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Solve Money Applications

Solve coin word problems

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.

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 $\text{\$0.10}$—that is the value of one dime. To find the total value of the pile of $17$ dimes, multiply $17$ by $\text{\$0.10}$ to get $\text{\$1.70}.$ This is the total value of all $17$ dimes.

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

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

Condensed — the full section is in OpenStax Prealgebra 2e.

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.

Example

Try it.

At a school concert, the total value of tickets sold was $\text{\$1,506}.$ Student tickets sold for $\text{\$6}$ each and adult tickets sold for $\text{\$9}$ 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	$\text{Number}$	$\text{Value (\$)}$	$\text{Total Value (\$)}$
Student		$6$
Adult		$9$
	$1,506$

Step 2. Identify what you are looking for.

$\ \text{We are looking for the number of student and adult tickets.}$

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

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

$\ \text{Let}\ s\ \text{be the number of student tickets.}$

$\ \text{Then}\ 3s-5\ \text{is the number of adult tickets.}$

$\ \text{Multiply the number times the value to get the total value of each type of ticket.}$

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

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.

\[\ \begin{array}{l}6s+27s-45=1506 \\ 33s-45=1506 \\ 33s=1551 \\ s=47\ \text{students}\end{array}\]

Substitute to find the number of adults.

Step 6. Check. There were $47$ student tickets at $\text{\$6}$ each and $136$ adult tickets at $\text{\$9}$ each. Is the total value $\text{\$1506}?$ 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.

\[\begin{array}{lll}47\cdot 6 & = & \ 282 \\ 136\cdot 9 & = & \underset{\text{_____}}{1224} \\ & & \ 1506✓\end{array}\]

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.

Condensed — the full section is in OpenStax Prealgebra 2e.

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:
  $\begin{array}{l}\text{number}\cdot \text{value}=\text{total value}\end{array}$
  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 $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$

Solve Money Applications

Solve Coin Word Problems

In the following exercises, solve the coin word problems.

Try it.

Jaime has $\text{\$2.60}$ 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 $\text{\$1.75}$ 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 $\text{\$3.50}.$ 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

Try it.

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

Try it.

Carolyn has $\text{\$2.55}$ 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

Try it.

Julio has $\text{\$2.75}$ 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 $\text{\$11.30}$ 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

Try it.

Tyler has $\text{\$9.70}$ 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?

Try it.

A cash box of $\text{\$1}$ and $\text{\$5}$ bills is worth $\text{\$45}.$ The number of $\text{\$1}$ bills is $3$ more than the number of $\text{\$5}$ bills. How many of each bill does it contain?

Solution

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

Try it.

Joe's wallet contains $\text{\$1}$ and $\text{\$5}$ bills worth $\text{\$47}.$ The number of $\text{\$1}$ bills is $5$ more than the number of $\text{\$5}$ bills. How many of each bill does he have?

Try it.

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

Solution

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

Try it.

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

Try it.

Mukul has $\text{\$3.75}$ 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 $\text{\$4.70}$ 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.

Try it.

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

Solution

30 child tickets, 50 adult tickets

Try it.

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

Try it.

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

Solution

110 child tickets, 50 adult tickets

Try it.

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

Try it.

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

Try it.

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

Try it.

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

Solution

30 at 49 cents, 10 at 21 cents

Try it.

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

Try it.

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

Solution

15 at $10 shares, 5 at $12 shares

Try it.

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

Condensed — the full section is in OpenStax Prealgebra 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 .

Onthul het antwoord

$3.5$
Simplify: $100(0.2+0.05n).$
If you missed this problem, review .

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$20+5\text{n}$
Solve: $0.25x+0.10(x+4)=2.5$
If you missed this problem, review .

Onthul het antwoord

$6$
Adalberto has $\text{\$2.25}$ in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?
Onthul het antwoord
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 $\text{\$0.10}$ and the value of a nickel is $\text{\$0.05}.$ The total value of all the coins is $\text{\$2.25}.$
Type $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
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.
  $\ \text{Let}\ d=\text{number of dimes.}$
  $\ d+9=\text{number of nickels}$
  Fill in the “number” column to help get everything organized.
Type $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
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 $\text{number}\cdot \text{value}$ and write the results in the Total Value column.
Type $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
Dimes $d$ $0.10$ $0.10d$
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.

