3 Quarters Is How Much
ix.3: Solve Money Applications
- Page ID
- 5006
- Solve coin discussion problems
- Solve ticket and stamp give-and-take problems
Before you get started, take this readiness quiz.
- Multiply: 14(0.25). If yous missed this problem, review Example 5.3.5.
- Simplify: 100(0.2 + 0.05n). If you missed this problem, review Example seven.four.6.
- Solve: 0.25x + 0.10(x + 4) = ii.5 If y'all missed this problem, review Example 8.6.8.
Solve Coin Word Problems
Imagine taking a handful of coins from your pocket or purse and placing them on your desk-bound. How would you make up one's mind the value of that pile of coins?
If yous tin can form a footstep-past-step program for finding the total value of the coins, it volition help you as you brainstorm solving coin word issues.
I way to bring some order to the mess of coins would be to separate the coins into stacks according to their value. Quarters would go with quarters, dimes with dimes, nickels with nickels, and and so on. To get the total value of all the coins, you would add the full value of each pile.
Figure \(\PageIndex{i}\) - To determine the full value of a stack of nickels, multiply the number of nickels times the value of one nickel.(Credit: Darren Hester via ppdigital)
How would you make up one's mind the value of each pile? Remember almost the dime pile—how much is it worth? If you count the number of dimes, you'll know how many you have—the number of dimes.
But this does non tell yous the value of all the dimes. Say you counted 17 dimes, how much are they worth? Each dime is worth $0.10 —that is the value of one dime. To observe the total value of the pile of 17 dimes, multiply 17 by $0.10 to get $1.70. This is the total value of all 17 dimes.
\[\begin{split} 17 \cdot \$0.x &= \$ 1.70 \\ number\; \cdot value &= full\; value \cease{split up}\]
For coins of the same type, the total value can be found every bit follows:
\[number\; \cdot value = total\; value\]
where number is the number of coins, value is the value of each money, and total value is the total value of all the coins.
Yous could continue this procedure for each blazon of coin, and so you would know the total value of each type of coin. To go the total value of all the coins, add the total value of each type of coin.
Let'southward look at a specific case. Suppose in that location are 14 quarters, 17 dimes, 21 nickels, and 39 pennies. Nosotros'll make a table to organize the information – the type of coin, the number of each, and the value.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Quarters | 14 | 0.25 | 3.50 |
| Dimes | 17 | 0.10 | 1.70 |
| Nickels | 21 | 0.05 | 1.05 |
| Pennies | 39 | 0.01 | 0.39 |
| half-dozen.64 |
The total value of all the coins is $6.64. Notice how Table \(\PageIndex{one}\) helped u.s. organize all the data. Permit's encounter how this method is used to solve a coin word problem.
Adalberto has $ii.25 in dimes and nickels in his pocket. He has nine more than nickels than dimes. How many of each blazon of coin does he have?
Solution
Pace 1. Read the problem. Make certain you empathise all the words and ideas.
- Make up one's mind the types of coins involved.
Think about the strategy we used to discover the value of the scattering of coins. The first thing you need is to find what types of coins are involved. Adalberto has dimes and nickels.
- Create a table to organize the information.
- Label the columns 'type', 'number', 'value', 'total value'.
- List the types of coins.
- Write in the value of each blazon of money.
- Write in the total value of all the coins.
We tin work this trouble all in cents or in dollars. Here we volition do information technology in dollars and put in the dollar sign ($) in the table as a reminder.
The value of a dime is $0.ten and the value of a nickel is $0.05. The total value of all the coins is $2.25.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Dimes | 0.10 | ||
| Nickels | 0.05 | ||
| 2.25 |
Step 2. Identify what you are looking for.
- We are asked to find the number of dimes and nickels Adalberto has.
Step iii. Name what you are looking for.
- Apply variable expressions to correspond the number of each type of money.
- Multiply the number times the value to get the total value of each type of coin. In this problem you cannot count each type of coin—that is what y'all are looking for—only yous have a inkling. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.
- Let d = number of dimes.
- d + 9 = number of nickels
- Fill in the "number" column to help get everything organized.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Dimes | d | 0.ten | |
| Nickels | d + nine | 0.05 | |
| ii.25 |
Now we accept all the information nosotros need from the trouble!
You multiply the number times the value to go the total value of each type of coin. While you do not know the bodily number, yous do take an expression to correspond it.
