Lesson Objectives
- Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
- Demonstrate an understanding of the six-step method used for solving applications of linear equations
- Learn how to solve coin word problems
How to Solve Bill and Coin Word Problems
In our last lesson, we reviewed the six-step method used to solve a word problem that involves a linear equation in one variable.
Example 1: Solve each word problem.
At the end of Jessica's shift, she counted down her register. She had a total of $14.25 in change (coins). This change consisted of nickels, dimes, and quarters only. There were three times as many quarters as dimes and two-thirds the number of nickels as quarters. How many of each type of coin did Jessica have in her register?
Step 1) After reading the problem, it is clear that we want to find the number of nickels, dimes, and quarters that were in Jessica's register
Step 2) When we have more than one unknown, we can let a variable represent one of the unknowns and then model the other unknowns based on our variable. In this case, quarters are used in both comparisons.
let x = # of quarters that were present in Jessica's register
Then (1/3)x = # of dimes that were present in Jessica's register (since the number of quarters was 3 times more than the number of dimes)
Then (2/3)x = # of nickels that were present in Jessica's register (since the number of nickels was 2/3 of the number of quarters)
Step 3) To make an equation, we have to think about the information given. Many people make the mistake of trying to just add the amounts we just came up with (1/3)x, (2/3)x, and x and set this equal to 14.25. This will not work as we are comparing a number to a value. In other words, $14.25 is the value of the coins, whereas the sum of (1/3)x, (2/3)x, and x would be the number of coins. In this case and with most coin or denomination of money problems, we have to take an extra step and multiply each coin by its value. Let's look at the information organized in a table:
To make things easier, we can use 1/20 in the place of .05, 1/10 in the place of .1, and 1/4 in the place of .25.
Total Value:
Nickels: $$\frac{1}{20}\cdot \frac{2}{3}x=\frac{1}{30}x$$ Dimes: $$\frac{1}{10}\cdot \frac{1}{3}x=\frac{1}{30}x$$ Quarters: $$\frac{1}{4}x$$ Let's set up our equation: $$\frac{1}{30}x + \frac{1}{30}x + \frac{1}{4}x=14.25$$ Step 4) Solve the equation: $$\frac{1}{30}x + \frac{1}{30}x + \frac{1}{4}x=14.25$$ $$\frac{2}{30}x + \frac{1}{4}x=14.25$$ $$\frac{1}{15}x + \frac{1}{4}x=14.25$$ Clear the fractions, multiply each side by 60: $$4x + 15x=855$$ $$19x=855$$ $$x=45$$ Step 5) Since x represented the number of quarters, this tells us Jessica had 45 quarters. She had 2/3 the number of nickels as quarters. This means she had 30 nickels (45 • 2/3). Lastly, she had 1/3 the number of dimes as quarters. This means she had 15 dimes (45 • 1/3). We can state our answer as:
Jessica had 30 nickels, 15 dimes, and 45 quarters in her register.
Step 6) We can check to see if the value matches.
Nickels: 30 • .05 = 1.5
Dimes: 15 • .1 = 1.5
Quarters: 45 • .25 = 11.25
Sum the amounts:
1.5 + 1.5 + 11.25 = 14.25
14.25 = 14.25 ✔
We can also check the number of coins: we are told there are three times as many dimes as quarters:
3 • 15 = 45
45 = 45 ✔
We are also told there are two-thirds the number of nickels as quarters:
45 • 2/3 = 30
30 = 30 ✔
Example 2: Solve each word problem.
A local farm stand collected $553 in $20 bills, $10 bills, $5 bills, and $1 bills. The number of $20 bills is 5 less than the number of $10 bills and 5 more than the number of $5 bills. Additionally, there are 12 more $20 bills than $1 bills. How much of each bill type ($1, $5, $10, and $20 bills) does the farm stand have?
Step 1) After reading the problem, it is clear that we want to find the number of $1, $5, $10, and $20 bills
Step 2) When we have more than one unknown, we can let a variable represent one of the unknowns and then model the other unknowns based on our variable. In this case, the number of $20 bills is used in all comparisons.
let x = # of $20 bills
Then x + 5 = # $10 bills (since the number of $20 bills was 5 less than the number of $10 bills)
Then x - 5 = # $5 bills (since the number of $20 bills was 5 more than the number of $5 bills)
Then x - 12 = # $1 bills (since the number of $20 bills was 12 more than the number of $1 bills)
Step 3) To make an equation, we have to think about the information given. Many people make the mistake of trying to just add the amounts we just came up with x, x + 5, x - 5, and x - 12, and set this equal to 553. This will not work as we are comparing a number to a value. In other words, $553 is the value of the bills, whereas the sum of x, x + 5, x - 5, and x - 12 would be the number of bills. In this case and with most coin or denomination of money problems, we have to take an extra step and multiply each bill by its value. Let's look at the information organized in a table:
Total Value:
$20 Bills: 20x
$10 Bills: 10(x + 5)
$5 Bills: 5(x - 5)
$1 Bills: x - 12
Let's set up our equation:
20x + 10(x + 5) + 5(x - 5) + x - 12 = 553
20x + 10x + 50 + 5x - 25 + x - 12 = 553
36x + 13 = 553
36x = 553 - 13
36x = 540
36x/36 = 540/36
x = 15
Step 5) Since x represented the number of $20 bills, we can state that we have 15 $20 bills, 20 $10 bills, 10 $5 bills, and 3 $1 bills.
