Problem 51 - Prealgebra and Introductory Algebra

Question

An express train traveled \(600 \mathrm{mi}\) in the same amount of time it took a freight train to travel \(360 \mathrm{mi}\). The rate of the express train was \(20 \mathrm{mph}\) faster than the rate of the freight train. Find the rate of each train.

Step-by-Step Solution

Answer

The speed of the freight train is around \( 45 \mathrm{mph} \) and the speed of the expresss train is \( 65 \mathrm{mph} \).

Step 1: Set Up the Express Train Speed Equation

Let's denoted the speed of the freight train as \( x \) mph. Then, according to the problem, the speed of the express train is \( x + 20 \) mph. Now, we know that Time = Distance/Speed. Here, distance = 600 miles and speed = \( x+20 \) mph. Therefore, the express train's time equation can be written as \( T = \frac{600}{x+20} \).

Step 2: Set Up the Frieght Train Speed Equation

The freight train also has Time = Distance/Speed for a distance = 360 miles and speed = \( x \) mph. So this equation can be written as \( T = \frac{360}{x} \).

Step 3: Equate and Solve

Since the time for both the trains is the same, equate the two equations \( \frac{600}{x+20} = \frac{360}{x} \). Cross-multiply and then solve for \( x \), which denotes the speed of the freight train.

Step 4: Calculate the Speed of the Express Train

Once you've found \( x \), simply add 20 to find the speed of the express train, as the problem states it's 20 mph faster.

Key Concepts

Distance, Rate, Time

Understanding the relationship between distance, rate, and time is crucial for solving word problems in algebra. This relationship can be summarized with the formula:

• Distance = Rate x Time
• Rate = Distance / Time
• Time = Distance / Rate

In problems involving trains, automobiles, or planes, where two objects travel from one place to another, this formula becomes your best friend. For instance, if you know the distance and rate, you can easily find the time taken by rearranging the formula to
\( \text{Time} = \frac{\text{Distance}}{\text{Rate}} \).
Understanding which variables you have and which you need is crucial for setting up the correct equations and ultimately solving the problem.

Linear Equations

Linear equations are fundamental to problem-solving in algebra. They involve equations that map out straight lines when plotted on a graph. In the context of our train problem, linear equations allow us to express the relationship between time, distance, and rate algebraically.
Here, we assume the variable \( x \) to be the speed of the freight train. We then express the speed of the express train in terms of \( x \), i.e., \( x + 20 \). By connecting the travel times for both trains, we create two separate linear equations:

• \( T = \frac{600}{x+20} \)
• \( T = \frac{360}{x} \)

Since both equations equal the same time \( T \), setting them equal to each other creates a single equation that we can solve to find \( x \). This is the heart of solving such word problems using linear equations.

Problem Solving Strategy

Solving algebra Word problems can be simplified by structuring a clear problem-solving strategy. Consider these steps:

• **Identify what is asked:** Determine what you need to find. In our exercise, these are the rates of the trains.
• **Define variables:** Assign variables to the unknowns. We defined \( x \) as the freight train's speed.
• **Build the equations:** Develop equations based on the relationship between distance, rate, and time. Here, we used \( \frac{360}{x} \) and \( \frac{600}{x+20} \) to represent the time each train spent traveling.
• **Solve and check:** Use algebraic methods to solve for the variable. After solving for \( x \), calculate the speed of both trains and verify if they satisfy the conditions given in the problem.

Breaking down the problem into smaller, manageable tasks makes it less daunting and helps ensure accuracy throughout the solution process.