Lesson 2.4: Solve Mixture and Uniform Motion Applications

New Concept

This lesson teaches you to translate real-world scenarios into algebraic equations. You'll master setting up and solving problems involving coins, tickets, mixtures, and motion by applying fundamental formulas like number × value = total value and `$D = rt$$.

What’s next

Now you have the big picture. Next, we'll dive into interactive examples for each problem type, starting with coin and ticket problems.

The Core Idea

To find the total value of a group of identical coins, you use a simple formula:

• number: How many coins you have.
• value: What one coin is worth (e.g., $0.25$ for a quarter).
• total value: The combined worth of all the coins.

How to Solve Coin Problems

1. Organize with a Table: Create columns for “Coin Type,” “Number,” “Value,” and “Total Value.” This helps keep track of everything.
2. Use Variables: Assign a variable (like $d$ for dimes) to the number of one coin type. Then, use the problem's clues to write expressions for the other coin types.
3. Find the Total Value for Each Coin Type: Multiply the number by the value for each row in your table.
4. Build Your Equation: Add the total value expressions for each coin type and set them equal to the grand total given in the problem.

Examples

• Problem: A wallet has 3.00 dollars in dimes and quarters. If there are 15 coins in total, how many dimes and quarters are there?

Solution: Let $d$ be the number of dimes. Since there are 15 coins total, the number of quarters is $15 - d$. The equation for the total value is:

Solving this gives: $0.10d + 3.75 - 0.25d = 3.00$, which simplifies to $-0.15d = -0.75$. Dividing by $-0.15$ gives $d = 5$.
Answer: There are 5 dimes and $15 - 5 = 10$ quarters.

Property

These problems use the same model as coin problems:

Examples

• A school play sold 150 tickets for a total of 950 dollars. Adult tickets cost 8 dollars and student tickets cost 5 dollars. Let $a$ be the number of adult tickets. The equation is $8a + 5(150 - a) = 950$. Solving gives $3a + 750 = 950$, so $3a = 200$, which is not an integer. Let's adjust. Total sales were 900 dollars. $8a + 5(150-a) = 900$, so $3a = 150$ and $a=50$. They sold 50 adult tickets and 100 student tickets.

• Maria spent $\$13.00$ on stamps. The number of 50-cent stamps was 10 less than twice the number of 80-cent stamps. Let $e$ be the number of 80-cent stamps. The equation is $0.80e + 0.50(2e - 10) = 13.00$. This gives $0.80e + 1.00e - 5 = 13.00$, so $1.80e = 18.00$ and $e=10$. She bought ten 80-cent stamps and ten 50-cent stamps.

Property

Mixture problems involve combining two or more items with different values to create a mixture with a new, specific value. The fundamental principle is that the sum of the values of the individual components equals the value of the final mixture.

This is organized in a table with rows for each component and the final mixture, and columns for Number of Units, Price per Unit, and Total Value.

Examples

• A coffee shop wants to make a 20-pound blend of coffee costing 15 dollars per pound. They mix a bean that costs 12 dollars per pound with another that costs 18 dollars per pound. Let $x$ be the pounds of the 12 dollar coffee. The equation is $12x + 18(20 - x) = 15(20)$. Solving gives $12x + 360 - 18x = 300$, so $-6x = -60$ and $x=10$. They should use 10 pounds of the 12 dollar coffee and 10 pounds of the 18 dollar coffee.

• A chemist needs 100 mL of a 30% acid solution. She has a 20% solution and a 50% solution available. Let $x$ be the mL of the 20% solution. The equation is $0.20x + 0.50(100 - x) = 0.30(100)$. This gives $0.20x + 50 - 0.50x = 30$, so $-0.30x = -20$ and $x = \frac{200}{3}$. She should mix $66.67$ mL of the 20% solution and $33.33$ mL of the 50% solution.

Property

Uniform motion problems are solved using the formula Distance = Rate × Time, or $D = rt$. The strategy involves drawing a diagram, creating a table with columns for Rate, Time, and Distance, and determining the relationship between the distances traveled.
Common scenarios:

1. Same Distance: Two objects travel the same path. The equation is $r_1t_1 = r_2t_2$.
2. Total Distance: Two objects travel towards or away from each other. The equation is $D_1 + D_2 = D_{total}$.

It is critical to ensure all units are consistent (e.g., if rate is in miles per hour, time must be in hours).

Examples

• Two trains leave the same station at the same time, traveling in opposite directions. One travels at 65 mph and the other at 75 mph. How long will it take for them to be 420 miles apart? Let $t$ be the time in hours. The equation is $65t + 75t = 420$. This gives $140t = 420$, so $t = 3$. It will take them 3 hours.

• Jim can bike to the lake in 2 hours. If he drives, it takes 30 minutes ($0.5$ hours). His driving speed is 24 mph faster than his biking speed. Find his biking speed. Let $r$ be his biking speed. The distance is the same, so $2r = 0.5(r + 24)$. This gives $2r = 0.5r + 12$, so $1.5r = 12$ and $r = 8$. His biking speed is 8 mph.