Question

The cost of renting a midsize car from Company A is \(\$ 279\) per week with no extra charge for mileage. The cost of renting a similar car from Company B is \(\$ 199\) per week, plus 32 cents for each mile driven. How many miles must you drive in a week to make the rental fee for Company \(\mathrm{B}\) greater than that for Company \(\mathrm{A}\) ?

Short answer

You need to drive more than 250 miles in a week for Company B's rental to be more expensive than Company A's.

Step-by-step solution

Identify the variables

Let's denote the number of miles driven in a week by a variable, say 'm'.

Form the inequality equation

Given that the cost of renting from Company A is $279 and from Company B it's $199 plus 32 cents per mile, we can form the following inequality to represent when Company B becomes more expensive than Company A: 199 + 0.32*m > 279.

Solve the inequality for 'm'

We can solve the above inequality for 'm' by subtracting 199 from both sides which gives 0.32m > 80.

Division step

Next, by dividing both sides by 0.32, we find 'm' > 250. This means that Company B's rental becomes more expensive than Company A's after driving more than 250 miles in a week.

Key concepts

Inequality

Inequalities are mathematical expressions that show the relationship between two values, indicating one is greater, less, or equal to the other. In this particular problem, we have an inequality that helps determine when the cost from Company B surpasses Company A. The inequality is represented as:

• Company A cost: \( 279 \)
• Company B cost: \( 199 + 0.32m \)
• Inequality condition: \( 199 + 0.32m > 279 \)

The aim is to find the number of miles (m) where Company B's cost exceeds that of Company A.
You set up inequalities by comparing expressions using symbols such as \( > \), \( < \), and \( = \). Solving an inequality involves similar steps as solving an equation, with some special rules. For example, if you multiply or divide by a negative number, the inequality sign flips.

Variables

Variables are symbols, usually letters, used to represent unknown or varying quantities in algebraic expressions and equations. Here, the variable \( m \) is used to signify the number of miles driven during the car rental period.
Making use of variables is a powerful way to simplify complex situations into manageable mathematical expressions. By replacing unknown quantities with variables, you can form equations or inequalities to solve specific problems.

• Choosing the right variable can simplify your calculations. In this exercise, \( m \) directly represents miles driven, making it intuitive.
• Having a clear meaning attached to your variable helps prevent errors, especially in word problems where context is crucial.

In essence, variables are placeholders for values you either know (as constants) or are trying to find in mathematical problems.

Algebraic Equations

Algebraic equations form the backbone of math problems that involve unknown values. They are statements showing that two expressions are equal, and solving them means finding the unknown variable(s).
In this exercise, we start with the calculation part:

• Calculate the cost difference needed: \( 279 - 199 = 80 \)
• Equation: \( 0.32m = 80 \)

The algebraic transformation then comes into play. We subtract the fixed costs of Company B from Company A and equate it against the cost per mile \( 0.32 \times m \). Solving this equation involves simple algebraic manipulation:

• Isolate the variable by division: \( m = \frac{80}{0.32} \)
• Compute: \( m = 250 \)

Thus, by solving the equation, we determine that more than 250 miles must be driven for Company B's rental cost to exceed that of Company A. Understanding these steps allows for clarity in setting up and solving algebraic equations in similar scenarios.