some jellybeans and some gumdrops. It sells for $10. What is the price of gumdrops?
C) A candy maker sells jellybeans for $2 more per pound than gumdrops. One of his most popular items is called a “beanbag,” which contains 0.5 lb of jellybeans and 1.5 lbs of gumdrops. It sells for $10. What is the price of gumdrops?
D) A candy maker sells jellybeans for $2 more per pound than gumdrops. One of his most popular items is called a “beanbag,” which contains 0.5 lb of jellybeans and 1.5 lbs of gumdrops. How much does it sell for?
Summary
Mixture problems are a subset of rate problems. To solve them, it is important to first recognize the context within which the problem occurs, and then to identify a formula that can be used to represent the different quantities (and rates at which those quantities occur) within the problem. Mixture problems can often be set up in terms of a single variable, although some people may prefer to set them up as twovariable systems.