Solve Money Applications

Solve Coin Word Problems

Imagine taking a handful of coins from your pocket or purse and placing them on your desk. How would you determine the value of that pile of coins?

If you can form a step-by-step plan for finding the total value of the coins, it will help you as you begin solving coin word problems.

One way to bring some order to the mess of coins would be to separate the coins into stacks 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, you would add the total value of each pile.

An image of a large stack of pennies, a large stack of nickels, a shorter stack of dimes, and a stack of quarters is shown. There are several coins in the background. — To determine the total value of a stack of nickels, multiply the number of nickels times the value of one nickel.(Credit: Darren Hester via ppdigital)

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

But this does not tell you the value of all the dimes. Say you 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.

\begin{matrix} 17 \cdot $0.10 = $1.70 \\ number \cdot value = total value \end{matrix}

You could continue this process for each type of coin, and then you would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.

Let's look at a specific case. Suppose there are $14$ quarters, $17$ dimes, $21$ nickels, and $39$ pennies. We'll make a table to organize the information – the type of coin, the number of each, and the value.

Type	$Number$	$Value ($)$	$Total Value ($)$
Quarters	$14$	$0.25$	$3.50$
Dimes	$17$	$0.10$	$1.70$
Nickels	$21$	$0.05$	$1.05$
Pennies	$39$	$0.01$	$0.39$
			$6.64$

The total value of all the coins is $$6.64 .$ Notice how Table 1 helped us organize all the information. Let's see how this method is used to solve a coin word problem.

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?

Solution

Step 1. Read the problem. Make sure you understand all the words and ideas.

Determine the types of coins involved.

Think about the strategy we used to find the value of the handful of coins. The first thing you need is to notice what types of coins are involved. Adalberto has dimes and nickels.

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.

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 .$

Type	$Value ($)$	$Total Value ($)$
Dimes	$0.10$
Nickels	$0.05$
		$2.25$

Step 2. Identify what you are looking for.

We are asked to find the number of dimes and nickels Adalberto has.

Step 3. Name what you are looking for.

Use variable expressions to represent the number of each type of coin.
Multiply the number times the value to get the total value of each type of coin.
In this problem you cannot count each type of coin—that is what you are looking for—but you have a clue. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.
$Let d = number of dimes.$
$d + 9 = number of nickels$
Fill in the “number” column to help get everything organized.

Type	$Number$	$Value ($)$	$Total Value ($)$
Dimes	$d$	$0.10$
Nickels	$d + 9$	$0.05$
			$2.25$

Now we have all the information we need from the problem!

You multiply the number times the value to get the total value of each type of coin. While you do not know the actual number, you do have an expression to represent it.

And so now multiply $number \cdot value$ and write the results in the Total Value column.

Type	$Number$	$Value ($)$	$Total Value ($)$
Dimes	$d$	$0.10$	$0.10 d$
Nickels	$d + 9$	$0.05$	$0.05 (d + 9)$
			$2.25$

Step 4. Translate into an equation. Restate the problem in one sentence. Then translate into an equation.
The sentence “Sum of the value of the dimes and value of the nickels is total value of the coins,” is written. Below “value of the dimes” is 0.10d. Below “and” is a plus sign. Below “value of the nickels” is 0.05(d plus 9). Below “is” is an equal sign. Below “total value of the coins” is 2.25.

Step 5. Solve the equation using good algebra techniques.

Write the equation.
Distribute.
Combine like terms.
Subtract 0.45 from each side.
Divide to find the number of dimes.
The number of nickels is d + 9

Step 6. Check.

$\begin{matrix} 12 dimes: 12 (0.10) & = & 1.20 \\ 21 nickels: 21 (0.05) & = & \underset{_____}{1.05} \\ $2.25 ✓ \end{matrix}$

Step 7. Answer the question.

$Adalberto has twelve dimes and twenty-one nickels.$

If this were a homework exercise, our work might look like this:

How many of each type does he have?” Below this is a table with 4 rows and 4 columns. The first row is a header row. The headings are, “Type”, “Number”, “Value ($)”, and “Total Value ($)” Under the “Type” column are the entries dimes and nickels. Under the “Number”,column are d and d plus 9. Under the “Value”,column are the values 0.10 and 0.05. Under the “Total Value”,column are 0.10d and 0.05(d plus 9) followed by 2.25. Below the table is the word “nickels”,in bold. The equation 0.10d plus 0.05(d plus 9) equals 2.25 is shown. Below that are 2 columns. The left column says 0.10d plus 0.05d plus 0.45 equals 2.25, then 0.15d plus 0.45 equals 2.25, then 0.15d equals 1.80, then d equals 12 dimes. There is a red arrow pointing to the right column. The right column says d plus 9, then a red 12 plus 9, then 21 nickels.

Check:

$\begin{matrix} 12 dimes & 12 (0.10) & = & 1.20 \\ 21 nickels & 21 (0.05) & = & \underset{_____}{1.05} \\ $2.25 \end{matrix}$

Solve a coin word problem.