Step 6) We can check to see if the value matches.
$20 Bills: 15 • 20 = 300
$10 Bills: 20 • 10 = 200
$5 Bills: 10 • 5 = 50
$1 Bills: 3 • 1 = 3
Sum the amounts:
300 + 200 + 50 + 3 = 553
553 = 553 ✔
We can also check the number of bills:
We are told there are 5 less $20 bills than $10 bills:
20 - 5 = 15
15 = 15 ✔
We are also told there are 5 more $20 bills than $5 bills:
15 - 5 = 10
10 = 10 ✔
We are also told there are 12 more $20 bills than $1 bills:
15 - 12 = 3
3 = 3 ✔
Six-step method for Solving Word Problems with Linear Equations in One Variable
- Read the problem and determine what you are asked to find
- Assign a variable to represent the unknown
- If more than one unknown exists, we express the other unknowns in terms of this variable
- Write out an equation that describes the given situation
- Solve the equation
- State the answer using a nice clear sentence
- Check the result
- We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.
Solving Bill and Coin Word Problems
Essentially this type of problem will give us a value for all of the coins or bills in the problem. Our goal will be to find the individual number of coins or bills involved. Let's look at some examples.Example 1: Solve each word problem.
At the end of Jessica's shift, she counted down her register. She had a total of $14.25 in change (coins). This change consisted of nickels, dimes, and quarters only. There were three times as many quarters as dimes and two-thirds the number of nickels as quarters. How many of each type of coin did Jessica have in her register?
Step 1) After reading the problem, it is clear that we want to find the number of nickels, dimes, and quarters that were in Jessica's register
Step 2) When we have more than one unknown, we can let a variable represent one of the unknowns and then model the other unknowns based on our variable. In this case, quarters are used in both comparisons.
let x = # of quarters that were present in Jessica's register
Then (1/3)x = # of dimes that were present in Jessica's register (since the number of quarters was 3 times more than the number of dimes)
Then (2/3)x = # of nickels that were present in Jessica's register (since the number of nickels was 2/3 of the number of quarters)
Step 3) To make an equation, we have to think about the information given. Many people make the mistake of trying to just add the amounts we just came up with (1/3)x, (2/3)x, and x and set this equal to 14.25. This will not work as we are comparing a number to a value. In other words, $14.25 is the value of the coins, whereas the sum of (1/3)x, (2/3)x, and x would be the number of coins. In this case and with most coin or denomination of money problems, we have to take an extra step and multiply each coin by its value. Let's look at the information organized in a table:
| Coin | Number of Coins | Value of Each Coin | Total Value |
|---|---|---|---|
| Nickel | (2/3)x | .05 | (1/30)x |
| Dime | (1/3)x | .10 | (1/30)x |
| Quarter | x | .25 | .25x |
Total Value:
Nickels: $$\frac{1}{20}\cdot \frac{2}{3}x=\frac{1}{30}x$$ Dimes: $$\frac{1}{10}\cdot \frac{1}{3}x=\frac{1}{30}x$$ Quarters: $$\frac{1}{4}x$$ Let's set up our equation: $$\frac{1}{30}x + \frac{1}{30}x + \frac{1}{4}x=14.25$$ Step 4) Solve the equation: $$\frac{1}{30}x + \frac{1}{30}x + \frac{1}{4}x=14.25$$ $$\frac{2}{30}x + \frac{1}{4}x=14.25$$ $$\frac{1}{15}x + \frac{1}{4}x=14.25$$ Clear the fractions, multiply each side by 60: $$4x + 15x=855$$ $$19x=855$$ $$x=45$$ Step 5) Since x represented the number of quarters, this tells us Jessica had 45 quarters. She had 2/3 the number of nickels as quarters. This means she had 30 nickels (45 • 2/3). Lastly, she had 1/3 the number of dimes as quarters. This means she had 15 dimes (45 • 1/3). We can state our answer as:
Jessica had 30 nickels, 15 dimes, and 45 quarters in her register.
Step 6) We can check to see if the value matches.
Nickels: 30 • .05 = 1.5
Dimes: 15 • .1 = 1.5
Quarters: 45 • .25 = 11.25
Sum the amounts:
1.5 + 1.5 + 11.25 = 14.25
14.25 = 14.25 ✔
We can also check the number of coins: we are told there are three times as many dimes as quarters:
3 • 15 = 45
45 = 45 ✔
We are also told there are two-thirds the number of nickels as quarters:
45 • 2/3 = 30
30 = 30 ✔
Example 2: Solve each word problem.
