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Mixture Problems

Use linear systems to solve story problems

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Mixture Problems

Mixture Problems 

Systems of equations crop up frequently in problems that deal with mixtures of two things—chemicals in a solution, nuts and raisins, or even the change in your pocket! Let’s look at some examples of these.

Mixture Problem: Coins 

Janine empties her purse and finds that it contains only nickels (worth 5 cents each) and dimes (worth 10 cents each). If she has a total of 7 coins and they have a combined value of 55 cents, how many of each coin does she have?

Since we have 2 types of coins, let’s call the number of nickels \begin{align*}x\end{align*} and the number of dimes \begin{align*}y\end{align*}. We are given two key pieces of information to make our equations: the number of coins and their value.

\begin{align*}& \# \ \text{of coins equation:} && x + y = 7 && (number \ of \ nickels) + (number \ of \ dimes)\\ & \text{value equation:} && 5x + 10y = 55 && (since \ nickels \ are \ worth \ 5c \ and \ dimes \ 10c)\end{align*}

We can quickly rearrange the first equation to isolate \begin{align*}x\end{align*}:

\begin{align*}& x = 7 - y && now \ substitute \ into \ equation \ 2:\\ &5(7 - y) + 10y = 55 && distribute \ the \ 5:\\ &35 - 5y + 10y = 55 && collect \ like \ terms:\\ &35 + 5y = 55 && subtract \ 35 \ from \ both \ sides:\\ &5y = 20 && divide \ by \ 5:\\ &\underline{y = 4} && substitute \ back \ into \ equation \ 1:\\ & x + 4 = 7 && subtract \ 4 \ from \ both \ sides:\\ &\underline{x = 3}\end{align*}

Janine has 3 nickels and 4 dimes.

Sometimes a question asks you to determine (from concentrations) how much of a particular substance to use. The substance in question could be something like coins as above, or it could be a chemical in solution, or even heat. In such a case, you need to know the amount of whatever substance is in each part. There are several common situations where to get one equation you simply add two given quantities, but to get the second equation you need to use a product. Three examples are below.

Type of mixture First equation Second equation
Coins (items with $ value) total number of items \begin{align*}(n_1 + n_2)\end{align*} total value (item value \begin{align*}\times\end{align*} no. of items)
Chemical solutions total solution volume \begin{align*}(V_1 + V_2)\end{align*} amount of solute (vol \begin{align*}\times\end{align*} concentration)
Density of two substances total amount or volume of mix total mass (volume \begin{align*}\times\end{align*} density)

For example, when considering mixing chemical solutions, we will most likely need to consider the total amount of solute in the individual parts and in the final mixture. (A solute is the chemical that is dissolved in a solution. An example of a solute is salt when added to water to make a brine.) To find the total amount, simply multiply the amount of the mixture by the fractional concentration. To illustrate, let’s look at an example where you are given amounts relative to the whole.

Mixture Problem: Chemistry 

A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. She wishes to use a high concentration solution (60%) and dilute it with a low concentration solution (5%) in order to do this. How much of each solution should she use?

To set this problem up, we first need to define our variables. Our unknowns are the amount of concentrated solution \begin{align*}(x)\end{align*} and the amount of dilute solution \begin{align*}(y)\end{align*}. We will also convert the percentages (60%, 15% and 5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information are the final volume (500 ml) and the final amount of solute (15% of \begin{align*}500 \ ml = 75 \ ml\end{align*}). Our equations will look like this:

Volume equation: \begin{align*}x + y = 500\end{align*}

Solute equation: \begin{align*}0.6x + 0.05y = 75\end{align*}

To isolate a variable for substitution, we can see it’s easier to start with equation 1:

\begin{align*}& x + y = 500 && subtract \ y \ from \ both \ sides:\\ & x = 500 - y && now \ substitute \ into \ equation \ 2:\\ & 0.6(500 - y) + 0.05y = 75 && distribute \ the \ 0.6:\\ & 300 - 0.6y + 0.05y = 75 && collect \ like \ terms:\\ & 300 - 0.55y = 75 && subtract \ 300 \ from \ both \ sides:\\ & -0.55y = -225 && divide \ both \ sides \ by \ -0.55:\\ & \underline{y = 409 \ ml} && substitute \ back \ into \ equation \ for \ x:\\ & x = 500 - 409 = \underline{91 \ ml}\end{align*}

So the chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution.

Mixture Problem: Coffee 

A coffee company makes a product which is a mixture of two coffees, using a coffee that costs $10.20 per pound and another coffee that costs $6.80 per pound. In order to make 20 pounds of a mixture that costs $8.50 per pound, how much of each type of coffee should it use?

Let \begin{align*}m\end{align*} be the amount of the $10.20 coffee, and let \begin{align*}n\end{align*} be the amount needed of the $6.80 coffee. Since we want 20 pounds of coffee that costs $8.50 per pound, the total cost for all 20 pounds is \begin{align*}20\cdot \$8.50=\$170\end{align*}. The cost for the 20 pounds of mixture is equal to the cost of each type of coffee added together: \begin{align*} 10.20\cdot m + 6.8\cdot n =170\end{align*}.

Also, the amount of each type of coffee added together equals 20 pounds: \begin{align*}m+n=20\end{align*}.

