3 Simple Steps for Solving Mixture Problems

Mixture Problems Struggling to understand your algebra assignments? Breaking problems down into manageable steps can make them much easier to handle. Here, Tucson teacher Blake C. shares his simple, 3-step process for solving mixture problems with ease:


A source of endless confusion for many algebra students is the dreaded mixture problem. These are the problems that ask you things like, if you mix 10lbs of peanuts costing $1.50 per pound with cashews costing $2.50 per pound, how many pounds of cashews would you need to add so that the resulting mixture has a cost per pound of $1.95? Or, if you mix 10 liters of pure water with 15 liters of a 30% alcohol solution, what is the concentration of the resulting mixture?

Thankfully, essentially all mixture problems are exactly the same if we use the proper setup.

Step 1: The Set Up

Mixture problems have three amounts. Two of them are the amounts being mixed, and the third is the resulting mixture amount. Each amount has its own % strength or cost. So, the set up follows this logic exactly. I’ll give you one example for each of the two types.

Solution Problems:
(% 1) (amount 1) + (% 2)(amount 2)= (final %)(total amount)

Mixture Problems:
(cost 1)(amount 1) + (cost 2) (amount 2) = (final cost) (total amount)

Now, it’s important to realize that in these problems any one of these six pieces of information can be the unknown. Your job is to fill in all of the given information and figure out what the unknown is and replace it with “x”.

Helpful tips on setup:

  1. Though it does not matter whether you use 22 or .22 for 22% for solutions problems, it is ESSENTIAL that you stick with your choice for the entire problem.
  2. The concentration of pure acid is %100.
  3.  The concentration of pure water, is 0%.

Step 2: Identifying the “x”

Let’s look at a difficult one to show you how this works out in practice.

“You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay extra to have him make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?”

Okay, so this one has an extra little trick to it that may come up often in these types of problems. You may notice we’ve only been given the number value of one amount (10 liters). In this problem we have two unknowns: the amount needed in liters for both the 10% and the 30% percent solutions.

This might seem like a problem, but there’s a simple fix. Think of it this way. We must arrive at a total of 10 liters. Let’s call the amount of liters needed of our 10% solution “x”. So, how many liters do we need of the 30% solution? Well, there are 10 liters in total, x liters have been spoken for, so what remains of our allotted 10 liters then, is 10-x.

Using this simple trick we can express both of our unknowns in terms of a single variable. It doesn’t even matter which of the amounts we call x and which we call 10-x, provided we keep track of which is which.

Step 3: Working the Problem

Let’s take that same example problem listing only the most important bits.

“You decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?”

Okay, let’s put it into our setup for a solution problem. I’ll be putting percents in decimal form throughout as that is my preference.
.10(unknown amount) + .30(unknown amount 2) = .15(10)

Now, we talked about identifying the x above. We figured out that we must call one of the unknowns x, and the other 10-x. So now, let’s plug that in.
.10(x)+.30(10-x) = .15(10).

Now, let’s solve it!

Step 1: Foil
.10x+3-.3x= 1.5
Step 2: X’s on one side by subtracting 3 from both sides
.1x-.3x= -1.5
Step 3: Combine like terms
-.2x= -1.5
Step 4: Divide by -.2
x= 7.5
Step 5: Find the other unknown
10-x= 10-7.5= 2.5
Step 6: Interpret the result
Since x was used to fill the unknown amount of the %10 solution, we have 7.5 liters of the 10% solution, and 2.5 liters of the 30% solution to end up with 10 liters of our desired 15% solution.
Step 7: Check your work
.10 (7.5) + .30(2.5) = .15(10)
.75+.75= 1.5
This is a true statement.

So there you have it. Practice this method and mixture and solutions problems will be a walk in the park!

BlakeCBlake C. tutors in various subjects, including math, reading, and SAT prep, in Tucson, AZ and online. A Flinn Scholar, Blake C. graduated from The University of Arizona with a degree in Business Management in 2007 and later returned for a second Bachelor’s in Music Theory History and Criticism which was awarded in December of 2013. Learn more about Blake here!


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Photo by Robert Cudmore

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3 replies
  1. aajayunlimited
    aajayunlimited says:

    Here’s a way that you’ve probably never thought of to solve some mixture problems and these will be kind of tricky for some folks. Out of my college algebra book. Cross-multiplication.

    Question 1. A tank contains 20 gallons of a mixture of alcohol which is 40% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will 25% alcohol by volume?

    .40[alcohol]/.25= 20/20 -x[resulting expression common to mixture problems]. So, by cross multiplying, the equation becomes: .40(20 – x) = .25(20) | 8 – .40x = 5 | 40 – 2x = 25 | -2x = -15 | Answer: x = 7.5 gallons

    Question 2.
    What weight of water must be evaporated from 40 lb of a 20% salt solution to produce a 50% solution? All percents are by weight.

    .20/.50 = 40/40 – x | Cross- multiplying again: 20 – x/2 = 8 | 40 – x = 16 | -x = -24 | Answer: x = 24 lb.
    Just like the chart makes certain mixture problems far more simple, Cross-multiplying makes these types of problems easier. Note: must be lined up right to its corresponding part or you will get the wrong answer in all likelihood.


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