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.
$\begin{array}{lll}12\ \text{dimes:}\ 12(0.10) & = & \ 1.20 \\ 21\ \text{nickels:}\ 21(0.05) & = & \ \underset{\text{_____}}{1.05} \\ & & \ \text{\$2.25}✓\end{array}$
Step 7. Answer the question.
$\ \text{Adalberto has twelve dimes and twenty-one nickels.}$
If this were a homework exercise, our work might look like this:
Check:
$\begin{array}{llllll}\text{12 dimes} & & & 12(0.10) & = & \ 1.20 \\ \text{21 nickels} & & & 21(0.05) & = & \ \underset{\text{_____}}{1.05} \\ & & & & & \text{\$2.25}\end{array}$
Michaela has $\text{\$2.05}$ in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?

Onthul het antwoord

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

Onthul het antwoord

17 nickels, 5 quarters
Maria has $\text{\$2.43}$ in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each type does she have?
Onthul het antwoord
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 $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
Quarters $0.25$
Pennies $0.01$
$2.43$
Step 2. Identify what you are looking for.
$\ \text{We are looking for the number of quarters and pennies.}$
Step 3. Name: Represent the number of quarters and pennies using variables.
$\ \text{We know Maria has twice as many pennies as quarters. The number of pennies is defined in terms of quarters.}$
$\ \text{Let}\ q\ \text{represent the number of quarters.}$
$\ \text{Then the number of pennies is}\ 2q.$
Type $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
Quarters $q$ $0.25$
Pennies $2q$ $0.01$
$2.43$
Multiply the ‘number’ and the ‘value’ to get the ‘total value’ of each type of coin.
Type $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
Quarters $q$ $0.25$ $0.25q$
Pennies $2q$ $0.01$ $0.01(2q)$
$2.43$
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 $\text{\$2.43}?$
$\begin{array}{llllll}\text{9 quarters} & & & 9(0.25) & = & \ 2.25 \\ \text{18 pennies} & & & 18(0.01) & = & \ \underset{\text{_____}}{0.18} \\ \text{Total} & & & & & \text{\$2.43}✓\end{array}$
Step 7. Answer the question. Maria has nine quarters and eighteen pennies.
Sumanta has $\text{\$4.20}$ in nickels and dimes in her desk drawer. She has twice as many nickels as dimes. How many coins of each type does she have?

Onthul het antwoord

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

Onthul het antwoord

51 dimes, 17 quarters

Danny has $\text{\$2.14}$ 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?

Onthul het antwoord

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 $p=\text{number of pennies}$
The number of nickels is two more than then times the number of pennies.	$10p+2=\text{number of nickels}$

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

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

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 $\text{\$2.14}?$

\[\begin{array}{l} \\ 4(0.01)+42(0.05)\overset{?}{=}2.14 \\ 2.14=2.14✓\end{array}\]

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

Jesse has $\text{\$6.55}$ 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?

Onthul het antwoord

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

Onthul het antwoord

22 nickels, 59 dimes
At a school concert, the total value of tickets sold was $\text{\$1,506}.$ Student tickets sold for $\text{\$6}$ each and adult tickets sold for $\text{\$9}$ 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?
Onthul het antwoord
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 $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
Student $6$
Adult $9$
$1,506$
Step 2. Identify what you are looking for.
$\ \text{We are looking for the number of student and adult tickets.}$
Step 3. Name. Represent the number of each type of ticket using variables.
$\ \text{We know the number of adult tickets sold was}\ 5\ \text{less than three times the number of student tickets sold.}$
$\ \text{Let}\ s\ \text{be the number of student tickets.}$
$\ \text{Then}\ 3s-5\ \text{is the number of adult tickets.}$
$\ \text{Multiply the number times the value to get the total value of each type of ticket.}$
Type $\text{Number}$ $\text{Value (\$)}$ $\text{Total Value (\$)}$
Student $s$ $6$ $6s$
Adult $3s-5$ $9$ $9(3s-5)$
$1,506$
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.
\[\ \begin{array}{l}6s+27s-45=1506 \\ 33s-45=1506 \\ 33s=1551 \\ s=47\ \text{students}\end{array}\]
Substitute to find the number of adults.
Step 6. Check. There were $47$ student tickets at $\text{\$6}$ each and $136$ adult tickets at $\text{\$9}$ each. Is the total value $\text{\$1506}?$ 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.
\[\begin{array}{lll}47\cdot 6 & = & \ 282 \\ 136\cdot 9 & = & \underset{\text{_____}}{1224} \\ & & \ 1506✓\end{array}\]
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 $\text{\$17,610}.$ One-day passes sold for $\text{\$20}$ and tournament passes sold for $\text{\$30}.$ 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?

