Solve Money Applications

Solve coin word problems

Imagine taking a handful of coins from your pocket and placing them on your desk. How would you determine the value of that pile of coins? If you count the number of dimes, you’ll know how many you have — the number of dimes. But that does not tell you the value of all the dimes. Say you counted $17$ dimes. Each dime is worth $\text{\textdollar}0.10$ — that is the value of one dime. To find the total value of the pile, multiply $17$ by $\text{\textdollar}0.10$ to get $\text{\textdollar}1.70$.

You could continue this process for each type of coin, then add the total value of each type to get the total value of all the coins. Suppose there are $14$ quarters, $17$ dimes, $21$ nickels, and $39$ pennies. A table helps organize the type, number, and value of each:

The total value of all the coins is $\text{\textdollar}6.64$.

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

We create a table labeled type, number, value, and total value. Adalberto has dimes (worth $\text{\textdollar}0.10$) and nickels (worth $\text{\textdollar}0.05$), totaling $\text{\textdollar}2.25$. We cannot count each type directly, but we have a clue: there are nine more nickels than dimes. Let $d =$ number of dimes, so $d + 9 =$ number of nickels.

Translating: the value of the dimes plus the value of the nickels equals the total value of the coins:

Distributing and combining like terms:

Subtracting $0.45$ from each side and dividing by $0.15$ gives $d = 12$ dimes, so the number of nickels is $d + 9 = 21$. Checking: $12$ dimes is $12(0.10) = \text{\textdollar}1.20$, and $21$ nickels is $21(0.05) = \text{\textdollar}1.05$; together that’s $\text{\textdollar}2.25$. Adalberto has twelve dimes and twenty-one nickels.

Sometimes the relationship between the two types of coins involves a multiple rather than a difference.

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

Let $q =$ the number of quarters, so the number of pennies is $2q$.

Dividing by $0.27$ gives $q = 9$ quarters, so the number of pennies is $2(9) = 18$. Checking: $9$ quarters is $9(0.25) = \text{\textdollar}2.25$, and $18$ pennies is $18(0.01) = \text{\textdollar}0.18$; together that’s $\text{\textdollar}2.43$. Maria has nine quarters and eighteen pennies.

In the next example, the number of nickels is described as “two more than ten times” the number of pennies — a linear expression rather than a plain sum or multiple.

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

Let $p =$ number of pennies, so $10p + 2 =$ number of nickels.

The number of nickels is $10(4) + 2 = 42$. Checking: $4(0.01) + 42(0.05) = 0.04 + 2.10 = 2.14$. Danny has $4$ pennies and $42$ nickels.

Solve ticket and stamp word problems

The strategy used for coin problems applies easily to other problems too. Problems involving tickets or stamps are very similar to coin problems — like coins, tickets and stamps have different values, so we can organize the information in a table the same way.

Example. At a school concert, the total value of tickets sold was $\text{\textdollar}1{,}506$. Student tickets sold for $\text{\textdollar}6$ each and adult tickets sold for $\text{\textdollar}9$ 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?

Let $s =$ the number of student tickets, so $3s - 5$ is the number of adult tickets.

Substituting to find the number of adults: $3(47) - 5 = 136$ adults. Checking: $47$ student tickets at $\text{\textdollar}6$ each is $\text{\textdollar}282$, and $136$ adult tickets at $\text{\textdollar}9$ each is $\text{\textdollar}1{,}224$; together that’s $\text{\textdollar}1{,}506$. They sold $47$ student tickets and $136$ adult tickets.

Stamp problems work exactly like coin and ticket problems.

Example. Monica paid $\text{\textdollar}10.44$ for stamps she needed to mail 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?

Let $x =$ number of $8$-cent stamps, so $2x + 4 =$ number of $49$-cent stamps.

Monica bought $8$ eight-cent stamps. The number of $49$-cent stamps is $2(8) + 4 = 20$. Checking: $8(0.08) + 20(0.49) = 0.64 + 9.80 = 10.44$. Monica bought eight $8$-cent stamps and twenty $49$-cent stamps.

Key terms

coin word problem — a word problem that finds the number of coins of each type from clues about their relationship and their total value, using a table of number, value, and total value for each coin type. ticket (and stamp) word problem — a word problem structurally identical to a coin word problem, with ticket or stamp prices in place of coin values.

This section is adapted from Prealgebra 2e, Section 9.2: Solve Money Applications by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the number/value/total-value worksheets as tables; omitted the Be Prepared quiz, the coin photograph, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.