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Solve Mixture Applications

Solve coin word problems

Solve Coin Word Problems

In mixture problems, we will have two or more items with different values to combine together. The mixture model is used by grocers and bartenders to make sure they set fair prices for the products they sell. Many other professionals, like chemists, investment bankers, and landscapers also use the mixture model.

We will start by looking at an application everyone is familiar with—money!

Imagine that we take a handful of coins from a pocket or purse and place them on a desk. How would we determine the value of that pile of coins? If we can form a step-by-step plan for finding the total value of the coins, it will help us as we begin solving coin word problems.

So what would we do? To get some order to the mess of coins, we could separate the coins into piles according to their value. Quarters would go with quarters, dimes with dimes, nickels with nickels, and so on. To get the total value of all the coins, we would add the total value of each pile.

How would we determine the value of each pile? Think about the dime pile—how much is it worth? If we count the number of dimes, we’ll know how many we have—the number of dimes.

But this does not tell us the value of all the dimes. Say we counted 17 dimes, how much are they worth? Each dime is worth $0.10—that is the value of one dime. To find the total value of the pile of 17 dimes, multiply 17 by $0.10 to get $1.70. This is the total value of all 17 dimes. This method leads to the following model.

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

\[\begin{array}{lll}number\cdot value & = & total\ value \\ 17\cdot \$0.10 & = & \text{\$}1.70\end{array}\]

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Solve Ticket and Stamp Word Problems

Problems involving tickets or stamps are very much like coin problems. Each type of ticket and stamp has a value, just like each type of coin does. So to solve these problems, we will follow the same steps we used to solve coin problems.

Example

Try it.

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?

Solution

Step 1. Read the problem.

Determine the types of tickets involved. There are student tickets and adult tickets.
Create a table to organize the information.

Step 2. Identify what we are looking for.

We are looking for the number of student and adult tickets.

Step 3. Name. Represent the number of each type of ticket using variables.

We know the number of adult tickets sold was five less than three times the number of student tickets sold.

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

Multiply the number times the value to get the total value of each type of ticket.

Step 4. Translate. Write the equation by adding the total values of each type of ticket.

\[6s+9(3s-5)=1506\]

Step 5. Solve the equation.

$\begin{array}{lll}6s+27s-45 & = & 1506 \\ 33s-45 & = & 1506 \\ 33s & = & 1551 \\ s & = & 47\ \text{student tickets}\end{array}$

$\begin{array}{llll} & & & 3s-5 \\ & & & 3(47)-5\end{array}$

136 adult tickets

Step 6. Check the answer.

There were 47 student tickets at $6 each and 136 adult tickets at $9 each. Is the total value $1,506? We find the total value of each type of ticket by multiplying the number of tickets times its value then add to get the total value of all the tickets sold.

\[\begin{array}{lll}47\cdot 6 & = & \ 282 \\ 136\cdot 9 & = & \underset{\text{_____}}{1,224} \\ & & 1,506✓\end{array}\]

Step 7. Answer the question. They sold 47 student tickets and 136 adult tickets.

We have learned how to find the total number of tickets when the number of one type of ticket is based on the number of the other type. Next, we’ll look at an example where we know the total number of tickets and have to figure out how the two types of tickets relate.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Solve 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

Try it.

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.

\[\begin{array}{l}\text{Let}\ x=\text{number of pounds of raisins.} \\ 10-x=\text{number of pounds of nuts}\end{array}\]

We enter the price per pound for each item.

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

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

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

Total Value of Coins For the same type of coin, the total value of a number of coins is found by using the model.
$number\cdot value=total\ value$ where number is the number of coins and value is the value of each coin; total value is the total value of all the coins
Problem-Solving Strategy—Coin Word Problems
1. Read the problem. Make all the words and ideas are understood. Determine the types of coins involved.
  - Create a table to organize the information.
  - Label the columns type, number, value, total value.
  - List the types of coins.
  - Write in the value of each type of coin.
  - Write in the total value of all the coins.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
  Use variable expressions to represent the number of each type of coin and write them in the table.
  Multiply the number times the value to get the total value of each type of coin.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.
  Write the equation by adding the total values of all the types of coins.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Solve Mixture Applications

Solve Coin Word Problems

In the following exercises, solve each coin word problem.

