➕Pre-Algebra Unit 9 Review

Money word problems show up constantly in pre-algebra, and they all follow the same core idea: total value = number of items × value per item. Once you see that pattern, every coin, ticket, and stamp problem works the same way.

Coin and Bill Word Problems

The key to coin problems is knowing what each coin is worth:

Pennies: $\$0.01$ each
Nickels: $\$0.05$ each
Dimes: $\$0.10$ each
Quarters: $\$0.25$ each

Bills come in denominations of $\$1$ , $\$5$ , $\$10$ , $\$20$ , $\$50$ , and $\$100$ .

To find the total value of a mix of coins or bills, multiply the number of each type by its value, then add everything up.

Example: You have 6 quarters and 4 dimes. Total = $6(0.25) + 4(0.10) = 1.50 + 0.40 = \$1.90$

When a problem gives you the total but not the number of one type of coin, you set up an equation:

Pick a variable for the unknown quantity (say $x$ = number of quarters).
Write expressions for the value of each type of coin.
Set the sum of those expressions equal to the total.
Solve for $x$ .

Example: You have 3 times as many dimes as quarters, and the total value is $\$1.10$ . If $x$ = number of quarters, then $3x$ = number of dimes. $0.25x + 0.10(3x) = 1.10$ $0.25x + 0.30x = 1.10$ $0.55x = 1.10$ $x = 2$ That's 2 quarters and 6 dimes.

Coin and bill word problems, Solve Mixture Applications with Systems of Equations · Intermediate Algebra

Ticket and Stamp Value Calculations

These work exactly like coin problems, just with different values. Tickets might be priced at $\$12$ for adults and $\$8$ for children. Stamps might cost $\$0.68$ for first-class and $\$1.55$ for priority mail.

The setup is the same every time:

Multiply the quantity of each type by its price.
Add them up to get the total cost.
If one quantity is unknown, use a variable and solve.

Example: A group buys adult tickets at $\$12$ and child tickets at $\$8$ . They buy 15 tickets total and spend $\$148$ . How many of each? Let $x$ = number of adult tickets. Then $15 - x$ = number of child tickets.

$12x + 8(15 - x) = 148$

$12x + 120 - 8x = 148$

$4x + 120 = 148$ $4x = 28$ $x = 7$ That's 7 adult tickets and 8 child tickets.

Watch for extra steps in some problems: you may need to apply a discount before totaling, or add tax at the end.

Coin and bill word problems, Why It Matters: Polynomials and Rational Expressions | College Algebra

Algebraic Equations for Money Quantities

Any time a money problem has an unknown, you're really just translating words into an equation. Here's the process:

Identify what you don't know and assign it a variable.
Translate the word problem into an equation using that variable.
Solve the equation.
Check that your answer makes sense in context.

Example: An employee earns $\$15$ per hour and made $\$120$ total. How many hours did they work? Let $x$ = hours worked. $15x = 120$ $x = \frac{120}{15} = 8$ The employee worked 8 hours.

The "check your answer" step matters. If you get a negative number of coins or a fractional number of tickets, something went wrong. Go back and re-read the problem.

Financial Calculations

These are extensions of the same idea, applied to real-world money situations:

Exchange rates: Multiply the amount in one currency by the exchange rate to convert. If $1$ USD = $0.85$ EUR, then $\$50$ USD = $50 \times 0.85 = 42.50$ EUR.
Simple interest: Use $I = P \times r \times t$ , where $P$ is the starting amount, $r$ is the annual rate (as a decimal), and $t$ is time in years.
Profit: Subtract total costs from total revenue. If revenue is $\$500$ and costs are $\$320$ , profit = $\$180$ .
Budgeting: Add up all income, add up all expenses, and subtract expenses from income to see what's left.

➕Pre-Algebra Unit 9 Review