Problem 50 - Elementary Algebra

Question

A runner is training for a marathon. She wants to reduce the time it takes her to run 6 miles by 3 minutes. She plans to reduce her time by 12 seconds every week. Find the number of weeks needed for her to meet her goal. Use negative numbers to represent loss of time.

Step-by-Step Solution

Answer

15 weeks

Step 1: Understand the goal

The goal is to reduce the running time for a 6-mile distance by 3 minutes. This means a total reduction of 3 minutes, which is equivalent to 3 * 60 = 180 seconds.

Step 2: Recognize the reduction per week

The runner plans to reduce her time by 12 seconds every week. Therefore, each week she loses 12 seconds from her running time.

Step 3: Set up the equation

Let the number of weeks needed be represented by W. Each week she reduces her time by 12 seconds, so the total reduction after W weeks is 12W seconds. According to the goal from Step 1, this total reduction needs to be 180 seconds.

Step 4: Formulate the equation

The equation representing the problem is: 12W = 180.

Step 5: Solve for W

Solve the equation 12W = 180 for W. Divide both sides by 12: W = 180 / 12 W = 15

Step 6: Interpret the result

It will take the runner 15 weeks to reduce her 6-mile running time by 3 minutes if she reduces her time by 12 seconds every week.

Key Concepts

Linear Equations

Linear equations are fundamental in algebra. They represent relationships where the change in one variable directly affects another variable in a constant way. For example, in the problem given, the runner reduces her time by a consistent 12 seconds every week. We can represent this relationship with a linear equation: \[12W = 180\]

• **Variable:** The number of weeks (W) is what we need to find.
• **Constant Term:** 12 seconds per week is the constant reduction.
• **Result:** 180 seconds is the total reduction goal.

To interpret a linear equation like this, divide the total change by the consistent change per unit of the other variable. This gives us: \(W = \frac{180}{12} = 15\)So, it will take 15 weeks for the reduction.

Time Conversion

Time conversion helps in transforming minutes into seconds, making it easier to work with simpler units. In our exercise, the conversion of minutes into seconds was crucial.
The runner wants to reduce her time by 3 minutes. To convert minutes to seconds, remember that 1 minute equals 60 seconds.

Step-by-Step Conversion

• Multiply the number of minutes by 60 seconds.
• 3 minutes * 60 seconds per minute = 180 seconds.

This formula helps simplify our equation as we deal only with seconds. No mixed units!
Understanding these conversions is practical in daily problem-solving and various algebra exercises.

Problem Solving Steps

Approaching algebra word problems effectively means breaking them into manageable steps. Let's revisit each step in our running exercise:

Step 1: Understand the Goal

Clearly define what needs to be achieved; here reducing the running time by 3 minutes.

Step 2: Recognize the Reduction per Week

Identify the rate of change; here, it is 12 seconds per week.

Step 3: Set up the Equation

Translate the word problem into a mathematical equation: \[12W = 180\]

Step 4: Formulate the Equation

Define the equation’s terms: Total reduction = weekly reduction * number of weeks.

Step 5: Solving the Equation

Solve for the unknown variable. In our case, divide 180 by 12 to find W.

Step 6: Interpret the Result

Conclude by explaining the solution; here, W equals 15 weeks.

Negative Numbers

Negative numbers might seem tricky, but they add depth to algebra problems by representing decreases or losses. In our running scenario, we use negative numbers to represent the weekly reduction in time.

• **Reduction in Time:** Each week the runner 'loses' 12 seconds; mathematically, this is -12 seconds per week.
• **Total Reduction:** Over several weeks, the losses accumulate: -12W = -180 seconds.

Understanding how to work with negative numbers is crucial in scenarios involving reductions or deficits. It helps represent real-world problems more accurately and opens up a broader range of mathematical solutions.