And so now multiply number • value and write the results in the Total Value column.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Dimes | d | 0.10 | 0.10d |
| Nickels | d + 9 | 0.05 | 0.05(d + 9) |
| ii.25 |
Footstep 4. Interpret into an equation. Restate the problem in one sentence. Then translate into an equation.
Stride 5. Solve the equation using good algebra techniques.
| Write the equation. | $$0.10d + 0.05(d + 9) = 2.25$$ |
| Distribute. | $$0.10d + 0.05d + 0.45 = 2.25$$ |
| Combine similar terms. | $$0.15d + 0.45 = ii.25$$ |
| Subtract 0.45 from each side. | $$0.15d = one.eighty$$ |
| Carve up to find the number of dimes. | $$d = 12$$ |
| The number of nickels is d + ix. | $$\begin{split} d + 9& \\ \textcolor{red}{12} + 9& \\ 21& \stop{split}$$ |
Step 6. Check.
\[\begin{carve up} 12\; dimes:\; 12(0.10) &= one.twenty \\ 21\; nickels:\; 21(0.05) &= 1.05 \\ \hline &\quad \$ 2.25\; \checkmark \end{split}\]
Pace 7. Answer the question.
Adalberto has twelve dimes and xx-one nickels.
If this were a homework practice, our work might await similar this:
Bank check:
\[\brainstorm{split} 12\; dimes \quad 12(0.10) &= 1.20 \\ 21\; nickels \quad 21(0.05) &= ane.05 \\ \hline &\quad \$ 2.25 \finish{separate}\]
Michaela has $2.05 in dimes and nickels in her change purse. She has vii more dimes than nickels. How many coins of each type does she have?
- Answer
-
9 nickels, xvi dimes
Liliana has $ii.ten in nickels and quarters in her backpack. She has 12 more nickels than quarters. How many coins of each type does she have?
- Reply
-
17 nickels, five quarters
Step one. Read the trouble. Make sure you understand all the words and ideas, and create a table to organize the information.
Step 2. Identify what you are looking for.
Step 3. Name what you are looking for. Choose a variable to represent that quantity.
- Use variable expressions to represent the number of each blazon of coin and write them in the tabular array.
- Multiply the number times the value to get the full value of each blazon of coin.
Pace iv. Translate into an equation. Write the equation by adding the total values of all the types of coins.
Step 5. Solve the equation using good algebra techniques.
Pace vi. Bank check the answer in the trouble and brand sure it makes sense.
Pace 7. Answer the question with a consummate sentence.
You may find information technology helpful to put all the numbers into the table to make sure they bank check.
| Type | Number | Value ($) | Total Value ($) | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
Maria has $two.43 in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each blazon does she accept?
Solution
Step ane. Read the trouble.
- Determine the types of coins involved. We know that Maria has quarters and pennies.
- Create a table to organize the information.
- Characterization the columns type, number, value, total value.
- Listing the types of coins.
- Write in the value of each type of money.
- Write in the total value of all the coins.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Quarters | 0.25 | ||
| Pennies | 0.01 | ||
| 2.43 |
Footstep 2. Identify what you are looking for.
- We are looking for the number of quarters and pennies.
Step 3. Name: Stand for the number of quarters and pennies using variables.
- We know Maria has twice as many pennies as quarters. The number of pennies is defined in terms of quarters.
- Let q represent the number of quarters. So the number of pennies is 2q.
| Type | Number | Value ($) | Full Value ($) |
|---|---|---|---|
| Quarters | q | 0.25 | |
| Pennies | 2q | 0.01 | |
| 2.43 |
Multiply the 'number' and the 'value' to get the 'total value' of each type of coin.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Quarters | q | 0.25 | 0.25q |
| Pennies | 2q | 0.01 | 0.01(2q) |
| 2.43 |
Step four. Translate. Write the equation by adding the 'total value' of all the types of coins.
Footstep 5. Solve the equation.
| Write the equation. | $$0.25q + 0.01(2q) = 2.43$$ |
| Multiply. | $$0.25q + 0.02q = 2.43$$ |
| Combine like terms. | $$0.27q = 2.43$$ |
| Divide by 0.27. | $$q = 9\; quarters$$ |
| The number of pennies is 2q. | $$\begin{dissever} &2q \\ &2\; \cdot\; \textcolor{red}{9} \\ &18\; pennies \stop{carve up}$$ |
Step half dozen. Cheque the respond in the problem. Maria has 9 quarters and eighteen pennies. Does this make $2.43?
\[\begin{separate} ix\; quarters \quad 9(0.25) &= ii.25 \\ 18\; pennies \quad 18(0.01) &= 0.18 \\ \hline Total \qquad \qquad \qquad \quad &\quad \$ ii.43\; \checkmark \stop{separate}\]
Step 7. Answer the question. Maria has nine quarters and eighteen pennies.