Try it.

Jaime has $2.60 in dimes and nickels. The number of dimes is 14 more than the number of nickels. How many of each coin does he have?

Solution

8 nickels, 22 dimes

Try it.

Lee has $1.75 in dimes and nickels. The number of nickels is 11 more than the number of dimes. How many of each coin does he have?

Try it.

Ngo has a collection of dimes and quarters with a total value of $3.50. The number of dimes is seven more than the number of quarters. How many of each coin does he have?

Solution

15 dimes, 8 quarters

Try it.

Connor has a collection of dimes and quarters with a total value of $6.30. The number of dimes is 14 more than the number of quarters. How many of each coin does he have?

Try it.

A cash box of $1 and $5 bills is worth $45. The number of $1 bills is three more than the number of $5 bills. How many of each bill does it contain?

Solution

10 at $1, 7 at $5

Try it.

Joe’s wallet contains $1 and $5 bills worth $47. The number of $1 bills is five more than the number of $5 bills. How many of each bill does he have?

Try it.

Rachelle has $6.30 in nickels and quarters in her coin purse. The number of nickels is twice the number of quarters. How many coins of each type does she have?

Solution

18 quarters, 36 nickels

Try it.

Deloise has $1.20 in pennies and nickels in a jar on her desk. The number of pennies is three times the number of nickels. How many coins of each type does she have?

Try it.

Harrison has $9.30 in his coin collection, all in pennies and dimes. The number of dimes is three times the number of pennies. How many coins of each type does he have?

Solution

30 pennies, 90 dimes

Try it.

Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type does he have?

Try it.

In a cash drawer there is $125 in $5 and $10 bills. The number of $10 bills is twice the number of $5 bills. How many of each are in the drawer?

Solution

10 at $10, 5 at $5

Try it.

John has $175 in $5 and $10 bills in his drawer. The number of $5 bills is three times the number of $10 bills. How many of each are in the drawer?

Try it.

Carolyn has $2.55 in her purse in nickels and dimes. The number of nickels is nine less than three times the number of dimes. Find the number of each type of coin.

Solution

12 dimes and 27 nickels

Try it.

Julio has $2.75 in his pocket in nickels and dimes. The number of dimes is 10 less than twice the number of nickels. Find the number of each type of coin.

Try it.

Chi has $11.30 in dimes and quarters. The number of dimes is three more than three times the number of quarters. How many of each are there?

Solution

63 dimes, 20 quarters

Try it.

Tyler has $9.70 in dimes and quarters. The number of quarters is eight more than four times the number of dimes. How many of each coin does he have?

Try it.

Mukul has $3.75 in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and nine more nickels than quarters. How many of each coin are in his pocket?

Solution

16 nickels, 12 dimes, 7 quarters

Try it.

Vina has $4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin are in her purse?

Solve Ticket and Stamp Word Problems

In the following exercises, solve each ticket or stamp word problem.

Try it.

The school play sold $550 in tickets one night. The number of $8 adult tickets was 10 less than twice the number of $5 child tickets. How many of each ticket were sold?

Solution

30 child tickets, 50 adult tickets

Try it.

If the number of $8 child tickets is seventeen less than three times the number of $12 adult tickets and the theater took in $584, how many of each ticket were sold?

Try it.

The movie theater took in $1,220 one Monday night. The number of $7 child tickets was ten more than twice the number of $9 adult tickets. How many of each were sold?

Solution

110 child tickets, 50 adult tickets

Try it.

The ball game sold $1,340 in tickets one Saturday. The number of $12 adult tickets was 15 more than twice the number of $5 child tickets. How many of each were sold?

Try it.

The ice rink sold 95 tickets for the afternoon skating session, for a total of $828. General admission tickets cost $10 each and youth tickets cost $8 each. How many general admission tickets and how many youth tickets were sold?

Solution

34 general, 61 youth

Try it.

For the 7:30 show time, 140 movie tickets were sold. Receipts from the $13 adult tickets and the $10 senior tickets totaled $1,664. How many adult tickets and how many senior tickets were sold?