Onthul het antwoord

330 day passes, 367 tournament passes
At the movie theater, the total value of tickets sold was $\text{\$2,612.50}.$ Adult tickets sold for $\text{\$10}$ each and senior/child tickets sold for $\text{\$7.50}$ 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?

Onthul het antwoord

112 adult tickets, 199 senior/child tickets

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

Onthul het antwoord

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

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

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

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

Write the equation from the total values.	$0.49(2x+4)+0.08x=10.44$
Solve the equation.	$0.98x+1.96+0.08x=10.44$ $1.06x+1.96=10.44$ $1.06x=8.48$ $x=8$
Monica bought 8 eight-cent stamps.
Find the number of 49-cent stamps she bought by evaluating.	$2x+4\ \text{for}\ x=8.$
	$2x+4$ $2⋅8+4$ $16+4$ $20$
Check. $8(0.08)+20(0.49)\overset{?}{=}10.44$ $0.64+9.80\overset{?}{=}10.44$ $10.44=10.44✓$

Monica bought eight $\text{8-cent}$ stamps and twenty $\text{49-cent}$ stamps.

Eric paid $\text{\$16.64}$ for stamps so he could mail thank you notes for his wedding gifts. The number of $\text{49-cent}$ stamps was eight more than twice the number of $\text{8-cent}$ stamps. How many $\text{49-cent}$ stamps and how many $\text{8-cent}$ stamps did Eric buy?

Onthul het antwoord

32 at 49 cents, 12 at 8 cents
Kailee paid $\text{\$14.84}$ for stamps. The number of $\text{49-cent}$ stamps was four less than three times the number of $\text{21-cent}$ stamps. How many $\text{49-cent}$ stamps and how many $\text{21-cent}$ stamps did Kailee buy?

Onthul het antwoord

26 at 49 cents, 10 at 21 cents
Jaime has $\text{\$2.60}$ in dimes and nickels. The number of dimes is $14$ more than the number of nickels. How many of each coin does he have?

Onthul het antwoord

8 nickels, 22 dimes
Lee has $\text{\$1.75}$ 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 $\text{\$3.50}.$ The number of dimes is $7$ more than the number of quarters. How many of each coin does he have?

Onthul het antwoord

15 dimes, 8 quarters
Connor has a collection of dimes and quarters with a total value of $\text{\$6.30}.$ The number of dimes is $14$ more than the number of quarters. How many of each coin does he have?
Carolyn has $\text{\$2.55}$ 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.

Onthul het antwoord

12 dimes and 27 nickels
Julio has $\text{\$2.75}$ 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 $\text{\$11.30}$ 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?

Onthul het antwoord

63 dimes, 20 quarters
Tyler has $\text{\$9.70}$ 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 $\text{\$1}$ and $\text{\$5}$ bills is worth $\text{\$45}.$ The number of $\text{\$1}$ bills is $3$ more than the number of $\text{\$5}$ bills. How many of each bill does it contain?

Onthul het antwoord

10 of the $1 bills, 7 of the $5 bills
Joe's wallet contains $\text{\$1}$ and $\text{\$5}$ bills worth $\text{\$47}.$ The number of $\text{\$1}$ bills is $5$ more than the number of $\text{\$5}$ bills. How many of each bill does he have?
In a cash drawer there is $\text{\$125}$ in $\text{\$5}$ and $\text{\$10}$ bills. The number of $\text{\$10}$ bills is twice the number of $\text{\$5}$ bills. How many of each are in the drawer?

Onthul het antwoord

10 of the $10 bills, 5 of the $5 bills
John has $\text{\$175}$ in $\text{\$5}$ and $\text{\$10}$ bills in his drawer. The number of $\text{\$5}$ bills is three times the number of $\text{\$10}$ bills. How many of each are in the drawer?
Mukul has $\text{\$3.75}$ 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?

