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
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 = 10
Substitute 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 = 20
nickels = d – 10 = 10
quarters = d + 22 = 42
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