Try it.

The box office sold 360 tickets to a concert at the college. The total receipts were $4170. General admission tickets cost $15 and student tickets cost $10. How many of each kind of ticket was sold?

Solution

114 general, 246 student

Try it.

Last Saturday, the museum box office sold 281 tickets for a total of $3954. Adult tickets cost $15 and student tickets cost $12. How many of each kind of ticket was sold?

Try it.

Julie went to the post office and bought both $0.41 stamps and $0.26 postcards. She spent $51.40. The number of stamps was 20 more than twice the number of postcards. How many of each did she buy?

Solution

40 postcards, 100 stamps

Try it.

Jason went to the post office and bought both $0.41 stamps and $0.26 postcards and spent $10.28. The number of stamps was four more than twice the number of postcards. How many of each did he buy?

Try it.

Maria spent $12.50 at the post office. She bought three times as many $0.41 stamps as $0.02 stamps. How many of each did she buy?

Solution

30 at $0.41, 10 at $0.02

Try it.

Hector spent $33.20 at the post office. He bought four times as many $0.41 stamps as $0.02 stamps. How many of each did he buy?

Try it.

Hilda has $210 worth of $10 and $12 stock shares. The numbers of $10 shares is five more than twice the number of $12 shares. How many of each does she have?

Solution

15 $10 shares, 5 $12 shares

Try it.

Mario invested $475 in $45 and $25 stock shares. The number of $25 shares was five less than three times the number of $45 shares. How many of each type of share did he buy?

Solve Mixture Word Problems

In the following exercises, solve each mixture word problem.

Try it.

Lauren in making 15 liters of mimosas for a brunch banquet. Orange juice costs her $1.50 per liter and champagne costs her $12 per liter. How many liters of orange juice and how many liters of champagne should she use for the mimosas to cost Lauren $5 per liter?

Solution

5 liters champagne, 10 liters orange juice

Try it.

Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. Macadamia nuts cost $9 per pound and almonds cost $5.25 per pound. How many pounds of macadamia nuts and how many pounds of almonds should Macario use for the mixture to cost $6.50 per pound to make?

Try it.

Kaapo is mixing Kona beans and Maui beans to make 25 pounds of coffee blend. Kona beans cost Kaapo $15 per pound and Maui beans cost $24 per pound. How many pounds of each coffee bean should Kaapo use for his blend to cost him $17.70 per pound?

Solution

7.5 lbs Maui beans, 17.5 Kona beans

Try it.

Estelle is making 30 pounds of fruit salad from strawberries and blueberries. Strawberries cost $1.80 per pound and blueberries cost $4.50 per pound. If Estelle wants the fruit salad to cost her $2.52 per pound, how many pounds of each berry should she use?

Try it.

Carmen wants to tile the floor of his house. He will need 1000 square feet of tile. He will do most of the floor with a tile that costs $1.50 per square foot, but also wants to use an accent tile that costs $9.00 per square foot. How many square feet of each tile should he plan to use if he wants the overall cost to be $3 per square foot?

Solution

800 at $1.50, 200 at $9.00

Try it.

Riley is planning to plant a lawn in his yard. He will need nine pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs $3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound?

Try it.

Vartan was paid $25,000 for a cell phone app that he wrote and wants to invest it to save for his son’s education. He wants to put some of the money into a bond that pays 4% annual interest and the rest into stocks that pay 9% annual interest. If he wants to earn 7.4% annual interest on the total amount, how much money should he invest in each account?

Solution

$8000 at 4%, $17,000 at 9%

Try it.

Vern sold his 1964 Ford Mustang for $55,000 and wants to invest the money to earn him 5.8% interest per year. He will put some of the money into Fund A that earns 3% per year and the rest in Fund B that earns 10% per year. How much should he invest into each fund if he wants to earn 5.8% interest per year on the total amount?

Try it.

Stephanie inherited $40,000. She wants to put some of the money in a certificate of deposit that pays 2.1% interest per year and the rest in a mutual fund account that pays 6.5% per year. How much should she invest in each account if she wants to earn 5.4% interest per year on the total amount?

Solution

$10,000 in CD, $30,000 in mutual fund

Try it.

