Problem 44 - Algebra A Combined Function

Question

Solve the following. A marketing manager travels 1080 miles in a corporate jet and then an additional 240 miles by car. If the car ride takes one hour longer than the jet ride takes, and if the rate of the jet is 6 times the rate of the car, find the time the manager travels by jet and find the time the manager travels by car.

Step-by-Step Solution

Answer

The jet travel time is 3 hours and the car travel time is 4 hours.

Step 1: Define Variables

Let \( r \) be the rate of the car in miles per hour. Then the rate of the jet is \( 6r \) miles per hour, as given in the problem statement.

Step 2: Write Travel Time Equations

Use the formula \( \text{time} = \frac{\text{distance}}{\text{rate}} \). For the jet:\[ t_j = \frac{1080}{6r} \]For the car:\[ t_c = \frac{240}{r} \]We know from the problem that the car travels for one hour longer than the jet, i.e., \( t_c = t_j + 1 \).

Step 3: Set Up the Equation

Substitute the expressions for \( t_j \) and \( t_c \) into the equation \( t_c = t_j + 1 \):\[ \frac{240}{r} = \frac{1080}{6r} + 1 \]

Step 4: Solve for \( r \)

Start by clearing the fractions by multiplying through by \( 6r \):\[ 6r \times \frac{240}{r} = 6r \times \frac{1080}{6r} + 6r \times 1 \]Simplifying, we get:\[ 1440 = 1080 + 6r \]Rearrange to solve for \( r \):\[ 1440 - 1080 = 6r \]\[ 360 = 6r \]\[ r = 60 \]

Step 5: Find Travel Times

With \( r = 60 \), find the travel times.For the jet:\[ t_j = \frac{1080}{6 \times 60} = \frac{1080}{360} = 3 \text{ hours} \]For the car:\[ t_c = \frac{240}{60} = 4 \text{ hours} \]

Step 6: Verify the Solution

Check that the car ride takes one hour longer than the jet ride:Jet time is 3 hours and car time is 4 hours, which satisfies \( t_c = t_j + 1 \). The solution is verified as correct.

Key Concepts

Distance, Rate, and Time Problems

In algebra word problems involving distance, rate, and time, the fundamental relationship is "distance = rate × time."
This equation helps in setting up scenarios where a person or object travels a certain distance at a particular speed (rate), either to find the missing variable (rate, time, or distance) or verify a condition.
For solving such problems, it's crucial to understand a few concepts:

• Distance: The total length traveled, typically measured in miles or kilometers.
• Rate: The speed of travel, usually in terms of miles per hour or kilometers per hour.
• Time: The duration of travel, typically measured in hours or minutes.

In the problem given, the marketing manager's journey involves both a jet and a car. Knowing the relationship between these variables lets us accurately model the situation using equations. The differences in time and rate for the jet and car are exactly what makes this problem solvable with algebra.
By systematically applying the distance-rate-time formula, you can break down the trip into two clear segments, making it easier to define equations and solve for the unknowns.

Equation Solving

Solving an equation is about finding the unknown variable that makes the equation true.
The key steps often involve isolating the variable and simplifying the equation.
To solve algebraic equations arising from word problems, here is a simple approach:

• Identify given information and relationship: Understand the relations between known and unknown values in the question.
• Translate the situation into an equation: Use algebraic expressions to transform the word problem into a mathematical equation.
• Manipulate the equation: Perform mathematical operations that simplify the equation, like addition, subtraction, multiplying, or dividing both sides by the same number.
• Solve for the unknown: Once the variable is isolated, check back with the problem to make sure the solution makes sense.
• Verify the solution: Substitute the value back into the original context to ensure it solves the problem conditions.

In the given exercise, manipulating the equations involves simple operations such as multiplying to clear fractions and collecting like terms to isolate the variable. This step-by-step simplification is the key to finding the car's rate and ensuring the solution holds true upon verification.

Variable Definition

Defining variables is a critical step in algebra to represent unknown quantities.
Proper variable definition helps in easily formulating equations from word problems.
Here's how to effectively define variables:

• Identify what you need to find: Understand what the problem is hinting at as the unknown.
• Label the variables: Assign a letter (often the most common ones are x, y, r, etc.) to each unknown quantity you intend to solve for.
• Write equations involving these variables: Using the hints and relationships in the problem, express the situation in terms of these variables.

In this exercise, the problem defines:- Letting the car's rate as \( r \)- Knowing that the jet's rate is 6 times that of the car, it becomes \( 6r \)These variable definitions are tied directly to the relations described in the problem and are crucial for shaping correct equations for solving. Writing them down clearly prevents confusion and lays the groundwork for efficient problem solving.

Fractions in Equations

Dealing with fractions in equations can be tricky, but they can often offer a precise way to express relationships in algebraic terms.
Here's how to manage fractions effectively:

• Understanding the fraction components: A fraction \( \frac{a}{b} \) in an equation can represent division operations or proportions based on the context.
• Clearing fractions for simplicity: Multiply every term in the equation by the least common multiple (LCM) of the denominators to simplify calculations.
• Checking back with the original problem: Always ensure that after clearing fractions, the equation still represents the original problem accurately.

In the original solution, the travel times \( t_j \) and \( t_c \) are expressed using fractions such as \( \frac{1080}{6r} \) and \( \frac{240}{r} \). These fractions provide a clear quantitative relationship between the distances and rates. To solve the equation efficiently, the fractions are eliminated by multiplying throughout, leading to a simpler form that is easier to manipulate and solve. Understanding this handling of fractions is essential for anyone dealing with complex algebraic equations.