Sumanta has $iv.xx in nickels and dimes in her desk drawer. She has twice equally many nickels as dimes. How many coins of each type does she take?
- Reply
-
42 nickels, 21 dimes
Alison has three times every bit many dimes as quarters in her bag. She has $9.35 birthday. How many coins of each type does she accept?
- Answer
-
51 dimes, 17 quarters
In the next example, we'll prove merely the completed table—make sure y'all sympathise how to make full it in step past pace.
Danny has $2.fourteen 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?
Solution
| Step 1: Read the trouble. | |
| Determine the types of coins involved. Create a table. | Pennies and nickels |
| Write in the value of each type of money. | Pennies are worth $0.01. Nickels are worth $0.05. |
| Footstep ii: Identify what you lot are looking for. | the number of pennies and nickels |
| Step 3: Proper noun. Stand for the number of each type of coin using variables. The number of nickels is defined in terms of the number of pennies, so start with pennies. | Let p = number of pennies |
| The number of nickels is two more then times the number of pennies. | 10p + 2 = number of nickels |
Multiply the number and the value to go the total value of each type of coin.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| pennies | p | 0.01 | 0.01p |
| nickels | 10p + 2 | 0.05 | 0.05(10p + 2) |
| $2.xiv |
Step 4. Translate: Write the equation by adding the total value of all the types of coins.
Footstep 5. Solve the equation.
| $$0.01p + 0.50p + 0.10 = 2.xiv$$ | |
| $$0.51p + 0.ten = 2.14$$ | |
| $$0.51p = 2.04$$ | |
| $$p = 4\; pennies$$ | |
| How many nickels? | $$10p + two$$ |
| $$10(\textcolor{cherry}{4}) + 2$$ | |
| $$42\; nickels$$ |
Step half-dozen. Check. Is the total value of iv pennies and 42 nickels equal to $two.14?
\[\begin{dissever} 4(0.01) + 42(0.05) &\stackrel{?}{=} 2.14 \\ two.fourteen &= two.14\; \checkmark \end{split}\]
Step 7. Answer the question. Danny has four pennies and 42 nickels.
Jesse has $6.55 worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse take?
- Respond
-
41 nickels, 18 quarters
Elaine has $vii.00 in dimes and nickels in her coin jar. The number of dimes that Elaine has is seven less than iii times the number of nickels. How many of each coin does Elaine take?
- Reply
-
22 nickels, 59 dimes
Solve Ticket and Postage Word Problems
The strategies we used for coin problems tin exist hands applied to some other kinds of problems as well. Problems involving tickets or stamps are very similar to money problems, for example. Similar coins, tickets and stamps have different values; so nosotros can organize the information in tables much similar nosotros did for money problems.
At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $vi each and adult tickets sold for $9 each. The number of adult tickets sold was 5 less than three times the number of pupil tickets sold. How many student tickets and how many adult tickets were sold?
Solution
Step 1: Read the trouble.
- Make up one's mind the types of tickets involved. There are educatee tickets and developed tickets.
- Create a tabular array to organize the information.
| Type | Number | Value ($) | Full Value ($) |
|---|---|---|---|
| Student | half dozen | ||
| Adult | 9 | ||
| 1,506 |
Step two. Identify what you lot are looking for. Nosotros are looking for the number of student and adult tickets.
Step iii. Name. Correspond the number of each type of ticket using variables.
- We know the number of adult tickets sold was 5 less than three times the number of pupil tickets sold. Permit due south be the number of student tickets.
- Then 3s − five is the number of adult tickets.
