Problem 24 - Elementary and Intermediate Algebra

Question

Physical Fitness. A woman can bicycle 28 miles in the same time as it takes her to walk 8 miles. She can ride 10 mph faster than she can walk. How fast can she walk?

Step-by-Step Solution

Answer

The woman can walk at 4 mph.

Step 1: Understanding the Problem

We need to determine the walking speed of a woman who walks and bicycles over given distances in the same time. She bicycles 28 miles and walks 8 miles. The bicycle speed is 10 mph faster than her walking speed.

Step 2: Defining Variables

Let the walking speed be \( x \) mph. Therefore, the bicycling speed is \( x + 10 \) mph, because she rides 10 mph faster.

Step 3: Setting Up Equations

The time taken for any activity is given by the distance divided by speed. Thus, for walking, the time is \( \frac{8}{x} \) hours, and for bicycling it is \( \frac{28}{x+10} \) hours. Since the times are equal, we set up the equation \( \frac{8}{x} = \frac{28}{x+10} \).

Step 4: Cross-Multiplying

We eliminate the fractions by cross-multiplying: \[ 8(x + 10) = 28x \]. This simplifies the equation to an easily solvable form.

Step 5: Solving for Walking Speed

Expand and rearrange the equation: \[ 8x + 80 = 28x \]. Subtract \( 8x \) from both sides: \[ 80 = 20x \]. Divide by 20 to find \( x \): \[ x = 4 \].

Step 6: Conclusion

The woman can walk at a speed of 4 mph. This satisfies the condition where she bicycles 10 mph faster than she walks, and both activities take the same time.

Key Concepts

Distance-Time Relationship

Understanding the relationship between distance and time is essential in solving many algebra word problems. The general formula used is

• Time = Distance / Speed

This relationship helps to find out how long an activity takes or solve for unknown distances or speeds when time is constant, like in our problem. Here, the woman travels two distances – bicycling 28 miles and walking 8 miles – in the same time, making it crucial to apply this formula effectively.
Remember, when two different modes of transportation take the same time, it means

• The walking time = The cycling time

This allows us to equate their time expressions, forming an equation that we can solve for the unknown speed.
Understanding this relationship might seem simple, but it is the foundation of calculating accurate results in a variety of real-world scenarios.

Equation Solving

Solving equations involving unknown variables is a powerful tool in mathematics that allows us to find values that satisfy given conditions. In our problem, we establish an equation after determining the time expressions for walking and bicycling.
The set equation is

• \(\frac{8}{x} = \frac{28}{x+10}\)

Here, the variable \(x\) represents the woman's walking speed. By correctly defining variables and setting up equations, we transform the word problem into a mathematical problem.
The goal is to isolate \(x\) on one side of the equation, making it easier to identify the woman's walking speed. This involves logical simplification techniques like isolating terms through subtraction and division.

Cross-Multiplication Techniques

Cross-multiplying is an effective technique when dealing with equations involving fractions. It essentially eliminates the fractions, converting them into an equation form that's easier to handle.
In this problem, we use cross-multiplication on the equation

• \(\frac{8}{x} = \frac{28}{x+10}\)

This gives us:

• 8(x + 10) = 28x

Cross-multiplication means multiplying the numerator of one fraction by the denominator of the other and setting them equal on opposite sides.
The new equation

• 8x + 80 = 28x

becomes straightforward to solve, letting us simplify and solve for \(x\) easily.
Mastery of cross-multiplication can greatly enhance problem-solving efficiency, especially when dealing with rational expressions.