To determine the overall monetary worth of a collection of identical coins, the physical count of the coins must be multiplied by the monetary value of a single coin. For example, to find the total value of a pile containing $$23$$ dimes, first identify that the value of one dime is $$\$0.10$$. Then, multiply the number of dimes by the value of each dime to calculate the total amount: $$23 \cdot \$0.10 = \$2.30$$. This specific example illustrates the fundamental principle that counting coins yields the physical quantity, but multiplying by the unit value is necessary to find the monetary total.

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For a collection of coins that are all the same type, the **total value** of those coins is found by multiplying the count of coins by the value of a single coin:

$$number \cdot value = total value$$

- $$number$$ is how many coins of that type are present
- $$value$$ is the monetary worth of one such coin
- $$total value$$ is the combined worth of all those coins

For example, if there are $$23$$ dimes and each dime is worth $$\$0.10$$, then the total value of the dimes is $$23 \cdot \$0.10 = \$2.30$$. When a collection contains several coin types (quarters, dimes, nickels, pennies), this model is applied separately to each type and the results are added to obtain the overall total value.

Total Value of Coins Model

https://openstax.org/details/books/intermediate-algebra-2e

OpenStax Intermediate Algebra

Apply the seven-step problem-solving strategy and the total value of coins model to a coin mixture problem in which the number of each coin type is unknown but the coin types are related by a stated condition.

**Problem:** Adalberto has \$2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?

1. **Read** the problem and identify the coin types involved: dimes (worth \$0.10 each) and nickels (worth \$0.05 each). The total value of all coins is \$2.25.
2. **Identify** what to find: the number of dimes and the number of nickels.
3. **Name** the unknowns using a single variable. Let $$d$$ = the number of dimes. Because there are nine more nickels than dimes, the number of nickels is $$d + 9$$. Organize everything in a table:

| Type | Number | Value (\$) | Total Value (\$) |
|---|---|---|---|
| Dimes | $$d$$ | 0.10 | $$0.10d$$ |
| Nickels | $$d + 9$$ | 0.05 | $$0.05(d + 9)$$ |
| | | | 2.25 |

4. **Translate** into an equation by adding the total values for each coin type and setting the sum equal to the overall total:

$$0.10d + 0.05(d + 9) = 2.25$$

5. **Solve** the equation:

- Distribute $$0.05$$: $$0.10d + 0.05d + 0.45 = 2.25$$
- Combine like terms: $$0.15d + 0.45 = 2.25$$
- Subtract $$0.45$$ from both sides: $$0.15d = 1.80$$
- Divide both sides by $$0.15$$: $$d = 12$$

Find the number of nickels: $$d + 9 = 12 + 9 = 21$$.

6. **Check**: $$12$$ dimes are worth $$12(0.10) = 1.20$$, and $$21$$ nickels are worth $$21(0.05) = 1.05$$. Adding: $$1.20 + 1.05 = 2.25$$ $$\checkmark$$
7. **Answer**: Adalberto has twelve dimes and twenty-one nickels.

This example extends the total value model from situations with known coin counts to ones where the counts are unknown. The relationship "nine more nickels than dimes" allows both quantities to be expressed through a single variable ($$d$$ and $$d + 9$$). Multiplying each variable expression by the coin's unit value and then summing produces an equation with decimal coefficients that requires distributing, combining like terms, and dividing — standard linear-equation techniques applied in a real-world context.

Solving a Coin Mixture Problem: Dimes and Nickels

Apply the seven-step problem-solving strategy and the total value of coins model to a coin mixture problem in which the relationship between the two coin counts involves both multiplication and addition.

**Problem:** Danny has \$2.14 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?

1. **Read** the problem and identify the coin types involved: pennies (worth \$0.01 each) and nickels (worth \$0.05 each). The total value of all coins is \$2.14.
2. **Identify** what to find: the number of pennies and the number of nickels.
3. **Name** the unknowns using a single variable. Since the number of nickels is defined in terms of the number of pennies, start with pennies. Let $$p$$ = the number of pennies. The phrase "two more than ten times" combines multiplication by $$10$$ with adding $$2$$, so the number of nickels is $$10p + 2$$. Organize everything in a table:

| Type | Number | Value (\$) | Total Value (\$) |
|---|---|---|---|
| Pennies | $$p$$ | 0.01 | $$0.01p$$ |
| Nickels | $$10p + 2$$ | 0.05 | $$0.05(10p + 2)$$ |
| | | | 2.14 |

