Problem 46 - Algebra for College Students

Question

Suppose that Wendy rides her bicycle 30 miles in the same time that it takes Kim to ride her bicycle 20 miles. If Wendy rides 5 miles per hour faster than Kim, find the rate of each.

Step-by-Step Solution

Answer

Kim's speed is 10 mph, and Wendy's speed is 15 mph.

Step 1: Define Variables

Let's denote Kim's speed as \( x \) miles per hour. Then Wendy's speed, which is 5 miles per hour faster, will be \( x + 5 \) miles per hour.

Step 2: Set Up the Equation Using Time Formula

The time taken to travel a certain distance is the distance divided by speed. For Kim, the time taken is \( \frac{20}{x} \) hours. For Wendy, it is \( \frac{30}{x+5} \) hours. Since the times are equal, set up the equation: \[ \frac{20}{x} = \frac{30}{x+5} \]

Step 3: Cross Multiply to Eliminate Fractions

Cross multiply the equation to eliminate the fractions: \[ 20(x + 5) = 30x \]

Step 4: Simplify the Equation

Distribute and simplify the equation: \[ 20x + 100 = 30x \]

Step 5: Solve for x

Rearrange the equation to solve for \( x \): \[ 100 = 30x - 20x \] simplifying gives \[ 100 = 10x \]. Dividing both sides by 10, we find \( x = 10 \).

Step 6: Find Wendy's Speed

Since Wendy's speed is 5 miles per hour more than Kim's, calculate Wendy's speed as \( x + 5 = 10 + 5 = 15 \) miles per hour.

Key Concepts

Rate Problems

Rate problems are a common type of algebra problem that involve understanding and calculating the rate at which something occurs. In this scenario, the rate of interest is the speed at which Wendy and Kim ride their bicycles. Knowing how to interpret rates helps us solve practical problems, like determining how fast each person is going.
Rates tell us how one quantity changes in relation to another. Here, we know that Wendy's speed is "5 miles per hour faster than Kim." This information lets us define the rate at which Wendy and Kim are biking.

• We define Kim's speed as a variable, say, \( x \) miles per hour.
• Since Wendy is biking faster, her speed is \( x + 5 \) miles per hour.

By expressing the speeds in terms of variables, we can use them in equations to find their actual values. Tackling rate problems often involves translating words into mathematical equations that represent the relationship between distances, speeds, and times.
Understanding rate problems is essential because it builds a foundation for dealing with more complex situations where other factors might influence rate.

Solving Equations

Solving equations is a crucial part of algebra, and rate problems often boil down to creating and solving equations. Here, the goal is to find Kim's speed first, and we do this by setting up an equation that reflects the problem's conditions.
In our exercise, we used the equation \[ \frac{20}{x} = \frac{30}{x+5} \] which arises from the condition that Wendy and Kim take the same amount of time to travel different distances. The equation equates the two expressions for time, each derived from the distance-speed-time formula.
The process of solving involves:

• Cross-multiplying to get rid of the fractions.
• Distributing and simplifying the equation to isolate the variable \( x \).
• Performing arithmetic operations to solve for \( x \).

We've simplified our initial equation into a more straightforward form \( 20x + 100 = 30x \). By arranging terms and solving, we found that \( x = 10 \), meaning Kim's speed is 10 miles per hour. Solving equations teaches us to manipulate algebraic expressions to find unknown values effectively.

Distance-Speed-Time Relationship

The relationship between distance, speed, and time is a fundamental concept in algebra and physics, central to understanding rate problems like the one in our exercise. This relationship is often summarized by the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]This equation shows us how these variables relate to one another:

• If we know the distance and speed, we can find time by rearranging the equation: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).
• If we know distance and time, we calculate speed as: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \).
• Lastly, if we know speed and time, distance is computed with \( \text{Distance} = \text{Speed} \times \text{Time} \).

In our scenario, this relationship is used to set up equations that help us solve for unknowns. Using the time formula for both Wendy and Kim rides, we form the equation that determines their speeds.
Grasping this distance-speed-time relationship is crucial for accurately interpreting and solving practical and theoretical problems across many fields like engineering, physics, and even in daily life scenarios.