Solving Mixture Word Problems

Now we’ll solve some more general applications of the mixture model. Grocers and bartenders use the mixture model to set a fair price for a product made from mixing two or more ingredients. Financial planners use the mixture model when they invest money in a variety of accounts and want to find the overall interest rate. Landscape designers use the mixture model when they have an assortment of plants and a fixed budget, and event coordinators do the same when choosing appetizers and entrees for a banquet.

Our first mixture word problem will be making trail mix from raisins and nuts.

Example

Henning is mixing raisins and nuts to make 10 pounds of trail mix. Raisins cost $2 a pound and nuts cost $6 a pound. If Henning wants his cost for the trail mix to be $5.20 a pound, how many pounds of raisins and how many pounds of nuts should he use?

Solution

As before, we fill in a chart to organize our information.

The 10 pounds of trail mix will come from mixing raisins and nuts.

$\begin{array}{c}\text{Let}\phantom{\rule{0.2em}{0ex}}x=\text{number of pounds of raisins.}\hfill \\ 10-x=\text{number of pounds of nuts}\hfill \end{array}$

We enter the price per pound for each item.

We multiply the number times the value to get the total value.

$This table has four rows and four columns. The top row is a header row that reads from left to right Type, Number of pounds, Price per pound (\$), and Total Value (\$). The second row reads raisins, x, 2, and 2x. The third row reads nuts, 10 minus x, 6, and 6 times the quantity (10 minus x). The fourth row reads trail mix, 10, 5.20, and 10 times 5.20.$

Notice that the last line in the table gives the information for the total amount of the mixture.

We know the value of the raisins plus the value of the nuts will be the value of the trail mix.

Write the equation from the total values.
Solve the equation.


Find the number of pounds of nuts.

	8 pounds of nuts
Check. $\begin{array}{ccc}\hfill 2\left(\$2\right)+8\left(\$6\right)& \stackrel{?}{=}& 10\left(\$5.20\right)\hfill \\ \$4+\$48& \stackrel{?}{=}& \$52\hfill \\ \hfill \$52& =& \$52✓\hfill \end{array}$
	Henning mixed two pounds of raisins with eight pounds of nuts.

We can also use the mixture model to solve investment problems using simple interest. We have used the simple interest formula, $I=Prt,$ where $t$ represented the number of years. When we just need to find the interest for one year, $t=1,$ so then $I=Pr.$

Example

Stacey has $20,000 to invest in two different bank accounts. One account pays interest at 3% per year and the other account pays interest at 5% per year. How much should she invest in each account if she wants to earn 4.5% interest per year on the total amount?

Solution

We will fill in a chart to organize our information. We will use the simple interest formula to find the interest earned in the different accounts.

The interest on the mixed investment will come from adding the interest from the account earning 3% and the interest from the account earning 5% to get the total interest on the $20,000.

$\begin{array}{ccc}\hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x& =\hfill & \text{amount invested at 3%.}\hfill \\ \hfill 20,000-x& =\hfill & \text{amount invested at 5%}\hfill \end{array}$

The amount invested is the principal for each account.

We enter the interest rate for each account.

We multiply the amount invested times the rate to get the interest.

This table has four rows and four columns. The top row is a header row that reads from left to right Type, Amount invested, Rate, and Interest. The second row reads 3%, x, 0.03, and 0.03x. The third row reads 5%, 20,000 minus x, 0.05, and 0.05 times the quantity (20,000 minus x). The fourth row reads 4.5%, 20,000, 0.045, and 0.045 times 20,000.

Notice that the total amount invested, 20,000, is the sum of the amount invested at 3% and the amount invested at 5%. And the total interest, $0.045\left(20,000\right),$ is the sum of the interest earned in the 3% account and the interest earned in the 5% account.

As with the other mixture applications, the last column in the table gives us the equation to solve.

Write the equation from the interest earned. Solve the equation.	$\begin{array}{ccc}\hfill 0.03x+0.05\left(20,000-x\right)& =\hfill & 0.045\left(20,000\right)\hfill \\ \hfill 0.03x+1,000-0.05x& =\hfill & 900\hfill \\ \hfill -0.02x+1,000& =\hfill & 900\hfill \\ \hfill -0.02x& =\hfill & -100\hfill \\ \hfill x& =\hfill & 5,000\hfill \end{array}$ amount invested at 3%
Find the amount invested at 5%.
Check. $\begin{array}{}\\ \hfill 0.03x+0.05\left(15,000+x\right)& \stackrel{?}{=}\hfill & 0.045\left(20,000\right)\hfill \\ \hfill 150+750& \stackrel{?}{=}\hfill & 900\hfill \\ \hfill 900& =\hfill & 900✓\hfill \end{array}$
	Stacey should invest $5,000 in the account that earns 3% and $15,000 in the account that earns 5%.

Write the equation from the interest earned. Solve the equation.	\(\begin{array}{ccc}\hfill 0.03x+0.05\left(20,000-x\right)& =\hfill & 0.045\left(20,000\right)\hfill \\ \hfill 0.03x+1,000-0.05x& =\hfill & 900\hfill \\ \hfill -0.02x+1,000& =\hfill & 900\hfill \\ \hfill -0.02x& =\hfill & -100\hfill \\ \hfill x& =\hfill & 5,000\hfill \end{array}\) amount invested at 3%
Find the amount invested at 5%.
Check. \(\begin{array}{}\\ \hfill 0.03x+0.05\left(15,000+x\right)& \stackrel{?}{=}\hfill & 0.045\left(20,000\right)\hfill \\ \hfill 150+750& \stackrel{?}{=}\hfill & 900\hfill \\ \hfill 900& =\hfill & 900✓\hfill \end{array}\)
	Stacey should invest $5,000 in the account that earns 3% and $15,000 in the account that earns 5%.

Solving Mixture Word Problems