4. **Translate** into an equation by adding the total values and setting the sum equal to the overall total:

$$0.01p + 0.05(10p + 2) = 2.14$$

5. **Solve** the equation:

- Distribute $$0.05$$: $$0.01p + 0.50p + 0.10 = 2.14$$
- Combine like terms: $$0.51p + 0.10 = 2.14$$
- Subtract $$0.10$$ from both sides: $$0.51p = 2.04$$
- Divide both sides by $$0.51$$: $$p = 4$$

Find the number of nickels: $$10(4) + 2 = 42$$.

6. **Check**: $$4(0.01) + 42(0.05) \stackrel{?}{=} 2.14$$ → $$0.04 + 2.10 = 2.14$$ $$\checkmark$$
7. **Answer**: Danny has four pennies and 42 nickels.

This example extends the coin mixture technique to a case where the relationship between the two coin counts involves **both multiplication and addition** — "two more than ten times" translates to $$10p + 2$$ — rather than a simple additive relationship like "nine more than." The expression $$10p + 2$$ introduces a larger coefficient ($$10$$) inside the parentheses, so distributing $$0.05$$ across $$10p + 2$$ produces $$0.50p$$ rather than a small decimal term. After distribution, combining $$0.01p + 0.50p$$ yields the non-obvious coefficient $$0.51$$, making the final division step ($$2.04 \div 0.51$$) more challenging than in simpler coin problems.

Solving a Coin Mixture Problem: Pennies and Nickels

Apply the seven-step problem-solving strategy and the total-value model to a ticket mixture problem where each ticket type has a whole-dollar value and the relationship between ticket counts involves both multiplication and subtraction.

**Problem:** At a school concert, the total value of tickets sold was \$1,506. Student tickets sold for \$6 each and adult tickets sold for \$9 each. The number of adult tickets sold was five less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

1. **Read** the problem and identify the types involved: student tickets (worth \$6 each) and adult tickets (worth \$9 each). The total value of all tickets is \$1,506.
2. **Identify** what to find: the number of student tickets and the number of adult tickets.
3. **Name** the unknowns using a single variable. Let $$s$$ = the number of student tickets. The phrase "five less than three times" combines multiplication by $$3$$ with subtracting $$5$$, so the number of adult tickets is $$3s - 5$$. Organize in a table:

| Type | Number | Value (\$) | Total Value (\$) |
|---|---|---|---|
| Student | $$s$$ | 6 | $$6s$$ |
| Adult | $$3s - 5$$ | 9 | $$9(3s - 5)$$ |
| | | | 1,506 |

4. **Translate** into an equation by adding the total values and setting the sum equal to the overall total:

$$6s + 9(3s - 5) = 1506$$

5. **Solve** the equation:

- Distribute $$9$$: $$6s + 27s - 45 = 1506$$
- Combine like terms: $$33s - 45 = 1506$$
- Add $$45$$ to both sides: $$33s = 1551$$
- Divide both sides by $$33$$: $$s = 47$$

Find the number of adult tickets: $$3(47) - 5 = 141 - 5 = 136$$.

6. **Check**: $$47 \cdot 6 = 282$$ and $$136 \cdot 9 = 1{,}224$$. Adding: $$282 + 1{,}224 = 1{,}506$$ $$\checkmark$$
7. **Answer**: They sold 47 student tickets and 136 adult tickets.

This example demonstrates that ticket and stamp problems follow the same total-value model used for coin problems — each type of ticket has a fixed value, just as each type of coin does. Here the relationship "five less than three times" translates to $$3s - 5$$, and distributing the \$9 ticket price across $$3s - 5$$ produces whole-number coefficients ($$27s$$ and $$-45$$) that are easier to work with than the decimal coefficients typical of coin problems. After distribution, combining $$6s + 27s$$ into $$33s$$ and solving the resulting two-step equation yields the answer.

Solving a Ticket Mixture Problem: Student and Adult Tickets

Apply the seven-step problem-solving strategy and the total-value model to a stamp mixture problem where each stamp type has a decimal value and the relationship between the two stamp counts involves both multiplication and addition.

**Problem:** Monica paid \$8.36 for stamps. The number of 41-cent stamps was four more than twice the number of two-cent stamps. How many 41-cent stamps and how many two-cent stamps did Monica buy?

