Problem 20 - Elementary Algebra

Question

A passenger car averages 16 miles per hour faster than the bus. If the bus travels 56 miles in the same time it takes the passenger car to travel 84 miles, then what is the speed of each?

Step-by-Step Solution

Answer

The bus travels at 32 mph, while the passenger car travels at 48 mph.

Step 1: Understand the Relationship Between Speeds

We know that the speed of the passenger car is 16 miles per hour faster than the speed of the bus. Let's denote the speed of the bus as \( x \) mph. Then, the speed of the passenger car is \( x + 16 \) mph.

Step 2: Set Up the Equation Based on Distance and Time

To find the time it takes for each vehicle to travel a certain distance, we use the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). Since both vehicles travel for the same amount of time, we set their time expressions equal to each other:\[\frac{56}{x} = \frac{84}{x + 16}\]Here, 56 miles is the distance that the bus travels, and 84 miles is the distance that the passenger car travels.

Step 3: Solve the Equation for x

Cross-multiply the fractions to eliminate them:\[ 56(x + 16) = 84x \]Expand the equation:\[ 56x + 896 = 84x \]Bring all terms involving \( x \) to one side:\[ 896 = 84x - 56x \]which simplifies to:\[ 896 = 28x \]

Step 4: Calculate the Speed of the Bus

Solve for \( x \) by dividing both sides by 28:\[ x = \frac{896}{28} \]This gives:\[ x = 32 \].Hence, the speed of the bus is 32 mph.

Step 5: Calculate the Speed of the Passenger Car

Using the speed of the bus, find the speed of the passenger car:\[ x + 16 = 32 + 16 = 48 \].Thus, the speed of the passenger car is 48 mph.

Key Concepts

Speed and Distance Problems

Speed and distance problems often involve understanding how fast something is moving and how far it travels. The fundamental concept is the relationship between speed, distance, and time. This relationship is captured by the formula:

• Speed = Distance / Time
• Distance = Speed × Time
• Time = Distance / Speed

In these problems, we typically know two of these values and need to find the third. For example, if you know the speed at which a bus travels and the distance it covers, you can find how long it takes to travel that distance. In our scenario, we are dealing with a passenger car and a bus. The key is to understand that even if distances are different, the time can be the same for both if their speeds align proportionately to their distances.

Solving Linear Equations

Solving linear equations is a crucial skill in algebra. These equations have variables, usually represented as letters like \( x \), and we need to solve for them. They often take the form \( ax + b = c \). The goal is to find the value of \( x \) that makes the equation true. Here are the steps to solve such equations:

• Isolate the variable \( x \) by performing inverse operations.
• Combine like terms on both sides of the equation if necessary.
• Use steps such as addition, subtraction, multiplication, or division to get \( x \) by itself.

In our exercise, you set up an equation where both sides equal the time both the bus and the passenger car take to travel different distances. By simplifying this equation, you can eventually find \( x \), which is the speed of the bus.

Cross Multiplication

Cross multiplication is a technique used to solve equations involving fractions. It involves multiplying the numerator of one fraction by the denominator of another. This is useful in equations set up as proportions or ratios. To cross-multiply:

• Multiply the numerator of the first fraction by the denominator of the second fraction.
• Multiply the denominator of the first fraction by the numerator of the second fraction.
• Set these two products equal to each other and solve for the variable.

In the exercise, you cross-multiply the equation \( \frac{56}{x} = \frac{84}{x + 16} \), which allows you to solve for the speed of the bus. By using cross-multiplication, you eliminate the fractions, making the equation simpler to solve.

Rates and Ratios

Rates and ratios are shortcuts to comparing quantities in math. A rate is a special type of ratio where two different units are compared, like miles per hour or price per item. Ratios, on the other hand, express how many times one number contains another. Understanding rates and ratios involves grasping how different values relate to one another in proportional reasoning.

• A rate might express how fast something moves over some time.
• A ratio gives a comparison, such as between the speeds of two vehicles.

In the problem, we are comparing the speeds of a bus and a passenger car, utilizing the fact that the car's speed is 16 miles per hour faster than the bus's speed. Knowing these relationships helps us set up and solve the problem efficiently.