Problem 26 - Algebra 2

Question

For Exercises 25 and \(26,\) use the following information. Namid is examining the calling card portion of his phone bill. A 4 -minute call at the night rate cost \(\$ 2.65 .\) A 10 -minute call at the night rate cost \(\$ 4.75\) . How much would it cost to talk for half an hour at the night rate?

Step-by-Step Solution

Answer

The cost for a 30-minute call is \(\$11.75\).

Step 1: Identify Known Values

We know the cost of a 4-minute call and a 10-minute call at the night rate. Specifically, a 4-minute call costs \(2.65\) and a 10-minute call costs \(4.75\).

Step 2: Establish the Relationships

Let \(r\) be the rate per minute and \(b\) be any fixed cost (base cost for connecting the call). We have two equations: \(4r + b = 2.65\) and \(10r + b = 4.75\).

Step 3: Solve for Rate per Minute and Fixed Cost

Subtract the first equation from the second to eliminate \(b\): \(10r + b - (4r + b) = 4.75 - 2.65\), resulting in \(6r = 2.10\). Then, solve for \(r\) by dividing both sides by 6 to get \(r = \frac{2.10}{6} = 0.35\). Substitute \(r = 0.35\) back into the first equation \(4r + b = 2.65\) to find \(b\). This gives \(4(0.35) + b = 2.65\) or \(1.40 + b = 2.65\), leading to \(b = 2.65 - 1.40 = 1.25\).

Step 4: Calculate the Cost for Half an Hour

A half-hour call is 30 minutes. Using the cost structure, the total cost is \(30 \times r + b\). Substitute \(r = 0.35\) and \(b = 1.25\): \(30 \times 0.35 + 1.25 = 10.5 + 1.25 = 11.75\).

Key Concepts

Rate Calculation

The concept of rate calculation is crucial to understanding how costs accumulate over time for services billed on a per-time-unit basis. In this exercise, each minute of phone use incurs a specific charge, represented as the rate per minute. This rate helps determine the total cost of calls of varying lengths. This is derived by dividing the difference in total call costs by the difference in minutes between two calls.
In our example:

• A 4-minute call costs \(2.65\).
• A 10-minute call costs \(4.75\).

By using these two calls, we can set up equations to solve for the rate per minute. The difference in the cost of the calls over the difference in time is used to calculate this rate. After solving the equations, we found a rate of \(0.35\) or \(35\) cents per minute, meaning every additional minute of call costs an extra \(0.35\) dollars.

Cost Structure

Understanding cost structure is fundamental to assessing how different elements contribute to the total cost of a service. In the context of phone billing, the cost structure might include both a dynamic cost tied to usage and a fixed cost.
For Namid’s phone calls:

• The dynamic cost is based on the rate per minute (\(r = 0.35\)).
• The fixed cost is a base charge (\(b = 1.25\)).

The fixed cost can be seen as the charge for connecting the call, regardless of the call's duration. When calculating the total cost, both components must be considered. Therefore, the total cost for any call duration is the sum of the dynamic and fixed costs, so for a 30-minute call, the total cost is calculated by adding the product of the call duration and the rate per minute to the base cost.

Solving Equations

Solving linear equations is a fundamental skill in mathematics that helps in finding values of unknown variables. These are typically represented by symbols such as \(x\) or \(r\). When given multiple linear equations, one efficient method to find unknown values is to use the elimination or substitution method.
In our exercise, we were given:

• \(4r + b = 2.65\)
• \(10r + b = 4.75\)

To eliminate the variable \(b\), which represents the fixed cost, we subtracted the first equation from the second. This resulted in an equation with a single variable, \(r\). By determining \(r\), we then substituted it back into the initial equation to solve for \(b\). Understanding these methods enables us to decode real-world problems presented as mathematical equations and find practical solutions.