Algebra: Coin Problems

In this algebra lesson, we will discuss coin problems are a category of word problems that involve pennies, nickels, dimes, quarters or half dollars. Examples and solutions are shown in the Problems Involving Coins.

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.

Be careful to distinguish between the value of the items and the quantity of the items. A table is useful for distinguishing between quantity and value in coin problems.

It is useful to learn how to solve coin problems as we commonly handle these items in everyday life.

Problems Involving Coins

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)

Use Substitution Method

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.

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

Answer: John received 10 nickels, 20 dimes and 42 quarters.

Have a look at the following video for another example:

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