Avery and Caden have saved $27,000 towards a down payment on a house. They want to keep some of the money in a bank account that pays 2.4% annual interest and the rest in a stock fund that pays 7.2% annual interest. How much should they put into each account so that they earn 6% interest per year?

Try it.

Dominic pays 7% interest on his $15,000 college loan and 12% interest on his $11,000 car loan. What average interest rate does he pay on the total $26,000 he owes? (Round your answer to the nearest tenth of a percent.)

Solution

9.1%

Try it.

Liam borrowed a total of $35,000 to pay for college. He pays his parents 3% interest on the $8,000 he borrowed from them and pays the bank 6.8% on the rest. What average interest rate does he pay on the total $35,000? (Round your answer to the nearest tenth of a percent.)

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Multiply: 14(0.25).
If you missed this problem, review .

Fi àwọn àgbèwọlé hàn

$3.5$
Solve: $0.25x+0.10(x+4)=2.5.$
If you missed this problem, review .

Fi àwọn àgbèwọlé hàn

$x=6$
The number of dimes is three more than the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes.
If you missed this problem, review .

Fi àwọn àgbèwọlé hàn

$d=q+3$
Adalberto has $2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?
Fi àwọn àgbèwọlé hàn
Step 1. Read the problem. Make sure all the words and ideas are understood.
- Determine the types of coins involved.
  Think about the strategy we used to find the value of the handful of coins. The first thing we need is to notice what types of coins are involved. Adalberto has dimes and nickels.
- Create a table to organize the information. See chart below.
  Label the columns “type,” “number,” “value,” “total value.”
  List the types of coins.
  Write in the value of each type of coin.
  Write in the total value of all the coins.
  We can work this problem all in cents or in dollars. Here we will do it in dollars and put in the dollar sign ($) in the table as a reminder.
  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. The table below shows this information.
Step 2. Identify what we are looking for.
- We are asked to find the number of dimes and nickels Adalberto has.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
- Use variable expressions to represent the number of each type of coin and write them in the table.
- Multiply the number times the value to get the total value of each type of coin.
Next we counted the number of each type of coin. In this problem we cannot count each type of coin—that is what you are looking for—but we have a clue. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.
\[\begin{array}{lll}\text{Let}\ d & = & \text{number of dimes.} \\ d+9 & = & \text{number of nickels}\end{array}\]
Fill in the “number” column in the table to help get everything organized.
Now we have all the information we need from the problem!
We multiply the number times the value to get the total value of each type of coin. While we do not know the actual number, we do have an expression to represent it.
And so now multiply $number\cdot value=total\ value.$ See how this is done in the table below.
Notice that we made the heading of the table show the model.
Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence. Translate the English sentence into an algebraic equation.
Write the equation by adding the total values of all the types of coins.
Step 5. Solve the equation using good algebra techniques.
Now solve this equation.
Distribute.
Combine like terms.
Subtract 0.45 from each side.
Divide.
So there are 12 dimes.
The number of nickels is $d+9$.
21$\$
Step 6. Check the answer in the problem and make sure it makes sense.
Does this check?
12 dimes $12(0.10)=1.20$
21 nickels $\begin{array}{l}21(0.05)=\underset{\text{____}}{1.05} \\ \$2.25✓\end{array}$
Step 7. Answer the question with a complete sentence.
- Adalberto has twelve dimes and twenty-one nickels.
If this were a homework exercise, our work might look like the following.
Michaela has $2.05 in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?

Fi àwọn àgbèwọlé hàn

9 nickels, 16 dimes
Liliana has $2.10 in nickels and quarters in her backpack. She has 12 more nickels than quarters. How many coins of each type does she have?

Fi àwọn àgbèwọlé hàn

17 nickels, 5 quarters

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?

Fi àwọn àgbèwọlé hàn

Step 1. Read the problem.

Determine the types of coins involved.

We know that Maria has quarters and pennies.

Create a table to organize the information.

Label the columns “type,” “number,” “value,” “total value.”
List the types of coins.
Write in the value of each type of coin.
Write in the total value of all the coins.

Step 2. Identify what you are looking for.