1. **Read** the problem and identify the types involved: 41-cent stamps (worth \$0.41 each) and two-cent stamps (worth \$0.02 each). The total value of all stamps is \$8.36.
2. **Identify** what to find: the number of 41-cent stamps and the number of two-cent stamps.
3. **Name** the unknowns using a single variable. Let $$x$$ = the number of two-cent stamps. The phrase "four more than twice" combines multiplication by $$2$$ with adding $$4$$, so the number of 41-cent stamps is $$2x + 4$$. Organize in a table:

| Type | Number | Value (\$) | Total Value (\$) |
|---|---|---|---|
| 41-cent stamps | $$2x + 4$$ | 0.41 | $$0.41(2x + 4)$$ |
| 2-cent stamps | $$x$$ | 0.02 | $$0.02x$$ |
| | | | 8.36 |

4. **Translate** into an equation by adding the total values and setting the sum equal to the overall total:

$$0.41(2x + 4) + 0.02x = 8.36$$

5. **Solve** the equation:

- Distribute $$0.41$$: $$0.82x + 1.64 + 0.02x = 8.36$$
- Combine like terms: $$0.84x + 1.64 = 8.36$$
- Subtract $$1.64$$ from both sides: $$0.84x = 6.72$$
- Divide both sides by $$0.84$$: $$x = 8$$

Find the number of 41-cent stamps: $$2(8) + 4 = 16 + 4 = 20$$.

6. **Check**: $$8(0.02) + 20(0.41) \stackrel{?}{=} 8.36$$ → $$0.16 + 8.20 = 8.36$$ $$\checkmark$$
7. **Answer**: Monica bought eight two-cent stamps and twenty 41-cent stamps.

This example demonstrates that the total-value model applies to stamps just as it does to coins and tickets — each stamp type has a fixed per-unit value, and the product of count times value gives the total value for that type. Here the relationship "four more than twice" translates to $$2x + 4$$, and distributing the decimal value $$0.41$$ across $$2x + 4$$ produces $$0.82x + 1.64$$. After combining $$0.82x + 0.02x$$ into $$0.84x$$, the final division step $$6.72 \div 0.84$$ requires careful decimal arithmetic — a skill reinforced by working with non-standard item values like $$0.41$$ rather than the rounder values typical of coins.

Solving a Stamp Mixture Problem: 41-Cent and 2-Cent Stamps

Apply the seven-step problem-solving strategy, the mixture model, and the known-total technique to a mixture word problem involving food products measured by weight.

Problem: Henning is mixing raisins and nuts to make $$25$$ pounds of trail mix. Raisins cost \$4.50 a pound and nuts cost \$8 a pound. If Henning wants his cost for the trail mix to be \$6.60 a pound, how many pounds of raisins and how many pounds of nuts should he use?

1. Read the problem and identify the two ingredient types: raisins (worth \$4.50 per pound) and nuts (worth \$8 per pound). The total mixture is $$25$$ pounds at a target cost of \$6.60 per pound.
2. Identify what to find: the number of pounds of raisins and the number of pounds of nuts.
3. Name the unknowns using a single variable. Because the total weight is known, use the known-total technique. Let $$x =$$ the number of pounds of raisins. Then the number of pounds of nuts is $$25 - x$$. Organize the information in a table:

| Type | Number of Pounds | Price per pound (\$) | Total Value (\$) |
|---|---|---|---|
| Raisins | $$x$$ | $$4.50$$ | $$4.50x$$ |
| Nuts | $$25 - x$$ | $$8$$ | $$8(25 - x)$$ |
| Trail mix | $$25$$ | $$6.60$$ | $$25(6.60) = 165$$ |

4. Translate into an equation. The value of the raisins plus the value of the nuts equals the value of the trail mix:
$$4.50x + 8(25 - x) = 165$$

5. Solve the equation:
- Distribute $$8$$: $$4.50x + 200 - 8x = 165$$
- Combine like terms: $$-3.50x + 200 = 165$$
- Subtract $$200$$ from both sides: $$-3.50x = -35$$
- Divide both sides by $$-3.50$$: $$x = 10$$
Find the number of pounds of nuts: $$25 - 10 = 15$$.

