Problem 58 - Intermediate Algebra

Question

Set up an equation and solve each problem. On a 135-mile bicycle excursion, Maria averaged 5 miles per hour faster for the first 60 miles than she did for the last 75 miles. The entire trip took 8 hours. Find her rate for the first 60 miles.

Step-by-Step Solution

Answer

Maria's rate for the first 60 miles was 20 mph.

Step 1: Define Variables

Let Maria's speed for the first 60 miles be \( r + 5 \) mph and for the last 75 miles be \( r \) mph. Here, \( r \) is her average speed for the last 75 miles.

Step 2: Set Up Equations for Time

Time is equal to distance divided by speed. The time for the first 60 miles is \( \frac{60}{r + 5} \) hours. The time for the last 75 miles is \( \frac{75}{r} \) hours.

Step 3: Create an Equation for Total Time

According to the problem, the entire trip took 8 hours. Set up the equation: \[ \frac{60}{r + 5} + \frac{75}{r} = 8 \]

Step 4: Solve for r (Her Speed for the Last 75 Miles)

To solve for \( r \), multiply through by \( r(r+5) \) to clear the fractions: \( 60r + 75(r+5) = 8r(r+5) \). Expand and simplify: \( 60r + 75r + 375 = 8r^2 + 40r \). This gives the quadratic equation: \( 8r^2 + 40r - 135r - 375 = 0 \), which simplifies to \( 8r^2 - 95r - 375 = 0 \).

Step 5: Use the Quadratic Formula

Use the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 8 \), \( b = -95 \), and \( c = -375 \). Calculate \( r = \frac{95 \pm \sqrt{95^2 + 4 \times 8 \times 375}}{16} \). This simplifies to \( r = \frac{95 \pm \sqrt{9025 + 12000}}{16} = \frac{95 \pm \sqrt{21025}}{16} = \frac{95 \pm 145}{16} \).

Step 6: Solve the Quadratic

Calculate the possible values for \( r \): \( r = \frac{240}{16} = 15 \) or \( r = \frac{-50}{16} \approx -3.125 \). Since speed cannot be negative, \( r = 15 \) mph.

Step 7: Calculate Maria's Speed for the First 60 Miles

Since \( r = 15 \) mph for the last part, her speed for the first 60 miles is \( r + 5 = 15 + 5 = 20 \) mph.

Key Concepts

Distance-Speed-Time Problems

Understanding distance-speed-time problems is key to solving real-life scenarios. These problems involve calculating one of these three variables when given the other two.
To solve such problems, use the formula:

• Distance = Speed × Time

For example, if a cyclist travels for a given distance and you know the speed, solving for time requires dividing distance by speed. Conversely, to find speed, you divide distance by time. In this specific exercise, the challenge was to find two different speeds over different segments of a journey that took a total of 8 hours.

Algebraic Problem Solving

Algebraic problem-solving is a technique used to translate real-world scenarios into mathematical equations.
It involves identifying relevant variables, setting up equations, and then solving these equations to find the desired values.
In the exercise, variables were defined for Maria's speed in two different portions of her trip. To solve for unknown speeds, equations were formed based on the descriptions of her journey and the total time. This process helped turn the word problem into an algebraic question that could be solved using mathematical tools.

Quadratic Formula

The quadratic formula is a fundamental tool used in algebra to solve quadratic equations. Quadratic equations are in the form \( ax^2 + bx + c = 0 \), and the solution is found using:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In the context of the exercise, the formula was used to determine Maria's speed for the trip. By substituting the coefficients from the equation \( 8r^2 - 95r - 375 = 0 \), the quadratic formula provided two possible values for \( r \), solving for the effective speed. Since speed cannot be negative, only the positive result was considered valid.

Defining Variables in Algebra

Defining variables is a crucial step in solving algebraic problems effectively. It involves choosing symbols to represent unknown quantities and clarifying their relationships.
This clarity is essential to accurately set up equations that reflect real-world scenarios.
In Maria's trip scenario, variables were set for her average speeds across different trip segments. It made setting up equations intuitive and straightforward, representing her speed for different distances accurately. The process involves thinking critically about the quantities involved and ensuring there is a logical representation between the variables and the situation.