Read the problem. Make sure you understand all the words and ideas, and create a table to organize the information.
Identify what you are looking for.
Name what you 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.
Translate into an equation. Write the equation by adding the total values of all the types of coins.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

You may find it helpful to put all the numbers into the table to make sure they check.

Type	Number	Value ($)	Total Value

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?

Solution

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.

Type	$Value ($)$	$Total Value ($)$
Quarters	$0.25$
Pennies	$0.01$
		$2.43$

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.

$We know Maria has twice as many pennies as quarters. The number of pennies is defined in terms of quarters.$

$Let q represent the number of quarters.$

$Then the number of pennies is 2 q .$

Type	$Number$	$Value ($)$	$Total Value ($)$
Quarters	$q$	$0.25$
Pennies	$2 q$	$0.01$
			$2.43$

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

Type	$Number$	$Value ($)$	$Total Value ($)$
Quarters	$q$	$0.25$	$0.25 q$
Pennies	$2 q$	$0.01$	$0.01 (2 q)$
			$2.43$

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

Step 5. Solve the equation.

Write the equation.
Multiply.
Combine like terms.
Divide by 0.27.
The number of pennies is 2q.

Step 6. Check the answer in the problem.

Maria has $9$ quarters and $18$ pennies. Does this make $$2.43 ?$

$\begin{matrix} 9 quarters & 9 (0.25) & = & 2.25 \\ 18 pennies & 18 (0.01) & = & \underset{_____}{0.18} \\ Total & $2.43 ✓ \end{matrix}$

Step 7. Answer the question. Maria has nine quarters and eighteen pennies.

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?

Solution

A step-by-step guide demonstrating how to identify variables, assign values, and set up equations for a word problem involving coins.
Step 1: Read the problem.
Determine the types of coins involved. Create a table.	Pennies and nickels
Write in the value of each type of coin.	Pennies are worth $$0.01.$ Nickels are worth $$0.05.$
Step 2: Identify what you 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 then times the number of pennies.	$10 p + 2 = number of nickels$

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

Type	$Number$	$Value ($)$	$Total Value ($)$
pennies	$p$	$0.01$	$0.01 p$
nickels	$10 p + 2$	$0.05$	$0.05 (10 p + 2)$
			$$2.14$

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. Is the total value of $4$ pennies and $42$ nickels equal to $$2.14 ?$

\begin{matrix} 4 (0.01) + 42 (0.05) \overset{?}{=} 2.14 \\ 2.14 = 2.14 ✓ \end{matrix}

Step 7. Answer the question. Danny has $4$ pennies and $42$ nickels.

Solve Ticket and Stamp Word Problems

The strategies we used for coin problems can be easily applied to some other kinds of problems too. Problems involving tickets or stamps are very similar to coin problems, for example. Like coins, tickets and stamps have different values; so we can organize the information in tables much like we did for coin problems.

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 $5$ 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.

Type	$Value ($)$	$Total Value ($)$
Student	$6$
Adult	$9$
		$1,506$

Step 2. Identify what you 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.

$We know the number of adult tickets sold was 5 less than three times the number of student tickets sold.$

$Let s be the number of student tickets.$

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

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

Type	$Number$	$Value ($)$	$Total Value ($)$
Student	$s$	$6$	$6 s$
Adult	$3 s - 5$	$9$	$9 (3 s - 5)$
			$1,506$

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

6 s + 9 (3 s - 5) = 1506

Step 5. Solve the equation.

\begin{matrix} 6 s + 27 s - 45 = 1506 \\ 33 s - 45 = 1506 \\ 33 s = 1551 \\ s = 47 students \end{matrix}

Substitute to find the number of adults.

The top line says 3s minus 5 equals number of adults. The bottom line shows 3 times a red 47 minus 5 equals 136 adults.

Step 6. Check. There were $47$ student tickets at $$6$ each and $136$ adult tickets at $$9$ each. Is the total value $$1506 ?$ We find the total value of each type of ticket by multiplying the number of tickets times its value; we then add to get the total value of all the tickets sold.

\begin{matrix} 47 \cdot 6 & = & 282 \\ 136 \cdot 9 & = & \underset{_____}{1224} \\ 1506 ✓ \end{matrix}

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

Now we'll do one where we fill in the table all at once.

Monica paid $$10.44$ for stamps she needed to mail the invitations to her sister's baby shower. The number of $49-cent$ stamps was four more than twice the number of $8-cent$ stamps. How many $49-cent$ stamps and how many $8-cent$ stamps did Monica buy?

Solution

The type of stamps are $49-cent$ stamps and $8-cent$ stamps. Their names also give the value.

“The number of $49$ cent stamps was four more than twice the number of $8$ cent stamps.”