- Multiply the number times the value to become the full value of each blazon of ticket.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Educatee | s | 6 | 6s |
| Developed | 3s - v | ix | 9(9s - 5) |
| i,506 |
Step 4. Translate: Write the equation past adding the total values of each blazon of ticket.
\[6s + 9(3s − 5) = 1506\]
Pace 5. Solve the equation.
\[\begin{dissever} 6s + 27s − 45 &= 1506 \\ 33s − 45 &= 1506 \\ 33s &= 1551 \\ s &= 47\; students \end{carve up}\]
Substitute to find the number of adults.
\[\begin{separate} 3s - 5 &= number\; of\; adults \\ 3(\textcolor{red}{47}) - 5 &= 136\; adults \end{dissever}\]
Footstep six. Bank check. In that location were 47 educatee tickets at $six each and 136 adult tickets at $9 each. Is the total value $1506? We find the total value of each type of ticket by multiplying the number of tickets times its value; we and then add together to become the total value of all the tickets sold.
\[\begin{split} 47 \cdot 6 &= 282 \\ 136 \cdot 9 &= 1224 \\ \hline &\quad 1506 \finish{split}\]
Step 7. Answer the question. They sold 47 educatee tickets and 136 adult tickets.
The kickoff day of a water polo tournament, the full value of tickets sold was $17,610. One-mean solar day passes sold for $xx and tournament passes sold for $xxx. The number of tournament passes sold was 37 more than than the number of day passes sold. How many day passes and how many tournament passes were sold?
- Reply
-
330 day passes, 367 tournament passes
At the movie theater, the full value of tickets sold was $ii,612.fifty. Adult tickets sold for $10 each and senior/ child tickets sold for $seven.50 each. The number of senior/child tickets sold was 25 less than twice the number of adult tickets sold. How many senior/child tickets and how many developed tickets were sold?
- Reply
-
112 adult tickets, 199 senior/child tickets
Now we'll exercise one where we make full in the tabular array all at once.
Monica paid $x.44 for stamps she needed to postal service the invitations to her sister's babe shower. The number of 49-cent stamps was four more than twice the number of eight-cent stamps. How many 49-cent stamps and how many viii-cent stamps did Monica buy?
Solution
The blazon of stamps are 49-cent stamps and viii-cent stamps. Their names also give the value. "The number of 49 cent stamps was 4 more than than twice the number of 8 cent stamps."
Let ten = number of 8-cent stamps
2x + four = number of 49-cent stamps
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| 49-cent stamps | 2x + 4 | 0.49 | 0.49(2x + 4) |
| viii-cent stamps | x | 0.08 | 0.08x |
| ten.44 |
| Write the equation from the full values. | $$0.49(2x + 4) + 0.08x = 10.44$$ |
| Solve the equation. | $$\brainstorm{split up} 0.98x + i.96 + 0.08x &= 10.44 \\ 1.06x + one.96 &= x.44 \\ 1.06x &= 8.48 \\ x &= eight \cease{split up}$$ |
| Monica bought 8 8-cent stamps. | |
| Find the number of 49-cent stamps she bought by evaluating. | 2x + 4 for x = 8.$$\begin{dissever} &2x + 4 \\ &two \cdot 8 + 4 \\ &16 + 4 \\ &20 \finish{split}$$ |
| Cheque. | $$\begin{split} 8(0.08) + 20(0.49) &\stackrel{?}{=} 10.44 \\ 0.64 + 9.80 &\stackrel{?}{=} x.44 \\ x.44 &= 10.44\; \checkmark \end{split}$$ |
Monica bought eight 8-cent stamps and xx 49-cent stamps.
Eric paid $16.64 for stamps so he could mail service thank y'all notes for his wedding ceremony gifts. The number of 49-cent stamps was 8 more than twice the number of 8-cent stamps. How many 49-cent stamps and how many eight-cent stamps did Eric buy?
- Reply
-
32 at 49 cents, 12 at 8 cents
Kailee paid $xiv.84 for stamps. The number of 49-cent stamps was 4 less than 3 times the number of 21-cent stamps. How many 49-cent stamps and how many 21-cent stamps did Kailee buy?
- Respond
-
26 at 49 cents, 10 at 21 cents
Practise Makes Perfect
Solve Coin Discussion Issues
In the following exercises, solve the coin word problems.
- Jaime has $two.threescore in dimes and nickels. The number of dimes is 14 more than the number of nickels. How many of each coin does he have?
- Lee has $1.75 in dimes and nickels. The number of nickels is 11 more than the number of dimes. How many of each coin does he have?
- Ngo has a drove of dimes and quarters with a full value of $3.l. The number of dimes is 7 more than the number of quarters. How many of each coin does he have?
- Connor has a collection of dimes and quarters with a full value of $6.thirty. The number of dimes is xiv more than than the number of quarters. How many of each coin does he have?
- Carolyn has $2.55 in her bag in nickels and dimes. The number of nickels is 9 less than iii times the number of dimes. Find the number of each type of coin.
