2.4 Solve Mixture and Uniform Motion Applications

Mixture and Uniform Motion Applications

Mixture and uniform motion problems ask you to translate real-world scenarios into linear equations and solve for an unknown. You'll work with coins, tickets, solution concentrations, and objects in motion. The key skill in every case is the same: organize the given information, set up an equation that captures the relationship, and solve.

Combinations of Coins and Currency

Coin and currency problems give you information about a collection of money and ask you to figure out how many of each type you have. The core idea is that the total value equals the sum of (number of each type) × (value of each type).

Common U.S. coin values to know:

• Pennies: $0.01
• Nickels: $0.05
• Dimes: $0.10
• Quarters: $0.25

How to solve a coin problem:

1. Read the problem and identify what's unknown. Let a variable represent that quantity (e.g., let $x$ = number of quarters).
2. Express the other quantities in terms of that same variable, using relationships given in the problem. For example, if you're told there are "twice as many dimes as quarters," the number of dimes is $2x$.
3. Write an equation using the total value. Multiply each coin count by its value and set the sum equal to the total.
4. Solve for $x$, then find any other quantities.
5. Check your answer by plugging the values back in to confirm the total value and total count match.

Example: You have 15 coins, all nickels and quarters, worth $2.75 total. How many of each?

• Let $x$ = number of quarters. Then the number of nickels = $15 - x$.
• Value equation: $0.25x + 0.05(15 - x) = 2.75$
• Simplify: $0.25x + 0.75 - 0.05x = 2.75$
• Combine: $0.20x + 0.75 = 2.75$
• Solve: $0.20x = 2.00$, so $x = 10$

You have 10 quarters and 5 nickels. Check: $10(0.25) + 5(0.05) = 2.50 + 0.25 = 2.75$. ✓

Scenarios with Tickets and Stamps

Ticket and stamp problems work exactly like coin problems. Different types have different prices, and you're given a total cost or total count (or both).

How to solve a ticket/stamp problem:

1. Identify the types and their prices (e.g., adult tickets at $10, child tickets at $5).
2. Let $x$ represent the unknown quantity. Use any stated relationship to express the other quantity in terms of $x$.
3. Write a value equation: (number of type 1) × (price of type 1) + (number of type 2) × (price of type 2) = total cost.
4. Solve for $x$ and find the remaining quantities.
5. Verify by checking both the total number and total cost.

Example: A theater sold 200 tickets. Adult tickets cost $10 and child tickets cost $5. Total revenue was $1,400. How many of each were sold?

• Let $x$ = adult tickets. Then child tickets = $200 - x$.
• $10x + 5(200 - x) = 1400$
• $10x + 1000 - 5x = 1400$
• $5x = 400$, so $x = 80$

80 adult tickets and 120 child tickets. Check: $80(10) + 120(5) = 800 + 600 = 1400$. ✓

Concentrations in Mixture Problems

Concentration problems involve combining two solutions (like saline or alcohol mixtures) to get a solution with a desired concentration. The principle: the amount of pure substance before mixing equals the amount of pure substance after mixing.

The mixture equation is:

$C_1 V_1 + C_2 V_2 = C_f V_f$

• $C_1, C_2$ = concentrations of the two solutions (written as decimals, e.g., 20% = 0.20)
• $V_1, V_2$ = volumes of the two solutions
• $C_f$ = final concentration of the mixture
• $V_f$ = final volume, which equals $V_1 + V_2$

How to solve a concentration problem:

1. Identify each solution's concentration and volume. Note which quantity is unknown.
2. Write the mixture equation. Remember that $V_f = V_1 + V_2$.
3. Solve for the unknown variable.
4. Check by computing the amount of pure substance on each side of the equation.

Example: How many liters of a 40% acid solution must be mixed with 6 liters of a 10% acid solution to get a 20% acid solution?

• Let $x$ = liters of 40% solution. Final volume = $x + 6$.
• $0.40x + 0.10(6) = 0.20(x + 6)$
• $0.40x + 0.60 = 0.20x + 1.20$
• $0.20x = 0.60$, so $x = 3$

You need 3 liters of the 40% solution. Check: $0.40(3) + 0.10(6) = 1.20 + 0.60 = 1.80$, and $0.20(9) = 1.80$. ✓

Linear Equations for Uniform Motion

Uniform motion means traveling at a constant speed. Every problem of this type uses one formula:

$D = RT$

• $D$ = distance
• $R$ = rate (speed)
• $T$ = time

You can rearrange this to find any of the three: $R = \frac{D}{T}$ or $T = \frac{D}{R}$. Before setting up your equation, make sure your units are consistent (e.g., if speed is in miles per hour, time must be in hours, not minutes).

Single-object problems are straightforward: plug in the two known values and solve for the third.

Two-object problems are where it gets interesting. These typically involve two people or vehicles traveling toward each other, away from each other, or in the same direction. The setup depends on the scenario:

• Traveling toward each other (opposite directions, closing a gap): Their distances add up to the total distance between them. $D_1 + D_2 = \text{total distance}$
• Traveling in the same direction: The faster object covers more distance. The difference in distances equals the gap. $D_1 - D_2 = \text{gap}$

How to solve a two-object motion problem:

1. Draw a quick sketch showing starting positions and directions.
2. Set up a $D = RT$ equation for each object. Use the same variable $T$ if they travel for the same amount of time.
3. Write a relationship between the distances (they add up, subtract, or are equal, depending on the problem).
4. Substitute the $D = RT$ expressions into that relationship and solve.
5. Check that your answer makes sense (positive time, reasonable speed, etc.).

Example: Two cars start 300 miles apart and drive toward each other. One travels at 60 mph and the other at 40 mph. How long until they meet?

• Car 1: $D_1 = 60T$. Car 2: $D_2 = 40T$.
• Since they're closing the gap: $D_1 + D_2 = 300$
• $60T + 40T = 300$
• $100T = 300$, so $T = 3$ hours

Check: Car 1 travels $60(3) = 180$ miles, Car 2 travels $40(3) = 120$ miles. $180 + 120 = 300$. ✓

Algebraic Modeling and Problem Solving

Across all these problem types, the process follows the same pattern:

1. Read carefully. Identify what you know and what you need to find.

2. Define your variable. Write a clear statement like "Let $x$ = the number of dimes."

3. Express other unknowns in terms of that variable using relationships from the problem.

4. Write an equation based on a total (value, distance, amount of substance).

5. Solve using standard algebra techniques.

6. Answer the question that was asked. Sometimes $x$ isn't the final answer; you may need to compute $2x$ or $15 - x$.

7. Verify by substituting back into the original problem (not just the equation).

A common mistake is setting up the variable but forgetting to convert it into what the problem actually asks for. Always re-read the question before writing your final answer.