6. Check: $$4.50(10) + 8(15) \stackrel{?}{=} 25(6.60)$$
$$45 + 120 \stackrel{?}{=} 165$$
$$165 = 165$$ $$\checkmark$$

7. Answer: Henning mixed $$10$$ pounds of raisins with $$15$$ pounds of nuts.

This example demonstrates that the mixture model ($$	ext{amount} \cdot 	ext{value} = 	ext{total value}$$) applies to products sold by weight. The known-total technique produces the expression $$25 - x$$ for the second ingredient, and solving the equation involves working with decimals.

Solving a Mixture Word Problem: Trail Mix (Raisins and Nuts)

When solving a word problem that involves coins of different types, the general seven-step problem-solving strategy is adapted with organizational tools specific to coin contexts. The revised steps are:

1. **Read** the problem. Make sure all the words and ideas are understood. Determine which types of coins are involved.
   - 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 type of coin.
   - Write in the total value of all the coins.
2. **Identify** what you are looking for.
3. **Name** what you are looking for. Choose a variable to represent that quantity. Use variable expressions to represent the number of each type of coin and write them in the table. Multiply the number times the value to get the total value of each type of coin.
4. **Translate** into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then translate the sentence into an equation by adding the total values of all the types of coins.
5. **Solve** the equation using good algebra techniques.
6. **Check** the answer in the problem and make sure it makes sense.
7. **Answer** the question with a complete sentence.

The key adaptations for coin problems are in Step 1 and Step 3: the four-column table keeps the coin types, their counts, their per-unit values, and their total values organized in one place, making it straightforward to build the equation in Step 4 by summing the entries in the Total Value column.

Problem-Solving Strategy for Coin Word Problems

In a professional setting such as a bank or retail store, which formula is used to calculate the total value of a collection of identical coins?

Match each component of the coin value model to its correct description as used in professional cash handling and accounting.

When a retail associate is performing a drawer count at the end of a shift, they calculate the total value of a specific coin type by multiplying the number of coins by the ____ of a single coin.

In a retail cash-handling procedure, the total value of a specific denomination of coins in a register drawer is determined by multiplying the total number of those coins by the monetary value of a single coin.

A retail associate is performing an end-of-shift cash count and needs to determine the total value of the nickels in their drawer. Arrange the steps of the 'Total Value of Coins Model' in the correct logical sequence to complete this task.

In a professional business environment, such as a retail outlet or financial institution, managers often use a specific mathematical model to perform a cash audit of coin denominations. State the mathematical formula used to find the total value of a collection of identical coins, and define each of the three terms in that formula.

Mathematical Model for Cash Audits

Based on the professional standard described, identify the mathematical formula used in the 'Total Value of Coins Model' and provide the definition for each of the three components involved in the calculation.

Professional Cash Handling and Audit Procedures

During a professional cash audit or a retail drawer count, an associate must calculate the total value of each coin denomination (such as dimes or quarters) individually. Describe the mathematical model used to determine the total value of a collection of identical coins. In your description, identify the two specific variables required for the calculation and state the mathematical operation used to combine them.

Standardized Calculation for Cash Denominations

In a professional cash-handling environment, what is the standard procedure for determining the overall value of a collection that contains several different types of coins, such as quarters, dimes, and nickels?

A banking associate is reviewing the 'Total Value of Coins Model' training manual. According to the specific example provided in the model, if an associate counts 17 dimes and each dime is worth $0.10, what is the resulting 'total value' of that collection?

To find the total value of a collection containing different types of coins, apply the total-value model ($$number \cdot value = total value$$) to each coin type separately, then add the results. Organizing the information in a table with columns for Type, Number, Value, and Total Value keeps the work clear. For instance, given $$14$$ quarters, $$17$$ dimes, $$21$$ nickels, and $$39$$ pennies:

| Type | Number | Value | Total Value |
| --- | --- | --- | --- |
| Quarters | $$14$$ | $$\$0.25$$ | $$\$3.50$$ |
| Dimes | $$17$$ | $$\$0.10$$ | $$\$1.70$$ |
| Nickels | $$21$$ | $$\$0.05$$ | $$\$1.05$$ |
| Pennies | $$39$$ | $$\$0.01$$ | $$\$0.39$$ |

Adding the four total-value amounts gives $$\$3.50 + \$1.70 + \$1.05 + \$0.39 = \$6.64$$, so the entire collection is worth $$\$6.64$$.

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