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Mixture Problems

Use linear systems to solve story problems

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Mixture Problems

What if you had two types of grape drink, one with 5% real fruit juice and another with 10% real fruit juice? Suppose you wanted a gallon of grape drink with 6% real fruit juice. How much of the 5% drink and how much of the 10% drink should you mix together to produce it? 

Mixture Problems

Systems of equations arise in chemistry when mixing chemicals in solutions and can even be seen in things like mixing nuts and raisins or examining the change in your pocket!

By rearranging one sentence in an equation into \begin{align*}y=\end{align*} algebraic expression or \begin{align*}x=\end{align*} algebraic expression, you can use the Substitution Method to solve the system.

Let's solve the following problems using the Substitution Method:

  1. Nadia empties her purse and finds that it contains only nickels and dimes. 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?

Begin by choosing appropriate variables for the unknown quantities. Let \begin{align*}n=\end{align*} the number of nickels and \begin{align*}d=\end{align*} the number of dimes.

There are seven coins in Nadia’s purse: \begin{align*}n+d=7\end{align*}.

The total is 55 cents: \begin{align*}0.05n+0.10d=0.55\end{align*}.

The system is: \begin{align*}\begin{cases} \qquad \quad \ \ n+d=7\\ 0.05n+0.10d=0.55 \end{cases}\end{align*}.

We can quickly rearrange the first equation to isolate \begin{align*}d\end{align*}, the number of dimes: \begin{align*}d=7-n\end{align*}.

Using the Substitution Property, every \begin{align*}d\end{align*} can be replaced with the expression \begin{align*}7-n\end{align*}.

\begin{align*}0.05n+0.10(7-n)&=0.55\\ \text{Now solve for} \ n: \qquad 0.05n+0.70-0.10n&=0.55 \qquad \text{Distributive Property}\\ -0.05n+0.70&=0.55 \qquad \text{Add like terms.}\\ -0.05n&=-0.15 \quad \ \text{Subtract} \ 0.70.\\ n& =3 \qquad \quad \ \text{Divide by} \ -0.05.\end{align*}

Nadia has 3 nickels. There are seven coins in the purse; three are nickels, so four must be dimes.

Check to make sure this combination is 55 cents: \begin{align*}0.05(3)+ 0.10(4)= 0.15+0.40=0.55\end{align*}.

  1. A chemist has two containers, Mixture \begin{align*}A\end{align*} and Mixture \begin{align*}B\end{align*}. Mixture \begin{align*}A\end{align*} has a 60% copper sulfate concentration. Mixture \begin{align*}B\end{align*} has a 5% copper sulfate concentration. The chemist needs to have a mixture equaling 500 mL with a 15% concentration. How much of each mixture does the chemist need?

Although not explicitly stated, there are two equations involved in this situation.

  • Begin by stating the variables. Let \begin{align*}A = mixture \ A \ and \ B = mixture \ B\end{align*}.
  • The total mixture needs to have 500 mL of liquid.

Equation 1 (how much total liquid): \begin{align*}A+B=500\end{align*}.

  • The total amount of copper sulfate needs to be 15% of the total amount of solution (500 mL). \begin{align*}0.15 \cdot 500=75 \ ounces\end{align*}

Equation 2 (how much copper sulfate the chemist needs): \begin{align*}0.60A+0.05B=75\end{align*}

\begin{align*}\begin{cases} A+B=500\\ 0.60A+0.05B=75 \end{cases}\end{align*}

By rewriting equation 1, the Substitution Property can be used: \begin{align*}A=500-B\end{align*}.

Substitute the expression \begin{align*}500-B\end{align*} for the variable \begin{align*}A\end{align*} in the second equation.


Solve for \begin{align*}B\end{align*}.

\begin{align*}300-0.60B+0.05B& =75 && \text{Distributive Property}\\ 300-0.55B&=75 && \text{Add like terms.}\\ -0.55B&=-225 && \text{Subtract} \ 300.\\ B & \approx 409 \ mL\end{align*}

The chemist needs approximately 409 mL of mixture \begin{align*}B\end{align*}. To find the amount of mixture \begin{align*}A\end{align*}, use the first equation: \begin{align*}A+409=500\end{align*}

\begin{align*}A=91 \ mL\end{align*}

The chemist needs 91 milliliters of mixture \begin{align*}A\end{align*} and 409 milliliters of mixture \begin{align*}B\end{align*} to get a 500 mL solution with a 15% copper sulfate concentration.

  1. 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:


\begin{align*}10.20(20-n) + 6.8n &=170\\ \text{Now solve for} \ n: \qquad 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

Earlier, you were told that you have two types of grape drink, one with 5% real fruit juice and one with 10% real fruit juice. Suppose you wanted a gallon of a grape drink with 6% real fruit juice. How much of the 5% drink and how much of the 10% drink should you mix together to produce it?

Let \begin{align*}x\end{align*} be the amount of the 5% juice and \begin{align*}y\end{align*} be the amount of 10% juice. We want one gallon of the grape drink so one of our equations is \begin{align*}x+y=1\end{align*}. Our second equation is \begin{align*}.05x + .10y = .06\end{align*} since we have \begin{align*}x\end{align*} gallons of the 5% juice, \begin{align*}y\end{align*} gallons of the 10% juice, and we want 1 gallon of the 6% juice.

The system is:

\begin{align*} \begin{cases}x + y = 1\\ .05x + .1y = .06\end{cases}\end{align*}  

We can isolate \begin{align*}x\end{align*} in the first equation and use substitution to solve the system:

\begin{align*}x = 1-y\end{align*} 

\begin{align*}.05(1-y) +.1y &= .06\\ .05 -.05y + .1y &=.06\\ .05 +.05y &=.06\\ .05y &= .01\\ y &=.2\end{align*}

Now, substitute \begin{align*}y\end{align*} in to the equation to solve for \begin{align*}x\end{align*}.

\begin{align*}x &= 1-y\\ x &=1-.2\\ x &=.8\end{align*}

You need .8 gallons of the 5% real fruit juice drink and .2 gallons of the 10% real fruit juice drink to get one gallon of a 6% real fruit juice drink.

Example 2

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:


\begin{align*}0.20(15-y)+0.45y&=3.75\\ \text{Now solve for} \ x: \qquad 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 $15.00 and wish to buy five pounds of mixed nuts for a party. Peanuts cost $2.20 per pound. Cashews cost $4.70 per pound. How many pounds of each should I 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%. How many liters of each should be mixed to give the acid needed for the experiment?
  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 used was 10 cc and the volume of gold used was 20 cc. Find the combined density of her bracelet.
  4. Jeffrey wants to make jam. He needs a combination of raspberries and blackberries totaling six pounds. He can afford $11.60. How many pounds of each berry should he buy?
  5. A farmer has fertilizer in 5% and 15% solutions. How much of each type should he mix to obtain 100 liters of fertilizer in a 12% solution?

Review (Answers)

To see 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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