A local farm stand collected $553 in $20 bills, $10 bills, $5 bills, and $1 bills. The number of $20 bills is 5 less than the number of $10 bills and 5 more than the number of $5 bills. Additionally, there are 12 more $20 bills than $1 bills. How much of each bill type ($1, $5, $10, and $20 bills) does the farm stand have?
Step 1) After reading the problem, it is clear that we want to find the number of $1, $5, $10, and $20 bills
Step 2) When we have more than one unknown, we can let a variable represent one of the unknowns and then model the other unknowns based on our variable. In this case, the number of $20 bills is used in all comparisons.
let x = # of $20 bills
Then x + 5 = # $10 bills (since the number of $20 bills was 5 less than the number of $10 bills)
Then x - 5 = # $5 bills (since the number of $20 bills was 5 more than the number of $5 bills)
Then x - 12 = # $1 bills (since the number of $20 bills was 12 more than the number of $1 bills)
Step 3) To make an equation, we have to think about the information given. Many people make the mistake of trying to just add the amounts we just came up with x, x + 5, x - 5, and x - 12, and set this equal to 553. This will not work as we are comparing a number to a value. In other words, $553 is the value of the bills, whereas the sum of x, x + 5, x - 5, and x - 12 would be the number of bills. In this case and with most coin or denomination of money problems, we have to take an extra step and multiply each bill by its value. Let's look at the information organized in a table:
| Bill | Number of Bills | Value of Bill | Total Value |
|---|---|---|---|
| $20 | x | 20 | 20x |
| $10 | x + 5 | 10 | 10(x + 5) |
| $5 | x - 5 | 5 | 5(x - 5) |
| $1 | x - 12 | 1 | x - 12 |
$20 Bills: 20x
$10 Bills: 10(x + 5)
$5 Bills: 5(x - 5)
$1 Bills: x - 12
Let's set up our equation:
20x + 10(x + 5) + 5(x - 5) + x - 12 = 553
20x + 10x + 50 + 5x - 25 + x - 12 = 553
36x + 13 = 553
36x = 553 - 13
36x = 540
36x/36 = 540/36
x = 15
Step 5) Since x represented the number of $20 bills, we can state that we have 15 $20 bills, 20 $10 bills, 10 $5 bills, and 3 $1 bills.
Step 6) We can check to see if the value matches.
$20 Bills: 15 • 20 = 300
$10 Bills: 20 • 10 = 200
$5 Bills: 10 • 5 = 50
$1 Bills: 3 • 1 = 3
Sum the amounts:
300 + 200 + 50 + 3 = 553
553 = 553 ✔
We can also check the number of bills:
We are told there are 5 less $20 bills than $10 bills:
20 - 5 = 15
15 = 15 ✔
We are also told there are 5 more $20 bills than $5 bills:
15 - 5 = 10
10 = 10 ✔
We are also told there are 12 more $20 bills than $1 bills:
15 - 12 = 3
3 = 3 ✔
Skills Check:
Example #1
Solve each word problem.
Megan, a cashier at a local retail store has a total of $210 in her register. That amount comes from 5 dollar bills and 20 dollar bills only. If the number of 5 dollar bills is 2 more than the number of 20 dollar bills, how much of each type of bill is in her register?
Please choose the best answer.
A
10: $5 bills, 8: $20 bills
B
12: $5 bills, 6: $20 bills
C
9: $5 bills, 4: $20 bills
D
12: $5 bills, 10: $20 bills
E
12: $5 bills, 14: $20 bills
Example #2
Solve each word problem.
Charlotte has a piggy bank with nickels, quarters, and dimes only. The piggy bank has a total of $61.50. The number of dimes is 60 less than the number of nickels. Additionally, if the number of quarters was reduced by 10, the number of dimes would be three times the amount of quarters. How much of each type of coin is in Charlotte’s piggy bank?
Please choose the best answer.
A
n: 455, d: 100, q: 19
B
n: 300, d: 240, q: 90
C
n: 500, d: 440, q: 176
D
n: 132, d: 72, q: 54
E
n: 292, d: 177, q: 55
Example #3
Solve each word problem.
Chloe works at the local fair. In order to play the games, customers must exchange money for one of two coin types, a 20 unit coin, or a 3 unit coin. At the end of Chloe’s shift, her coin bag has a total value of 2057 units. If she has 11 more 3 unit coins than 20 unit coins, how much of each coin type does she have in her bag?
Please choose the best answer.
A
20 unit: 10, 3 unit: 21
B
20 unit: 88, 3 unit: 99
C
20 unit: 101, 3 unit: 112
D
20 unit: 50, 3 unit: 61
E
20 unit: 70, 3 unit: 81
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