Problem 72 - Algebra and Trigonometry

Question

Aerobic Exercise A woman keeps fit by bicycling and running every day. On Monday she spends \(\frac{1}{2}\) hour at each activity, covering a total of 12\(\frac{1}{2} \mathrm{mi}\) On Tuesday she runs for 12 \(\mathrm{min}\) and cycles for 45 \(\mathrm{min}\) , covering a total of 16 \(\mathrm{mi}\) . Assuming that her running and cycling speeds don't change from day to day, find these speeds.

Step-by-Step Solution

Answer

The running speed is approximately 12.27 mph and the cycling speed is approximately 12.73 mph.

Step 1: Define Variables

Let \( r \) represent the running speed in miles per hour, and \( c \) represent the cycling speed in miles per hour.

Step 2: Create Equations from Monday's Data

On Monday, she spends \( \frac{1}{2} \) hour running and \( \frac{1}{2} \) hour cycling. Thus, the total distance covered can be expressed as: \[ \frac{1}{2}r + \frac{1}{2}c = 12.5 \] Simplify this equation to: \[ r + c = 25 \]

Step 3: Create Equations from Tuesday's Data

On Tuesday, she runs for 12 minutes (or \( \frac{1}{5} \) hour) and cycles for 45 minutes (or \( \frac{3}{4} \) hour), resulting in the equation: \[ \frac{1}{5}r + \frac{3}{4}c = 16 \] Simplify and convert to the same denominator to combine terms: \[ \frac{3}{15}r + \frac{9}{12}c = 16 \] Which further simplifies to: \[ 4r + 15c = 240 \]

Step 4: Solve the System of Equations

We now have the following system of equations: \[ r + c = 25 \] \[ 4r + 15c = 240 \] Solve the first equation for \( r \): \[ r = 25 - c \] Substitute this expression into the second equation: \[ 4(25 - c) + 15c = 240 \] Simplify and solve for \( c \): \[ 100 - 4c + 15c = 240 \] \[ 11c = 140 \] \[ c = \frac{140}{11} \approx 12.73 \text{ mph} \]

Step 5: Calculate Running Speed

Substitute \( c = \frac{140}{11} \) back into \( r + c = 25 \): \[ r + \frac{140}{11} = 25 \] \[ r = 25 - \frac{140}{11} \] Calculating \( r \): \[ r = \frac{275}{11} - \frac{140}{11} = \frac{135}{11} \approx 12.27 \text{ mph} \]

Step 6: Verify the Solution

Double-check by substituting back into the initial Tuesday's equation: \[ \frac{1}{5}(12.27) + \frac{3}{4}(12.73) = 16 \] Calculate to ensure both sides equal approximately 16, confirming the solution is correct.

Key Concepts

System of Equations

A system of equations consists of two or more equations with the same variables. The main goal in solving a system of equations is to find the values of these variables that satisfy all given equations simultaneously. This can be particularly useful in real-world situations, such as calculating speeds in travel workouts, like running and cycling from our exercise example. In this scenario, the variables represent the speeds of running and cycling, and the system of equations involves the distances covered on different days.

• The first equation is based on the distance covered on Monday: \( r + c = 25 \)
• The second stems from Tuesday's data: \( 4r + 15c = 240 \)

Solving the system allows us to determine the running and cycling speeds. By understanding these systems, students learn how different pieces of data can be combined to find solutions to complex problems, reflecting numerous applications in science and engineering.

Variable Substitution

Variable substitution is a common method for solving systems of equations. In this process, one variable is expressed in terms of another variable or constants, and this expression is then substituted into the other equation(s). This strategy simplifies the problem, allowing for easier computation and solution finding. In our example, we express one variable in terms of another:

• From the equation \( r + c = 25 \), we solve for \( r \) which gives us \( r = 25 - c \).
• This expression is substituted into the second equation \( 4r + 15c = 240 \).

By replacing \( r \) with \( 25 - c \), algebra becomes simpler since it reduces to a single equation with one variable. This step is crucial in finding the final values of the variables. Variable substitution not only simplifies problems but is also foundational for learning more complex algebraic methods in higher mathematics.

Unit Conversion

Unit conversion involves changing the units of a measure into another unit while retaining the same quantity. It's a useful skill in mathematics to solve real-world problems where measurements might be given in various units. In the given exercise, the conversion from minutes to hours is essential to align time units across the equations. Here's how it works:

• Running time on Tuesday: 12 minutes is converted to hours by dividing by 60, which equals \( \frac{1}{5} \) hour.
• Cycling time on Tuesday: 45 minutes becomes \( \frac{3}{4} \) hour when converted.

Without these unit conversions, the formulation of the equations would be incorrect, leading to potential mistakes in the calculation of running and cycling speeds. Mastering unit conversion helps students gain precision in various fields such as physics, chemistry, and engineering, where consistent and accurate units are critical for problem-solving.