Problem 40 - Elementary Algebra

Question

In \(1 \frac{1}{4}\) hours more time, Rita, riding her bicycle at 12 miles per hour, rode 2 miles farther than Sonya, who was riding her bicycle at 16 miles per hour. How long did each girl ride?

Step-by-Step Solution

Answer

Sonya rode for 3.25 hours, and Rita rode for 4.5 hours.

Step 1: Define Variables

Let's define the variables needed for this problem. Let \( t \) represent the time, in hours, that Sonya rode her bicycle. Then, the time Rita rode is \( t + 1.25 \) because she rode for 1 1/4 hours longer.

Step 2: Calculate Distances

The distance each girl rode can be found using the formula \( \, ext{Distance} = ext{Speed} \times ext{Time} \, \). Thus, Sonya's distance is \( 16t \) miles, and Rita's distance is \( 12(t + 1.25) \) miles. We know Rita rode 2 miles more than Sonya, giving us the equation \( 12(t + 1.25) = 16t + 2 \).

Step 3: Set Up Equation

Using the information from Step 2, set up the equation that equates Rita's distance with Sonya's distance plus 2 miles: \[ 12(t + 1.25) = 16t + 2 \].

Step 4: Simplify the Equation

Distribute the 12 on the left side of the equation: \[ 12t + 15 = 16t + 2 \]. Now, simplify the equation by subtracting \( 12t \) from both sides: \[ 15 = 4t + 2 \].

Step 5: Solve for Sonya's Time

Subtract 2 from both sides to isolate the term with \( t \): \[ 13 = 4t \]. Now, divide by 4 to solve for \( t \): \[ t = \frac{13}{4} = 3.25 \] hours.

Step 6: Find Rita's Time

Since Rita rode for \( t + 1.25 \) hours, substitute \( t = 3.25 \) into the expression: \[ t + 1.25 = 3.25 + 1.25 = 4.5 \] hours.

Key Concepts

Variables in Algebra

When solving algebra word problems, defining variables is the fundamental step. A variable represents an unknown quantity that we want to determine. In this problem, Sonya's riding time is represented by the variable \( t \). It's chosen as a variable because Sonya's riding time is initially unknown.

By expressing Rita's riding time in terms of \( t \), it allows us to relate the two unknowns efficiently using equations. In this case, Rita's riding time is \( t + 1.25 \) because she rides for 1 hour and 15 minutes longer than Sonya.

Using variables makes complex problems more manageable. It transforms a word problem into an algebraic equation which can then be solved systematically.

Distance-Speed-Time Problems

Distance-Speed-Time problems often require us to find one of these three components given the others. The core concept here is the relationship between distance, speed, and time, which can be written as:

• Distance = Speed \( \times \) Time
• Time = \( \frac{\text{Distance}}{\text{Speed}} \)
• Speed = \( \frac{\text{Distance}}{\text{Time}} \)

In the given problem, we use the relationship to calculate the distance each girl traveled. Since Sonya rides at 16 miles per hour for \( t \) hours, her distance is \( 16t \) miles. Similarly, Rita's distance is \( 12(t + 1.25) \) miles because her speed is 12 miles per hour and her time is \( t + 1.25 \) hours.

These expressions for distance help us set up an equation to compare the two distances, which eventually allows us to find the solution to the problem.

Equation Solving Steps

Solving equations involves several important steps. Initially, you set up your equation based on the relationships or conditions given in the problem. Here, Rita rode 2 miles farther than Sonya, leading to the equation \( 12(t + 1.25) = 16t + 2 \).

The next step involves simplifying the equation. Distribute the 12 in \( 12(t + 1.25) \) to get \( 12t + 15 \). We then simplify by bringing like terms together: subtract \( 12t \) from both sides to isolate the \( t \) term, resulting in \( 15 = 4t + 2 \).

To solve for \( t \), we isolate it by subtracting 2 and then dividing by 4, giving \( t = 3.25 \). Finally, we can find any other related quantities, like Rita's time, by substituting back into related expressions: \( t + 1.25 = 4.5 \). Each step logically progresses from the previous, allowing us to arrive at the solution.