We are looking for the number of quarters and pennies.

Step 3. Name. Represent the number of quarters and pennies using variables.

We know Maria has twice as many pennies as quarters. The number of pennies is defined in terms of quarters.
Let q represent the number of quarters.
Then the number of pennies is 2q.

Multiply the ‘number’ and the ‘value’ to get the ‘total value’ of each type of coin.

Step 4. Translate. Write the equation by adding the ‘total value’ of all the types of coins.

Step 5. Solve the equation.	$\begin{array}{lll}0.25q+0.01(2q) & = & 2.43\end{array}$
Multiply.	$\begin{array}{lll}0.25q+0.02q & = & 2.43\end{array}$
Combine like terms.	$\begin{array}{lll}0.27q & = & 2.43\end{array}$
Divide by 0.27.	$\begin{array}{lll}q & = & 9\ \text{quarters}\end{array}$
The number of pennies is $2q$.	$\begin{array}{llll} & & & 2q \\ & & & 2\cdot 9 \\ & & & \text{18 pennies}\end{array}$
Step 6. Check the answer in the problem.
Maria has 9 quarters and 18 pennies. Does this make $2.43?
$\begin{array}{llllll}\text{9 quarters} & & & 9(0.25) & = & 2.25 \\ \text{18 pennies} & & & 18(0.01) & = & \underset{\text{____}}{0.18} \\ \text{Total} & & & & & \$2.43✓\end{array}$
Step 7. Answer the question.	Maria has nine quarters and eighteen pennies.

Sumanta has $4.20 in nickels and dimes in her piggy bank. She has twice as many nickels as dimes. How many coins of each type does she have?

Fi àwọn àgbèwọlé hàn

42 nickels, 21 dimes
Alison has three times as many dimes as quarters in her purse. She has $9.35 altogether. How many coins of each type does she have?

Fi àwọn àgbèwọlé hàn

51 dimes, 17 quarters

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?

Fi àwọn àgbèwọlé hàn

Step 1. Read the problem.
Determine the types of coins involved.	pennies and nickels
Create a table.
Write in the value of each type of coin.	Pennies are worth $0.01. Nickels are worth $0.05.
Step 2. Identify what we are looking for.	the number of pennies and nickels
Step 3. Name. Represent the number of each type of coin using variables.
The number of nickels is defined in terms of the number of pennies, so start with pennies.	Let $p=$ number of pennies.
The number of nickels is two more than ten times the number of pennies.	$10p+2=$ number of nickels.
Multiply the number and the value to get the total value of each type of coin.

Step 4. Translate. Write the equation by adding the total value of all the types of coins.
Step 5. Solve the equation.
How many nickels?
Step 6. Check the answer in the problem and make sure it makes sense Danny has four pennies and 42 nickels. Is the total value $2.14? $\begin{array}{lll}4(0.01)+42(0.05) & ≟ & 2.14 \\ 2.14 & = & 2.14✓\end{array}$
Step 7. Answer the question.	Danny has four pennies and 42 nickels.

Jesse has $6.55 worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse have?

Fi àwọn àgbèwọlé hàn

41 nickels, 18 quarters
Elane has $7.00 total in dimes and nickels in her coin jar. The number of dimes that Elane has is seven less than three times the number of nickels. How many of each coin does Elane have?

Fi àwọn àgbèwọlé hàn

22 nickels, 59 dimes

Fi àwọn àgbèwọlé hàn

Step 1. Read the problem.

Determine the types of tickets involved. There are student tickets and adult tickets.
Create a table to organize the information.

Step 2. Identify what we are looking for.

We are looking for the number of student and adult tickets.

Step 3. Name. Represent the number of each type of ticket using variables.

We know the number of adult tickets sold was five less than three times the number of student tickets sold.

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

Multiply the number times the value to get the total value of each type of ticket.

Step 4. Translate. Write the equation by adding the total values of each type of ticket.

\[6s+9(3s-5)=1506\]

Step 5. Solve the equation.

