3.3 Solve Mixture Applications

Solving Mixture Applications

Mixture applications ask you to combine different items or substances and figure out unknown quantities using algebra. These problems show up as coin puzzles, ticket sales, solution concentrations, and investment scenarios. The approach is the same every time: organize what you know, set up equations, and solve.

Coin Combination Calculations

Coin problems give you information about a collection of coins and ask you to find how many of each type there are. The key idea is that every coin has two properties: it counts as one coin, and it has a dollar value.

Common U.S. coin values:

• Penny: $\$0.01$
• Nickel: $\$0.05$
• Dime: $\$0.10$
• Quarter: $\$0.25$

How to solve a coin problem:

1. Define your variables. Let each variable represent the number of a coin type (e.g., let $x$ = number of quarters).
2. Write a quantity equation for the total number of coins (e.g., $x + y = 15$).
3. Write a value equation by multiplying each coin's value by its quantity (e.g., $0.25x + 0.10y = 2.50$).
4. Solve the system using substitution or elimination.

Example: You have 15 coins, all quarters and dimes, worth $\$2.50$ total. How many of each?

• Quantity equation: $x + y = 15$
• Value equation: $0.25x + 0.10y = 2.50$
• From the first equation: $y = 15 - x$
• Substitute into the value equation: $0.25x + 0.10(15 - x) = 2.50$
• Simplify: $0.25x + 1.50 - 0.10x = 2.50$, so $0.15x = 1.00$, giving $x = 6.67$

(In a real textbook problem, the numbers will work out to whole numbers. If you get a decimal for a number of coins, double-check your setup.)

Multi-Item Quantity and Pricing

These problems work exactly like coin problems, just with tickets, stamps, or other items instead of coins. You'll have two types of items at different prices, and you need to find how many of each were sold.

How to solve a ticket/pricing problem:

1. Define variables for each item type (e.g., let $a$ = adult tickets, $c$ = child tickets).
2. Write a quantity equation: $a + c = \text{total items sold}$
3. Write a value equation: $(\text{price}_1)(a) + (\text{price}_2)(c) = \text{total revenue}$
4. Solve the system.

Example: A theater sold 80 tickets. Adult tickets cost $\$12$ and child tickets cost $\$8$. Total revenue was $\$840$. How many of each were sold?

• $a + c = 80$
• $12a + 8c = 840$
• From the first equation: $c = 80 - a$
• Substitute: $12a + 8(80 - a) = 840$
• Simplify: $12a + 640 - 8a = 840$, so $4a = 200$, giving $a = 50$
• Then $c = 80 - 50 = 30$

So 50 adult tickets and 30 child tickets were sold.

Mixture Problem Solutions

Concentration mixture problems involve combining two solutions with different strengths to get a desired final concentration. The core principle: the amount of pure substance before mixing equals the amount of pure substance after mixing.

To find the amount of pure substance in a solution, multiply the volume by the concentration (as a decimal):

$\text{Amount of substance} = \text{Volume} \times \text{Concentration}$

How to solve a concentration problem:

1. Define variables for the unknown volumes (e.g., let $x$ = liters of the 20% solution).
2. Write a volume equation: $x + y = \text{total volume}$
3. Write a substance equation: $(\text{concentration}_1)(x) + (\text{concentration}_2)(y) = (\text{final concentration})(\text{total volume})$
4. Solve the system.

Example: You need 10 liters of a 38% alcohol solution. You have a 20% solution and a 50% solution. How much of each do you mix?

• $x + y = 10$
• $0.20x + 0.50y = 0.38(10)$
• Simplify the right side: $0.20x + 0.50y = 3.80$
• From the first equation: $x = 10 - y$
• Substitute: $0.20(10 - y) + 0.50y = 3.80$
• Simplify: $2.00 - 0.20y + 0.50y = 3.80$, so $0.30y = 1.80$, giving $y = 6$
• Then $x = 10 - 6 = 4$

You need 4 liters of the 20% solution and 6 liters of the 50% solution.

Simple Interest Mixture Models

Simple interest problems become mixture problems when someone splits an investment across two accounts with different interest rates. The formula for simple interest is:

$I = P \times r \times t$

where $P$ is the principal (amount invested), $r$ is the annual interest rate as a decimal, and $t$ is time in years.

How to solve an interest mixture problem:

1. Define variables for the amounts invested at each rate (e.g., let $x$ = amount at 5%, $y$ = amount at 8%).
2. Write a principal equation: $x + y = \text{total invested}$
3. Write an interest equation: $(\text{rate}_1)(x)(t) + (\text{rate}_2)(y)(t) = \text{total interest earned}$
4. Solve the system.

Example: You invest $\$5000$ total in two accounts. One pays 4% annual interest, the other pays 7%. After 1 year, you earn $\$260$ in total interest. How much is in each account?

• $x + y = 5000$
• $0.04x + 0.07y = 260$
• From the first equation: $y = 5000 - x$
• Substitute: $0.04x + 0.07(5000 - x) = 260$
• Simplify: $0.04x + 350 - 0.07x = 260$, so $-0.03x = -90$, giving $x = 3000$
• Then $y = 5000 - 3000 = 2000$

So $\$3000$ was invested at 4% and $\$2000$ at 7%.

Tips for All Mixture Problems

Every mixture problem follows the same two-equation pattern. You always get:

• One equation for total quantity (total coins, total liters, total dollars invested)
• One equation for total value (total dollar value, total pure substance, total interest)

A few things to watch for:

• Always convert percentages to decimals before plugging them into equations (e.g., 38% becomes 0.38).
• If your answer is a number of items (coins, tickets), it should be a whole number. A decimal answer means something went wrong.
• Label your variables clearly at the start. Mixture problems have multiple unknowns, and it's easy to lose track of what $x$ represents.
• Combining like terms and distributing are the algebra skills you'll use most when simplifying these equations.