It is useful to learn how to solve coin problems as we commonly handle these items in everyday life.
Other similar algebra word problems may involve items with specific values like stamps or tickets.
Examples and solutions of these are shown in Stamp Problems and Ticket Problems
Related Topics: More Algebra Word Problems
Example 1:
Jane bought a pencil and received change for $3 in 20 coins, all nickels and quarters. How many of each kind are given?
Solution:
Step 1: Set up a table with quantity and value.
quantity
value
total
nickels
quarters
together
Step 2: Fill in the table with information from the question.
Jane bought a pencil and received change for $3 in 20 coins, all nickels and quarters.
Let n = number of nickels q = number of quarters Total = quantity × value
quantity
value
total
nickels
n
5¢
5n
quarters
q
25¢
25q
together
20
300¢
Step 3: Add down each column to get the equations
n + q = 20 (equation 1)
5n + 25q = 300 (equation 2)Isolate variable n in equation 1
n = 20 – q (equation 3)
Substitute equation 3 into equation 2
5(20 – q) + 25q = 300
100 – 5q + 25q = 300
25q – 5q = 300 – 100
20q = 200
q = 10Substitute q = 10 into equation 1
n + 10 = 20
n = 10
Answer: Jane received 10 nickels and 10 quarters.
Example 2:
John received change worth $13. He received 10 more dimes than nickels and 22 more quarters than dimes. How many coins of each did he receive?
Solution:
Step 1: Set up a table with quantity and value.

quantity 
value 
total 
nickels 

dimes 

quarters 

together 
Step 2: Fill in the table with information from the question.
John received change worth $13. He received 10 more dimes than nickels and 22 more quarters than dimes.
Let d = number of dimes.
From the question, work out the relationship between dimes and the other types of coins.
nickels = dimes – 10 = d – 10
quarters = dimes + 22 = d + 22
Total = quantity × value
quantity
value
total
nickels
d – 10
5¢
5(d – 10)
dimes
d
10¢
10d
quarters
d + 22
25¢
25(d + 22)
together
1300¢
Step 3: Add down the total column to get the equation
5(d – 10) + 10d + 25(d + 22) = 1300
Use Distributive Property and Combine Like Terms
5d – 50 + 10d + 25d + 550 = 1300
5d + 10d + 25d = 1300 + 50 – 550
40d = 800
d = 20nickels = d – 10 = 10
quarters = d + 22 = 42
Answer: John received 10 nickels, 20 dimes and 42 quarters.
Martin has a total of 19 nickels and dimes worth $1.65. How many of each type of coin does he have?
Jim has quarters and nickels. He has twice as many quarters as nickels. If the value of the coins totals $4.40, how many quarters amd nickels does Jim have?
This video gives an example of using an equation to solve a problem involving coins.
Problem: Zach had $11.85 in his piggy bank. He only had nickels and quarters in the bank. The number of quarters is equal to one less than twice the number of nickels. How many of each kind of coin did he have?
Solving for an unknown number of coins given the total number of coins and the type of coins.
Problem: Lucy saves all of her quarters and dimes in a jar.She collected 170 coins that are worth $26. How many of each coin does she have?
Have a look at the following video for another example:
A mixturetype word problem (coins)
One of the easiest of all the mixture word problems to understand is the coin problem since all students have some understanding of coins.
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