Problem 17 - Introductory Algebra for College Students

Question

You are choosing between two telephone plans. Plan A has a monthly fee of 20 dollar with a charge of 0.05 dollar per minute for all calls. Plan B has a monthly fee of 5 dollar with a charge of 0.10 dollar per minute for all calls. a. For how many minutes of calls will the costs for the two plans be the same? What will be the cost for each plan? b. If you make approximately 10 calls per month, each averaging 20 minutes, which plan should you select? Explain your answer.

Step-by-Step Solution

Answer

The costs for Plan A and Plan B are both 35 dollars for 300 minutes. For 10 calls per month, each averaging 20 minutes, Plan B is cheaper, costing 25 dollars.

Step 1: Establish the equations

The linear equations for both plans can be established. For Plan A, which has a monthly fee of 20 dollar and a charge of 0.05 dollar per minute, the cost equation is: \( C_A = 20 + 0.05m \), where \(m\) is number of minutes. For Plan B, which has a monthly fee of 5 dollar and a charge of 0.10 dollar per minute, the cost equation is: \( C_B = 5 + 0.10m \)

Step 2: Find the number of minutes the costs are the same

To find where the cost of the two plans will be the same, you can set \(C_A = C_B \) and solve for \( m \). This gives: \( 20 + 0.05m = 5 + 0.10m \). Solving the equation, you get that \( m = 300 \). This means the costs for the two plans are the same when call minutes total to 300 minutes in a month.

Step 3: Calculate the costs at 300 minutes

Substituting \( m = 300 \) into either of the cost equations, you will get the total cost when the two plans are equal. For instance with \( C_A = 20 + 0.05*300 \), the cost will be \( 35 \) dollars.

Step 4: Determine the Suitable Plan for 10 Calls Averaging 20 Minutes

For 10 calls per month, averaging 20 minutes each, the total minutes would be 200 minutes. Substituting \( m = 200 \) into the cost equations for both Plans A and B, you can determine the cheaper plan. With Plan A, the cost would be \( 20 + 0.05*200 = 30 \) dollars. With Plan B, the cost would be \( 5 + 0.10*200 = 25 \) dollars. Therefore, for 10 calls per month, averaging 20 minutes each, Plan B would be the cheaper option.

Key Concepts

Cost Comparison

When trying to decide between two different options with varying costs, it helps to do a cost comparison. This involves examining both fixed and variable costs to determine which option is more economical for your needs.
In the example problem, we have two telephone plans: Plan A and Plan B. These plans charge:

• Plan A: A fixed monthly fee of $20 plus $0.05 per minute for calls.
• Plan B: A fixed monthly fee of $5 plus $0.10 per minute for calls.

To compare these plans, calculate the total cost of each plan using a fixed amount of usage. This method will help us visualize which plan offers a better deal based on how much you use it.
In situations where you know your average usage, such as the number of minutes you spend on calls each month, you can easily calculate which option is more affordable.

Problem-Solving

Solving this kind of problem requires a clear understanding of linear equations, alongside some problem-solving skills. The essential task is to develop a strategy to compute at what point two given plans will be equivalent and then determine the better choice.
First, we set up cost equations for both Plan A and Plan B as from the exercise. These equations are:

• Plan A: \( C_A = 20 + 0.05m \)
• Plan B: \( C_B = 5 + 0.10m \)

Here, "\( m \)" represents the number of minutes of calls.
After setting up the equations, the next step is solving these equations to find at what point the costs of both plans are equal, known as the break-even point. This involves setting the equations equal to each other and solving for "\( m \)." Reaching an answer of 300 minutes tells us that both plans will cost the same monthly amount when you use your phone for 300 minutes.
Problem-solving in this context often involves substitutions and algebraic simplifications. Following these systematic processes can guide you to an optimal and clear solution.

Systems of Equations

Here, the challenge is to solve systems of linear equations to find where two expressions are equal. A system of equations is essentially a set of two or more equations that you work on together.
In our telephone plan scenario, solving the system helps determine when the plans cost the same. The main equation is:\[ 20 + 0.05m = 5 + 0.10m \]This reflects the point where the cost balances for both plans.
By isolating the variable "\( m \)," you can solve for the exact number of minutes. Begin by subtracting the lower number from both sides and then dividing by the coefficient of \( m \). This helps find how many minutes equalize the cost of both plans.
Solving systems of equations is invaluable in various real-life situations, especially when it involves comparing costs like in this exercise. Understanding this principle enables you to apply logical solutions efficiently.