Onthul het antwoord

16 nickels, 12 dimes, 7 quarters
Vina has $\text{\$4.70}$ 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?
The play took in $\text{\$550}$ one night. The number of $8 adult tickets was $10$ less than twice the number of $\text{\$5}$ child tickets. How many of each ticket were sold?

Onthul het antwoord

30 child tickets, 50 adult tickets
If the number of $\text{\$8}$ child tickets is seventeen less than three times the number of $\text{\$12}$ adult tickets and the theater took in $\text{\$584},$ how many of each ticket were sold?
The movie theater took in $\text{\$1,220}$ one Monday night. The number of $\text{\$7}$ child tickets was ten more than twice the number of $\text{\$9}$ adult tickets. How many of each were sold?

Onthul het antwoord

110 child tickets, 50 adult tickets
The ball game took in $\text{\$1,340}$ one Saturday. The number of $\text{\$12}$ adult tickets was $15$ more than twice the number of $\text{\$5}$ child tickets. How many of each were sold?
Julie went to the post office and bought both $\text{\$0.49}$ stamps and $\text{\$0.34}$ postcards for her office's bills She spent $\text{\$62.60}.$ The number of stamps was $20$ more than twice the number of postcards. How many of each did she buy?

Onthul het antwoord

40 postcards, 100 stamps
Before he left for college out of state, Jason went to the post office and bought both $\text{\$0.49}$ stamps and $\text{\$0.34}$ postcards and spent $\text{\$12.52}.$ The number of stamps was $4$ more than twice the number of postcards. How many of each did he buy?
Maria spent $\text{\$16.80}$ at the post office. She bought three times as many $\text{\$0.49}$ stamps as $\text{\$0.21}$ stamps. How many of each did she buy?

Onthul het antwoord

30 at 49 cents, 10 at 21 cents
Hector spent $\text{\$43.40}$ at the post office. He bought four times as many $\text{\$0.49}$ stamps as $\text{\$0.21}$ stamps. How many of each did he buy?

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.

\approx

approximately equal
Equal to the precision shown, not exactly.

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

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

a \bmod n

remainder
What is left after dividing a by n.

per cent
Per hundred: 15% = 15/100.

a : b,\ \frac{a}{b}

ratio, fraction
a for every b; a divided by b.

How to: Solve Money Applications

Solve coin word problems
Solve ticket and stamp word problems
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.
Write in the total value of all the coins.
We are asked to find the number of dimes and nickels Adalberto has.

Questions people ask

Why does the order of operations matter?

Because 2 + 3 × 4 would otherwise be two different numbers. The convention (brackets, exponents, multiplication and division, addition and subtraction) exists so every reader gets the same value from the same expression.

How do I check an arithmetic answer?

Estimate first (round every number and compute roughly), then compare. If the estimate and the exact answer disagree by more than a little, one of them is wrong. The solver shows every operation, so you can find which line went astray.

Why are fractions harder than decimals?

They are not harder, they are more exact: 1/3 is a precise number, 0.333 is an approximation. Fractions need a common denominator to add, which is the one extra step people trip on.

Probeer je eigen

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

Meer in Arithmetic

Order of operations Fractions Percentages Powers and roots

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

Type	\(\text{Number}\)	\(\text{Value (\$)}\)	\(\text{Total Value (\$)}\)
Student		\(6\)
Adult		\(9\)
	\(1,506\)

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

Type	\(\text{Number}\)	\(\text{Value (\$)}\)	\(\text{Total Value (\$)}\)

Type	\(\text{Number}\)	\(\text{Value (\$)}\)	\(\text{Total Value (\$)}\)
Dimes		\(0.10\)
Nickels		\(0.05\)
	\(2.25\)

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

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

Write the equation from the total values.	\(0.49(2x+4)+0.08x=10.44\)
Solve the equation.	\(0.98x+1.96+0.08x=10.44\) \(1.06x+1.96=10.44\) \(1.06x=8.48\) \(x=8\)
Monica bought 8 eight-cent stamps.
Find the number of 49-cent stamps she bought by evaluating.	\(2x+4\ \text{for}\ x=8.\)
	\(2x+4\) \(2⋅8+4\) \(16+4\) \(20\)
Check. \(8(0.08)+20(0.49)\overset{?}{=}10.44\) \(0.64+9.80\overset{?}{=}10.44\) \(10.44=10.44✓\)