Problem 33 - Algebra 2

Question

CYCLING On a particular day, the wind added 3 kilometers per hour to Alfonso's rate when he was cycling with the wind and subtracted 3 kilometers per hour from his rate on his return trip. Alfonso found that in the same amount of time he could cycle 36 kilometers with the wind, he could go only 24 kilometers against the wind. What is his normal bicycling speed with no wind? Determine whether your answer is reasonable.

Step-by-Step Solution

Answer

Alfonso's normal bicycling speed with no wind is 15 km/h.

Step 1: Define the Variables

Let \( x \) be Alfonso's normal bicycling speed with no wind (in km/h). When cycling with the wind, his speed is enhanced by the wind, so his speed becomes \( x + 3 \) km/h. Against the wind, his speed decreases, becoming \( x - 3 \) km/h.

Step 2: Write the Formula for Time

Time is equal to distance divided by speed. For cycling with the wind, the time taken is \( \frac{36}{x+3} \). For cycling against the wind, the time taken is \( \frac{24}{x-3} \). Since the time taken for both trips is the same, we equate the two expressions.

Step 3: Set Up the Equation

The equation representing the time equality is: \[ \frac{36}{x+3} = \frac{24}{x-3} \].

Step 4: Solve the Equation

Cross-multiply to solve for \( x \):\[ 36(x - 3) = 24(x + 3) \]Expand both sides:\[ 36x - 108 = 24x + 72 \]Bring like terms together:\[ 36x - 24x = 72 + 108 \]\[ 12x = 180 \]Solve for \( x \):\[ x = \frac{180}{12} = 15 \].

Step 5: Verify the Solution

Substitute \( x = 15 \) back into the time equations:With wind: \( \frac{36}{15+3} = \frac{36}{18} = 2 \) hours.Against wind: \( \frac{24}{15-3} = \frac{24}{12} = 2 \) hours.Both times are equal, confirming the solution is correct.

Key Concepts

Distance-Rate-Time Relationships

In the world of algebra, the relationship between distance, rate, and time is fundamental. It's all about understanding how these three concepts interact. When we think of distance, we usually consider how far something travels. Rate, on the other hand, describes how fast this travel occurs, while time tells us the duration it takes to complete the journey. The formula that connects these three elements is:

• Distance = Rate × Time

This formula can be rearranged to solve for any of the three variables if you know the other two. For example:

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

In Alfonso's cycling scenario, this relationship helps us determine the time he spends cycling with and against the wind. Despite the change in his rate due to the wind, the time for both trips is the same, guiding us to form and solve an equation.

Equation Solving

Solving equations is a crucial skill in algebra, involving finding the value of an unknown variable. The problem with Alfonso's cycling challenge involves setting up and solving an equation based on the information provided. We begin by representing Alfonso's normal bicycling speed as the variable \( x \). The presence of the wind alters his speed, described by:

• With wind: \( x + 3 \)
• Against wind: \( x - 3 \)

The key to solving the problem lies in setting an equation that reflects that the time for both trips is the same. We commence by equating the time expressions for both scenarios:

• \( \frac{36}{x+3} = \frac{24}{x-3} \)

This equation can be solved to find \( x \), which reveals Alfonso's speed in the absence of wind. The solution involves balancing and simplifying, a fundamental practice in algebra.

Cross-Multiplication

Cross-multiplication is a mathematical technique used to solve equations involving fractions. When confronting fractional equations, cross-multiplying transforms the equation into a simpler, more manageable form without fractions. For Alfonso's cycling exercise, cross-multiplication is the key to solving the time equality equation:

• Start by multiplying the numerator of each side's fraction with the denominator of the opposing side.

This yields the equation:

• \( 36(x - 3) = 24(x + 3) \)

By cross-multiplying, the fractions are eliminated, allowing us to focus on solving a straightforward linear equation. Cross-multiplication is especially useful because it simplifies the process of finding solutions in real-world applications like this one.

Real-World Applications

Understanding algebraic concepts and their practical applications help in solving real-world problems effectively. The story of Alfonso's cycling highlights how distance-rate-time relationships and algebra are not confined to the classroom, but rather they are tools to comprehend everyday scenarios.
When Alfonso has the wind assisting him, it adds to his regular speed, while opposing wind reduces it. Instead of theorizing, we use algebra to calculate the real impact of the wind on his travel. By applying the principles and techniques discussed, we can solve for his normal speed, gaining insights into the extent external factors like weather can influence regular activities.
These applications showcase how valuable algebra is beyond theoretical exercises, emphasizing its role in practical problem-solving.