$\begin{array}{lll}6s+27s-45 & = & 1506 \\ 33s-45 & = & 1506 \\ 33s & = & 1551 \\ s & = & 47\ \text{student tickets}\end{array}$

$\begin{array}{llll} & & & 3s-5 \\ & & & 3(47)-5\end{array}$

136 adult tickets

Step 6. Check the answer.

\[\begin{array}{lll}47\cdot 6 & = & \ 282 \\ 136\cdot 9 & = & \underset{\text{_____}}{1,224} \\ & & 1,506✓\end{array}\]

Step 7. Answer the question. They sold 47 student tickets and 136 adult tickets.

The first day of a water polo tournament the total value of tickets sold was $17,610. One-day passes sold for $20 and tournament passes sold for $30. The number of tournament passes sold was 37 more than the number of day passes sold. How many day passes and how many tournament passes were sold?

Fi àwọn àgbèwọlé hàn

330 day passes, 367 tournament passes
At the movie theater, the total value of tickets sold was $2,612.50. Adult tickets sold for $10 each and senior/child tickets sold for $7.50 each. The number of senior/child tickets sold was 25 less than twice the number of adult tickets sold. How many senior/child tickets and how many adult tickets were sold?

Fi àwọn àgbèwọlé hàn

112 adult tickets, 199 senior/child tickets
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?
Fi àwọn àgbèwọlé hàn
Step 1. Read the problem.
- Determine the types of tickets involved. There are children tickets and adult tickets.
- Create a table to organize the information.
Step 2. Identify what we are looking for.
- We are looking for the number of children and adult tickets.
Step 3. Name. Represent the number of each type of ticket using variables.
- We know the total number of tickets sold was 810. This means the number of children’s tickets plus the number of adult tickets must add up to 810.
- Let c be the number of children tickets.
- Then $810-c$ is the number of adult tickets.
- Multiply the number times the value to get the total value of each type of ticket.
Step 4. Translate.
- Write the equation by adding the total values of each type of ticket.
Step 5. Solve the equation.
\[\begin{array}{lll}3c+5(810-c) & = & 2,820 \\ 3c+4,050-5c & = & 2,820 \\ -2c & = & -1,230 \\ c & = & 615\ \text{children tickets}\end{array}\]
How many adults?
\[810-c\]\[810-615\]\[195\ \text{adult tickets}\]
Step 6. Check the answer. There were 615 children’s tickets at $3 each and 195 adult tickets at $5 each. Is the total value $2,820?
\[\begin{array}{lll}615\cdot 3 & = & 1845 \\ 195\cdot 5 & = & \ \underset{\text{____}}{975} \\ & & 2,820✓\end{array}\]
Step 7. Answer the question. Galen sold 615 children’s tickets and 195 adult tickets.
During her shift at the museum ticket booth, Leah sold 115 tickets for a total of $1,163. Adult tickets cost $12 and student tickets cost $5. How many adult tickets and how many student tickets did Leah sell?

Fi àwọn àgbèwọlé hàn

84 adult tickets, 31 student tickets
A whale-watching ship had 40 paying passengers on board. The total collected from tickets was $1,196. Full-fare passengers paid $32 each and reduced-fare passengers paid $26 each. How many full-fare passengers and how many reduced-fare passengers were on the ship?

Fi àwọn àgbèwọlé hàn

26 full-fare, 14 reduced fare

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?

Fi àwọn àgbèwọlé hàn

The types of stamps are 41-cent stamps and two-cent stamps. Their names also give the value!

“The number of 41-cent stamps was four more than twice the number of two-cent stamps.”

\[\begin{array}{l}\text{Let}\ x=\text{number of 2-cent stamps.} \\ 2x+4=\text{number of 41-cent stamps}\end{array}\]

Write the equation from the total values.	$\begin{array}{l}0.41(2x+4)+0.02x=8.36\end{array}$
Solve the equation.	$\begin{array}{lll}0.82x+1.64+0.02x & = & 8.36 \\ 0.84x+1.64 & = & 8.36 \\ 0.84x & = & 6.72 \\ x & = & 8\end{array}$
Monica bought eight two-cent stamps.	$2x+4\ \text{for}\ x=8.$
Find the number of 41-cent stamps she bought be evaluating.	$\begin{array}{l}2x+4 \\ 2(8)+4 \\ 20\end{array}$
Check.	$\begin{array}{lll}8(0.02)+20(0.41) & \overset{?}{=} & 8.36 \\ 0.16+8.20 & \overset{?}{=} & 8.36 \\ 8.36 & = & 8.36✓\end{array}$
	Monica bought eight two-cent stamps and 20 41-cent stamps.