The system is: \begin{align*}\begin{cases} \qquad \quad \ \ m+n=20\\ 10.20\cdot m + 6.8\cdot n =170 \end{cases}\end{align*}.

We can isolate one variable and use substitution to solve the system:


Now solve for \begin{align*}n\end{align*}.

\begin{align*}10.20(20-n) + 6.8n &=170\\ 204 -10.20n+6.8n&=0170 \qquad \text{Distributive Property}\\ 204 -3.4n&=170 \qquad \text{Add like terms.}\\ -3.4n&=-34 \quad \ \text{Subtract} \ 204.\\ n& =10 \qquad \quad \ \text{Divide by} \ -3.4.\end{align*}

Since \begin{align*}n=10\end{align*}, we can plug that into \begin{align*}m+n=20\end{align*}.

\begin{align*}m+10=20 \Rightarrow m=10\end{align*}.

The coffee company needs to use 10 pounds of each type of coffee in order to have a 20 pound mixture that costs $8.50 per pound.


Example 1

A light green latex paint that is 20% yellow paint is combined with a darker green latex paint that is 45% yellow paint. How many gallons of each paint must be used to create 15 gallons of a green paint that is 25% yellow paint?

Let \begin{align*}x\end{align*} be the number of gallons of the 20% yellow paint and let \begin{align*}y\end{align*} be the number of gallons of the 40% yellow paint. This means that we want those two numbers to add up to 15: \begin{align*}x+y=15\end{align*}

Now if we want 15 gallons of 25% yellow paint, that means we want \begin{align*}0.25 \cdot 15=3.75\end{align*} gallons of pure yellow pigment. The expression \begin{align*}0.20\cdot x\end{align*} represents the amount of pure yellow pigment in the \begin{align*} x \end{align*} gallons of 20% yellow paint. The expression \begin{align*}0.45\cdot y\end{align*} represents the amount of pure yellow pigment in the \begin{align*} y \end{align*} gallons of 45% yellow paint. Combing the last two adds up to the 3.75 gallons of pure pigment in the final mixture:

\begin{align*} 0.20x+0.40y=3.75\end{align*}

The system is: \begin{align*}\begin{cases} \qquad \quad \ \ x+y=15\\ 0.20x+0.45y=3.75 \end{cases}\end{align*}.

We can isolate one variable and use substitution to solve the system:


Now solve for \begin{align*}y\end{align*}.

\begin{align*}0.20(15-y)+0.45y&=3.75\\ 3-0.20y+0.45y&=3.75 \qquad \text{Distributive Property}\\ 3+0.2y&=3.75 \qquad \text{Add like terms.}\\ 0.25y&=0.75\quad \ \text{Subtract} \ 3.\\ y& =3 \qquad \quad \ \text{Divide by} \ 0.25.\end{align*}

Now we can plug in \begin{align*} y=3 \end{align*} into \begin{align*} x+y=15\end{align*}:

\begin{align*}x+y=15 \Rightarrow x+3=15 \Rightarrow x=12\end{align*}.

This means 12 gallons of 20% yellow paint should be mixed with 3 gallons of 45% yellow paint in order to get 15 gallons of 25% yellow paint.


  1. I have $15 and wish to buy five pounds of mixed nuts for a party. Peanuts cost $2.20 per pound. Cashews cost $4.70 per pound.
    1. How many pounds of each should I buy?
    2. If I suddenly realize I need to set aside $5 to buy chips, can I still buy 5 pounds of nuts with the remaining $10?
    3. What’s the greatest amount of nuts I can buy?
  2. A chemistry experiment calls for one liter of sulfuric acid at a 15% concentration, but the supply room only stocks sulfuric acid in concentrations of 10% and 35%.
    1. How many liters of each should be mixed to give the acid needed for the experiment?
    2. How many liters should be mixed to give two liters at a 15% concentration?
  3. Bachelle wants to know the density of her bracelet, which is a mix of gold and silver. Density is total mass divided by total volume. The density of gold is 19.3 g/cc and the density of silver is 10.5 g/cc. The jeweler told her that the volume of silver in the bracelet was 10 cc and the volume of gold was 20 cc. Find the combined density of her bracelet.
  4. Tickets to a show cost $10 in advance and $15 at the door. If 120 tickets are sold for a total of $1390, how many of the tickets were bought in advance?
  5. A light purple latex paint that is 40% blue paint is combined with a blue latex paint that is 100% blue paint. How many gallons of each paint must be used to create 15 gallons of a dark purple paint that is 60% blue paint?

In 6-10, the multiple-choice questions on a test are worth 2 points each, and the short-answer questions are worth 5 points each.

  1. If the whole test is worth 100 points and has 35 questions, how many of the questions are multiple-choice and how many are short-answer?
  2. If Kwan gets 31 questions right and ends up with a score of 86 on the test, how many questions of each type did she get right? (Assume there is no partial credit.)
  3. If Ashok gets 5 questions wrong and ends up with a score of 87 on the test, how many questions of each type did he get wrong? (Careful!)
  4. What are two ways you could have set up the equations for part c?
  5. How could you have set up part b differently?

Review (Answers)

To view the Review answers, open this PDF file and look for section 7.3. 

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substitute In algebra, to substitute means to replace a variable or term with a specific value.

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