Solve Mixture Applications

Solve coin word problems

In mixture problems, we have two or more items with different values to combine together. The mixture model is used by grocers and bartenders to set fair prices for the products they sell — chemists, investment bankers, and landscapers use it too.

We’ll start with an application everyone is familiar with: money. Imagine you take a handful of coins from a pocket and place them on a desk. To find the total value of the pile, you could separate the coins by type — quarters with quarters, dimes with dimes, and so on — and then add the total value of each pile.

How would you find the value of one pile, say the dimes? Counting the number of dimes tells you the number you have, but not the value. Each dime is worth $0.10 — that’s the value of one dime. To get the total value of a pile of 17 dimes, multiply 17 by $0.10 to get $1.70. This leads to the following model.

The number of dimes times the value of each dime equals the total value of the dimes:

We can repeat this for each type of coin, then add the total value of each type to get the total value of all the coins. Suppose there are 14 quarters, 17 dimes, 21 nickels, and 39 pennies:

The total value of all the coins is $6.64. Notice how the table organizes all the information — this is exactly how we’ll solve coin word problems.

Example. Adalberto has $2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?

We create a table with columns “type,” “number,” “value,” and “total value,” and fill in what we know. The value of a dime is $0.10 and the value of a nickel is $0.05; the total value of all the coins is $2.25.

Since the number of nickels is nine more than the number of dimes, let $d$ be the number of dimes; then the number of nickels is $d + 9$. We multiply number times value to get each row’s total value:

Adding the total values of all the types of coins gives the equation to solve:

So there are 12 dimes. The number of nickels is $d + 9 = 12 + 9 = 21$.

Check: $12$ dimes are worth $12(0.10) = 1.20$ dollars, and $21$ nickels are worth $21(0.05) = 1.05$ dollars; together that’s $1.20 + 1.05 = 2.25$ dollars. ✓ Adalberto has twelve dimes and twenty-one nickels.

Example. Maria has $2.43 in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each type does she have?

Maria has quarters and pennies, so we let $q$ represent the number of quarters; since she has twice as many pennies, the number of pennies is $2q$. Multiplying number by value:

Adding the total values gives the equation:

So Maria has 9 quarters, and the number of pennies is $2q = 2(9) = 18$.

Check: $9$ quarters are worth $9(0.25) = 2.25$ dollars and $18$ pennies are worth $18(0.01) = 0.18$ dollars; the total is $2.25 + 0.18 = 2.43$ dollars. ✓ Maria has nine quarters and eighteen pennies.

Example. Danny has $2.14 worth of pennies and nickels in his piggy bank. The number of nickels is two more than ten times the number of pennies. How many nickels and how many pennies does Danny have?

Let $p$ represent the number of pennies, so the number of nickels is $10p + 2$.

Danny has 4 pennies. The number of nickels is $10p + 2 = 10(4) + 2 = 42$.

Check: $4(0.01) + 42(0.05) = 0.04 + 2.10 = 2.14$ dollars. ✓ Danny has four pennies and 42 nickels.

Solve ticket and stamp word problems

Problems involving tickets or stamps work very much like coin problems — each type of ticket or stamp has a value, just like each type of coin does, so we follow the same steps.

Example. At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $6 each and adult tickets sold for $9 each. The number of adult tickets sold was five less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

Let $s$ be the number of student tickets; then the number of adult tickets is $3s - 5$.

There were 47 student tickets. The number of adult tickets is $3(47) - 5 = 136$.

Check: $47 \cdot 6 = 282$ and $136 \cdot 9 = 1{,}224$; the total is $282 + 1{,}224 = 1{,}506$ dollars. ✓ They sold 47 student tickets and 136 adult tickets.

Sometimes we know the total number of tickets and need to work out how the two types relate. Suppose 100 tickets are sold, each either an adult ticket or a child ticket. If 75 are child tickets, the number of adult tickets must be $100 - 75$, so 25 adult tickets were sold. If $x$ child tickets are sold, the same reasoning gives $100 - x$ adult tickets.

Example. Galen sold 810 tickets for his church’s carnival for a total of $2,820. Children’s tickets cost $3 each and adult tickets cost $5 each. How many children’s tickets and how many adult tickets did he sell?