Eric paid $13.36 for stamps. The number of 41-cent stamps was eight more than twice the number of two-cent stamps. How many 41-cent stamps and how many two-cent stamps did Eric buy?

Fi àwọn àgbèwọlé hàn

32 at $0.41, 12 at $0.02
Kailee paid $12.66 for stamps. The number of 41-cent stamps was four less than three times the number of 20-cent stamps. How many 41-cent stamps and how many 20-cent stamps did Kailee buy?

Fi àwọn àgbèwọlé hàn

26 at $0.41, 10 at $0.20

Fi àwọn àgbèwọlé hàn

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

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

\[\begin{array}{l}\text{Let}\ x=\text{number of pounds of raisins.} \\ 10-x=\text{number of pounds of nuts}\end{array}\]

We enter the price per pound for each item.

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

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

Orlando is mixing nuts and cereal squares to make a party mix. Nuts sell for $7 a pound and cereal squares sell for $4 a pound. Orlando wants to make 30 pounds of party mix at a cost of $6.50 a pound, how many pounds of nuts and how many pounds of cereal squares should he use?

Fi àwọn àgbèwọlé hàn

5 pounds cereal squares, 25 pounds nuts
Becca wants to mix fruit juice and soda to make a punch. She can buy fruit juice for $3 a gallon and soda for $4 a gallon. If she wants to make 28 gallons of punch at a cost of $3.25 a gallon, how many gallons of fruit juice and how many gallons of soda should she buy?

Fi àwọn àgbèwọlé hàn

21 gallons of fruit punch, 7 gallons of soda

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?

Fi àwọn àgbèwọlé hàn

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.

\[\begin{array}{lll}\text{Let}\ x & = & \text{amount invested at 3\%.} \\ 20,000-x & = & \text{amount invested at 5\%}\end{array}\]

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.

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, $0.045(20,000),$ 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.	$\begin{array}{lll}0.03x+0.05(20,000-x) & = & 0.045(20,000) \\ \\ \\ 0.03x+1,000-0.05x & = & 900 \\ -0.02x+1,000 & = & 900 \\ -0.02x & = & -100 \\ x & = & 5,000\end{array}$ amount invested at 3%
Find the amount invested at 5%.
Check. $\begin{array}{lll} \\ 0.03x+0.05(15,000+x) & ≟ & 0.045(20,000) \\ 150+750 & ≟ & 900 \\ 900 & = & 900✓\end{array}$
	Stacey should invest $5,000 in the account that earns 3% and $15,000 in the account that earns 5%.

Remy has $14,000 to invest in two mutual funds. One fund pays interest at 4% per year and the other fund pays interest at 7% per year. How much should she invest in each fund if she wants to earn 6.1% interest on the total amount?

Fi àwọn àgbèwọlé hàn

$4,200 at 4%, $9,800 at 7%
Marco has $8,000 to save for his daughter’s college education. He wants to divide it between one account that pays 3.2% interest per year and another account that pays 8% interest per year. How much should he invest in each account if he wants the interest on the total investment to be 6.5%?

Fi àwọn àgbèwọlé hàn

$2,500 at 3.2%, $5,500 at 8%
Jaime has $2.60 in dimes and nickels. The number of dimes is 14 more than the number of nickels. How many of each coin does he have?

Fi àwọn àgbèwọlé hàn

8 nickels, 22 dimes
Lee has $1.75 in dimes and nickels. The number of nickels is 11 more than the number of dimes. How many of each coin does he have?
Ngo has a collection of dimes and quarters with a total value of $3.50. The number of dimes is seven more than the number of quarters. How many of each coin does he have?

Fi àwọn àgbèwọlé hàn

15 dimes, 8 quarters
Connor has a collection of dimes and quarters with a total value of $6.30. The number of dimes is 14 more than the number of quarters. How many of each coin does he have?
A cash box of $1 and $5 bills is worth $45. The number of $1 bills is three more than the number of $5 bills. How many of each bill does it contain?

