Problem 19 - Precalculus Mathematics for Calculus

Question

A rental company charges \(\$ 65\) a day and 20 cents a mile for renting a truck. Michael rented a truck for 3 days, and his bill came to \(\$ 275 .\) How many miles did he drive?

Step-by-Step Solution

Answer

Michael drove 400 miles.

Step 1: Understand the Problem

The problem states that Michael rented a truck for 3 days at a base cost of $65 per day and an additional $0.20 per mile driven. His total bill amounted to $275. We need to find out how many miles he drove.

Step 2: Calculate Base Rental Cost for Days

Calculate the cost of renting the truck based solely on the number of days. Multiply the number of days (3) by the daily rate ($65): \[ \text{Base Cost} = 3 \times 65 = 195 \] This is the cost for renting the truck for 3 days alone.

Step 3: Determine Extra Cost for Miles Driven

Subtract the base cost calculated in Step 2 from the total bill to determine the cost associated with the miles driven: \[ \text{Extra Cost for Miles} = 275 - 195 = 80 \]

Step 4: Calculate Miles Driven

Since the extra cost is due to mileage, divide this cost by the cost per mile to find the number of miles: \[ \text{Miles Driven} = \frac{80}{0.20} = 400 \] Thus, Michael drove 400 miles.

Key Concepts

Linear Equations

Linear equations are a vital concept in algebra used to describe relationships between different variables. They are linear because they graph as straight lines. The general form is \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable.

When working with linear equations in real-world problems, like cost calculations, we often use them to find unknown quantities. For instance, if you know the total cost and the rate per unit, you can find the number of units. In this exercise, we set up a linear equation based on the cost structure:

• Base cost per day times the number of days
• Plus the cost per mile times the miles driven

This formed the equation: \( 65 \times 3 + 0.20x = 275 \). Solving for \( x \) helps find the miles driven.

Problem-Solving

Problem-solving in math often involves breaking down complex problems into manageable parts. The key is to understand what is given and what needs to be found.

In our example, the task was to determine the number of miles driven based on daily rental costs and mileage charges. By understanding:

• Michael's total days of rental and the per-day cost
• The bill's total amount
• The cost per mile driven

We systematically approached the problem, first finding fixed daily costs, then calculating the variable cost based on mileage, and finally solving for the unknown miles. Breaking down the problem helps in organizing thoughts and finding solutions efficiently.

Cost Calculation

Cost calculation involves understanding and computing the total cost by summing fixed and variable expenses.

In scenarios like renting a vehicle, there are typically fixed costs, such as a daily rental fee, and variable costs, like fees per mile.

• Fixed costs are easily calculated by multiplying the unit cost by the number of units.
• Variable costs depend on usage; thus, knowing the rate and the extent of use is crucial.

In Michael's case, his daily fixed cost was straightforward: \( 3 \times 65 = 195 \). The variable cost had to be calculated by subtracting this from the total bill, which was found to be an extra \( 80 \).

Properly discerning these elements can turn a seemingly complicated real-life problem into a simple computation.