- Julio has $two.75 in his pocket in nickels and dimes. The number of dimes is 10 less than twice the number of nickels. Find the number of each type of coin.
- Chi has $11.xxx in dimes and quarters. The number of dimes is 3 more iii times the number of quarters. How many dimes and nickels does Chi have?
- Tyler has $ix.lxx in dimes and quarters. The number of quarters is 8 more than four times the number of dimes. How many of each coin does he have?
- A greenbacks box of $1 and $5 bills is worth $45. The number of $1 bills is 3 more than the number of $5 bills. How many of each bill does it incorporate?
- Joe's wallet contains $1 and $5 bills worth $47. The number of $one bills is 5 more than the number of $5 bills. How many of each bill does he have?
- In a cash drawer there is $125 in $5 and $10 bills. The number of $10 bills is twice the number of $5 bills. How many of each are in the drawer?
- John has $175 in $5 and $x bills in his drawer. The number of $v bills is three times the number of $10 bills. How many of each are in the drawer?
- Mukul has $3.75 in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and ix more than nickels than quarters. How many of each money are in his pocket?
- Vina has $iv.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin are in her purse?
Solve Ticket and Stamp Word Problems
In the following exercises, solve the ticket and postage word problems.
- The play took in $550 one night. The number of $8 adult tickets was x less than twice the number of $5 kid tickets. How many of each ticket were sold?
- If the number of $8 child tickets is seventeen less than three times the number of $12 adult tickets and the theater took in $584, how many of each ticket were sold?
- The movie theatre took in $1,220 one Mon dark. The number of $7 child tickets was ten more than twice the number of $9 developed tickets. How many of each were sold?
- The ball game took in $one,340 1 Saturday. The number of $12 adult tickets was 15 more than than twice the number of $5 child tickets. How many of each were sold?
- Julie went to the post office and bought both $0.49 stamps and $0.34 postcards for her office's bills She spent $62.60. The number of stamps was xx more than twice the number of postcards. How many of each did she buy?
- Before he left for college out of state, Jason went to the post office and bought both $0.49 stamps and $0.34 postcards and spent $12.52. The number of stamps was iv more than than twice the number of postcards. How many of each did he buy?
- Maria spent $16.80 at the post office. She bought iii times as many $0.49 stamps every bit $0.21 stamps. How many of each did she purchase?
- Hector spent $43.twoscore at the post office. He bought 4 times as many $0.49 stamps as $0.21 stamps. How many of each did he buy?
- Hilda has $210 worth of $10 and $12 stock shares. The numbers of $10 shares is 5 more than than twice the number of $12 shares. How many of each does she accept?
- Mario invested $475 in $45 and $25 stock shares. The number of $25 shares was five less than three times the number of $45 shares. How many of each type of share did he buy?
Everyday Math
- Parent Volunteer Every bit the treasurer of her daughter's Girl Scout troop, Laney nerveless money for some girls and adults to get to a 3-twenty-four hour period camp. Each girl paid $75 and each adult paid $thirty. The total corporeality of money collected for camp was $765. If the number of girls is iii times the number of adults, how many girls and how many adults paid for army camp?
- Parent Volunteer Laurie was completing the treasurer's report for her son's Boy Scout troop at the end of the school twelvemonth. She didn't remember how many boys had paid the $24 full-yr registration fee and how many had paid a $16 partial-year fee. She knew that the number of boys who paid for a full-twelvemonth was ten more than the number who paid for a partial-year. If $400 was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?
Writing Exercises
- Suppose you take 6 quarters, 9 dimes, and four pennies. Explicate how you find the full value of all the coins.
- Practice you find information technology helpful to utilize a table when solving money problems? Why or why not?
- In the table used to solve money issues, one cavalcade is labeled "number" and some other cavalcade is labeled '"value." What is the difference between the number and the value?
- What similarities and differences did you come across betwixt solving the coin problems and the ticket and stamp problems?
Cocky Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this department.
(b) Subsequently reviewing this checklist, what will you lot do to get confident for all objectives?
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed nether Artistic Commons Attribution License v4.0 "Download for gratuitous at http://cnx.org/contents/fd53eae1-fa2...49835c3c@v.191."
3 Quarters Is How Much,
Source: https://math.libretexts.org/Bookshelves/PreAlgebra/Book%3A_Prealgebra_(OpenStax)/09%3A_Math_Models_and_Geometry/9.03%3A_Solve_Money_Applications
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