Since the total number of tickets sold was 810, if $c$ is the number of children’s tickets, then $810 - c$ is the number of adult tickets.

Galen sold 615 children’s tickets. The number of adult tickets is $810 - 615 = 195$.

Check: $615 \cdot 3 = 1{,}845$ and $195 \cdot 5 = 975$; the total is $1{,}845 + 975 = 2{,}820$ dollars. ✓ Galen sold 615 children’s tickets and 195 adult tickets.

Example. Monica paid $8.36 for stamps. The number of 41-cent stamps was four more than twice the number of two-cent stamps. How many 41-cent stamps and how many two-cent stamps did Monica buy?

The types of stamps are 41-cent stamps and two-cent stamps — their names give the value directly. Let $x$ represent the number of two-cent stamps; then the number of 41-cent stamps is $2x + 4$.

Monica bought eight two-cent stamps. The number of 41-cent stamps is $2(8) + 4 = 20$.

Check: $8(0.02) + 20(0.41) = 0.16 + 8.20 = 8.36$ dollars. ✓ Monica bought eight two-cent stamps and 20 forty-one-cent stamps.

Solve mixture word problems

Now we’ll solve more general applications of the mixture model. Grocers and bartenders use the mixture model to set a fair price for a product made by mixing two or more ingredients; financial planners use it when investing money across several accounts; landscape designers use it when mixing an assortment of plants on a fixed budget; and event coordinators use it when choosing appetizers and entrees for a banquet.

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?

The 10 pounds of trail mix will come from mixing raisins and nuts. Let $x$ be the number of pounds of raisins; then $10 - x$ is the number of pounds of nuts. We enter the price per pound for each item and multiply number times price to get the total value — the last row of the table gives the total amount of the mixture:

The value of the raisins plus the value of the nuts equals the value of the trail mix:

Henning should use 2 pounds of raisins. The number of pounds of nuts is $10 - 2 = 8$.

Check: $2(2) + 8(6) = 4 + 48 = 52$ dollars, and $10(5.20) = 52$ dollars. ✓ 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. Recall the simple interest formula $I = Prt$, where $t$ is the number of years; when we need the interest for just one year, $t = 1$, so $I = Pr$.

Example. Stacey has $20,000 to invest in two different bank accounts. One account pays interest at 3% per year and the other 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?

We fill in a table using the simple interest formula to find the interest earned in each account. Let $x$ be the amount invested at 3%; then $20{,}000 - x$ is the amount invested at 5%. The amount invested is the principal for each account. We multiply the amount invested by the rate to get the interest:

The total amount invested, $20,000, is the sum of the amounts invested at 3% and at 5%; the total interest, $0.045(20{,}000)$, is the sum of the interest earned in each account. As with the other mixture applications, the last column gives the equation to solve:

Stacey should invest $5,000 at 3%. The amount invested at 5% is $20{,}000 - 5{,}000 = 15{,}000$ dollars.

Check: $0.03(5{,}000) + 0.05(15{,}000) = 150 + 750 = 900$, and $0.045(20{,}000) = 900$. ✓ Stacey should invest $5,000 in the account that earns 3% and $15,000 in the account that earns 5%.

Key terms

mixture problem — a problem in which two or more items with different values are combined together. total value model — $\text{number} \cdot \text{value} = \text{total value}$: for a single type of coin, ticket, stamp, or ingredient, the number of items times the value per item gives the total value of that item. simple interest — interest computed once on the principal, using $I = Pr$ when the time is one year; in a mixture-model investment problem, the “value” of each account is its interest rate and the “total value” is the interest earned.

Practice

Solve coin word problems

Solve ticket and stamp word problems

Solve mixture word problems

Use the mixture model to solve investment problems using simple interest

This section is adapted from Elementary Algebra 2e, Section 3.3: Solve Mixture Applications by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the coin/ticket/stamp/mixture/investment tables as markdown tables; omitted the Manipulative Mathematics callout, the ticket-relationship summary table’s illustrative rows (kept as prose), and the Self Check checklist; converted the practice problems (“Try Its”) into interactive exercises with instant feedback; and adapted selected end-of-section exercises into the interactive Practice block, restating list answers in an explicit response order and splitting one two-quantity response into adjacent components.