Fi àwọn àgbèwọlé hàn

10 at $1, 7 at $5
Joe’s wallet contains $1 and $5 bills worth $47. The number of $1 bills is five more than the number of $5 bills. How many of each bill does he have?
Rachelle has $6.30 in nickels and quarters in her coin purse. The number of nickels is twice the number of quarters. How many coins of each type does she have?

Fi àwọn àgbèwọlé hàn

18 quarters, 36 nickels
Deloise has $1.20 in pennies and nickels in a jar on her desk. The number of pennies is three times the number of nickels. How many coins of each type does she have?
Harrison has $9.30 in his coin collection, all in pennies and dimes. The number of dimes is three times the number of pennies. How many coins of each type does he have?

Fi àwọn àgbèwọlé hàn

30 pennies, 90 dimes
Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type does he have?
In a cash drawer there is $125 in $5 and $10 bills. The number of $10 bills is twice the number of $5 bills. How many of each are in the drawer?

Fi àwọn àgbèwọlé hàn

10 at $10, 5 at $5
John has $175 in $5 and $10 bills in his drawer. The number of $5 bills is three times the number of $10 bills. How many of each are in the drawer?
Carolyn has $2.55 in her purse in nickels and dimes. The number of nickels is nine less than three times the number of dimes. Find the number of each type of coin.

Fi àwọn àgbèwọlé hàn

12 dimes and 27 nickels

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Mixture Applications

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
Determine the types of coins involved.
Label the columns “type,” “number,” “value,” “total value.”
List the types of coins.
Write in the value of each type of coin.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Wárá

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Diẹ̀ nínú Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Now solve this equation.
Distribute.
Combine like terms.
Subtract 0.45 from each side.
Divide.
So there are 12 dimes.
The number of nickels is \(d+9\).

	21\(\\)

12 dimes	\(12(0.10)=1.20\)
21 nickels	\(\begin{array}{l}21(0.05)=\underset{\text{____}}{1.05} \\ \$2.25✓\end{array}\)

Step 5. Solve the equation.	\(\begin{array}{lll}0.25q+0.01(2q) & = & 2.43\end{array}\)
Multiply.	\(\begin{array}{lll}0.25q+0.02q & = & 2.43\end{array}\)
Combine like terms.	\(\begin{array}{lll}0.27q & = & 2.43\end{array}\)
Divide by 0.27.	\(\begin{array}{lll}q & = & 9\ \text{quarters}\end{array}\)
The number of pennies is \(2q\).	\(\begin{array}{llll} & & & 2q \\ & & & 2\cdot 9 \\ & & & \text{18 pennies}\end{array}\)
Step 6. Check the answer in the problem.
Maria has 9 quarters and 18 pennies. Does this make $2.43?
\(\begin{array}{llllll}\text{9 quarters} & & & 9(0.25) & = & 2.25 \\ \text{18 pennies} & & & 18(0.01) & = & \underset{\text{____}}{0.18} \\ \text{Total} & & & & & \$2.43✓\end{array}\)
Step 7. Answer the question.	Maria has nine quarters and eighteen pennies.

Write the equation from the total values.	\(\begin{array}{l}0.41(2x+4)+0.02x=8.36\end{array}\)
Solve the equation.	\(\begin{array}{lll}0.82x+1.64+0.02x & = & 8.36 \\ 0.84x+1.64 & = & 8.36 \\ 0.84x & = & 6.72 \\ x & = & 8\end{array}\)
Monica bought eight two-cent stamps.	\(2x+4\ \text{for}\ x=8.\)
Find the number of 41-cent stamps she bought be evaluating.	\(\begin{array}{l}2x+4 \\ 2(8)+4 \\ 20\end{array}\)
Check.	\(\begin{array}{lll}8(0.02)+20(0.41) & \overset{?}{=} & 8.36 \\ 0.16+8.20 & \overset{?}{=} & 8.36 \\ 8.36 & = & 8.36✓\end{array}\)
	Monica bought eight two-cent stamps and 20 41-cent stamps.

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