$\begin{matrix} Let x = number of 8-cent stamps \\ 2 x + 4 = number of 49-cent stamps \end{matrix}$

Type	$Number$	$Value ($)$	$Total Value ($)$
$49-cent$ stamps	$2 x + 4$	$0.49$	$0.49 (2 x + 4)$
$8-cent$ stamps	$x$	$0.08$	$0.08 x$
			$10.44$

Solution steps for an algebraic word problem, including equation formulation, solving, and verification.
Write the equation from the total values.	$0.49 (2 x + 4) + 0.08 x = 10.44$
Solve the equation.	$0.98 x + 1.96 + 0.08 x = 10.44$ $1.06 x + 1.96 = 10.44$ $1.06 x = 8.48$ $x = 8$
Monica bought 8 eight-cent stamps.
Find the number of 49-cent stamps she bought by evaluating.	$2 x + 4 for x = 8 .$
	$2 x + 4$ $2 \cdot 8 + 4$ $16 + 4$ $20$
Check. $8 (0.08) + 20 (0.49) \overset{?}{=} 10.44$ $0.64 + 9.80 \overset{?}{=} 10.44$ $10.44 = 10.44 ✓$

Monica bought eight $8-cent$ stamps and twenty $49-cent$ stamps.

Key Concepts

Finding the Total Value for Coins of the Same Type
- For coins of the same type, the total value can be found as follows:
  $\begin{matrix} number \cdot value = total value \end{matrix}$
  where number is the number of coins, value is the value of each coin, and total value is the total value of all the coins.
Solve a Coin Word Problem
1. Read the problem. Make sure you understand all the words and ideas, and create a table to organize the information.
2. Identify what you are looking for.
3. Name what you 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. 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.
Type $Number$ $Value ($)$ $Total Value ($)$

Practice Makes Perfect

Solve Coin Word Problems

In the following exercises, solve the coin word problems.

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

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 $7$ more than the number of quarters. How many of each coin does he have?

Solution

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?

Carolyn has $$2.55$ in her purse in nickels and dimes. The number of nickels is $9$ less than three times the number of dimes. Find the number of each type of coin.

Solution

12 dimes and 27 nickels

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.

Chi has $$11.30$ in dimes and quarters. The number of dimes is $3$ more than three times the number of quarters. How many dimes and nickels does Chi have?

Solution

63 dimes, 20 quarters

Tyler has $$9.70$ in dimes and quarters. The number of quarters is $8$ more than four times the number of dimes. 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 $3$ more than the number of $$5$ bills. How many of each bill does it contain?

Solution

10 of the $1 bills, 7 of the $5 bills

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

Solution

10 of the $10 bills, 5 of the $5 bills

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?

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

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 the ticket and stamp word problems.

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

Solution

30 child tickets, 50 adult tickets

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?

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

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

Julie went to the post office and bought both $$0.49$ stamps and $$0.34$ postcards for her office's bills She spent $$62.60 .$ 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

Before he left for college out of state, Jason went to the post office and bought both $$0.49$ stamps and $$0.34$ postcards and spent $$12.52 .$ The number of stamps was $4$ more than twice the number of postcards. How many of each did he buy?

Maria spent $$16.80$ at the post office. She bought three times as many $$0.49$ stamps as $$0.21$ stamps. How many of each did she buy?

Solution

30 at 49 cents, 10 at 21 cents

Hector spent $$43.40$ at the post office. He bought four times as many $$0.49$ stamps as $$0.21$ stamps. How many of each did he buy?

Hilda has $$210$ worth of $$10$ and $$12$ stock shares. The numbers of $$10$ shares is $5$ more than twice the number of $$12$ shares. How many of each does she have?

Solution

15 at $10 shares, 5 at $12 shares

Mario invested $$475$ in $$45$ and $$25$ stock shares. The number of $$25$ shares was $5$ less than three times the number of $$45$ shares. How many of each type of share did he buy?

Everyday Math

Parent Volunteer As the treasurer of her daughter's Girl Scout troop, Laney collected money for some girls and adults to go to a $3-day$ camp. Each girl paid $$75$ and each adult paid $$30 .$ The total amount of money collected for camp was $$765 .$ If the number of girls is three times the number of adults, how many girls and how many adults paid for camp?

Solution

9 girls, 3 adults

Parent Volunteer Laurie was completing the treasurer's report for her son's Boy Scout troop at the end of the school year. She didn't remember how many boys had paid the $$24$ full-year registration fee and how many had paid a $$16$ partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If $$400$ was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?

Writing Exercises

Suppose you have $6$ quarters, $9$ dimes, and $4$ pennies. Explain how you find the total value of all the coins.

Solution

Answers will vary.

Do you find it helpful to use a table when solving coin problems? Why or why not?

In the table used to solve coin problems, one column is labeled “number” and another column is labeled ‘“value.” What is the difference between the number and the value?

Solution

Answers will vary.

What similarities and differences did you see between solving the coin problems and the ticket and stamp problems?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A self-assessment chart asking students to rate their ability to solve coin, ticket, and stamp word problems across three categories: Confidently, With some help, or No-I don't get it!

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?