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Chapter 3: Problem 31

Use a variation model to solve for the unknown value. The average daily cost to rent a car is inversely proportional to the number of miles driven. If \(100 \mathrm{mi}\) is driven, the average daily cost is \(\$ 0.80\) per mile. a. Find the average daily cost if \(200 \mathrm{mi}\) is driven. b. Find the average daily cost if \(300 \mathrm{mi}\) is driven. c. Find the average daily cost if \(400 \mathrm{mi}\) is driven. d. If the average cost is \(\$ 0.16\), how many miles were driven?

Short Answer

Expert verified
a: \)0.40 per mile, b: \(0.267 per mile, c: \)0.20 per mile, d: 500 miles.

Step by step solution

01

Identify the variation model

Since the cost is inversely proportional to the number of miles driven, we use the formula: \[ C = \frac{k}{m} \] where:\( C \) = cost per mile\( m \) = number of miles driven\( k \) = constant of proportionality
02

Determine the constant of proportionality

Using the given values \( C = 0.80 \) and \( m = 100 \):\[ 0.80 = \frac{k}{100} \]Solving for \( k \):\[ k = 0.80 \times 100 \]\[ k = 80 \]
03

Calculate the cost for 200 miles

Using the formula \( C = \frac{k}{m} \) and \( k = 80 \), for \( m = 200 \):\[ C = \frac{80}{200} \]\[ C = 0.40 \]The average daily cost for 200 miles is \( \$ 0.40 \) per mile.
04

Calculate the cost for 300 miles

Using the same formula for \( m = 300 \):\[ C = \frac{80}{300} \]\[ C = \approx 0.267 \]The average daily cost for 300 miles is \( \$ 0.267 \) per mile.
05

Calculate the cost for 400 miles

Using the same formula for \( m = 400 \):\[ C = \frac{80}{400} \]\[ C = 0.20 \]The average daily cost for 400 miles is \( \$ 0.20 \) per mile.
06

Find the number of miles for a cost of \(0.16

Using the formula and solving for \( m \) when \( C = 0.16 \):\[ 0.16 = \frac{80}{m} \]Multiplying both sides by \( m \):\[ 0.16m = 80 \]Dividing both sides by 0.16:\[ m = \frac{80}{0.16} \]\[ m = 500 \]The number of miles driven is 500.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

constant of proportionality
In this exercise, we come across the concept of the constant of proportionality. This constant is crucial when dealing with inverse variation problems. In simpler terms, it is a fixed value that helps us understand how one variable changes when another variable changes inversely. Here, the cost per mile (C) and the number of miles driven (m) are inversely proportional to each other. This means that as the number of miles increases, the cost per mile decreases and vice versa.
inverse relationship
An inverse relationship is one where two variables move in opposite directions. When one variable increases, the other decreases, and vice versa. In this problem, we observe an inverse relationship between the cost per mile and the number of miles driven. This is described using the equation \( C = \frac{k}{m} \). As seen in the exercise, if you drive more miles, the cost per mile decreases, following the principle of inverse variation.
variation model
To understand the variation model used in this exercise, we should look at how it helps us calculate unknown values. The model \( C = \frac{k}{m} \) allows us to find either the cost per mile or the number of miles driven if we know the constant of proportionality (k). We start by using given values to determine k, and then we can apply it to find other unknown values in similar inverse variation scenarios. This helps in solving multiple parts of the problem efficiently and accurately.
cost per mile calculation
Calculating the cost per mile in this problem involves using the inverse variation formula. For instance, if we know that driving 100 miles costs 0.80 per mile, we calculate the constant of proportionality k. We can then use k to determine the cost for different mileages. For example:
  • For 200 miles: \(C = \frac{80}{200} = 0.40\)
  • For 300 miles: \(C = \frac{80}{300} \approx 0.267\)
  • For 400 miles: \(C = \frac{80}{400} = 0.20\)
By following this method, students can easily determine the cost per mile for any given number of driven miles, demonstrating the practical application of inverse variation.

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Most popular questions from this chapter

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