Problem 20 - Introductory Algebra for College Students

Question

You are choosing between two plans at a discount warehouse. Plan A offers an annual membership fee of 300 dollar and you pay \(70 \%\) of the manufacturer's recommended list price. Plan B offers an annual membership fee of 40 dollar and you pay \(90 \%\) of the manufacturer's recommended list price. How many dollars of merchandise would you have to purchase in a year to pay the same amount under both plans? What will be the cost for each plan?

Step-by-Step Solution

Answer

You would have to purchase 1300 dollars worth of merchandise in a year to pay the same amount under both plans. The total cost for each plan would be 1210 dollars.

Step 1: Formulate Cost Equations for each Plan

Let \( x \) be the amount of merchandise purchased. The cost under Plan A is \( 300 + 0.7x \) (membership fee plus 70% of merchandise cost). The cost under Plan B is \( 40 + 0.9x \) (membership fee plus 90% of merchandise cost).

Step 2: Set the Two Equations Equal to Each Other

For the costs to be equal, we have to set the two expressions for costs equal to each other. Solve the following equation to find the amount of merchandise: \( 300 + 0.7x = 40 + 0.9x \).

Step 3: Solve for the Value of \( x \)

Rearrange the equation to get all the \( x \) terms on one side and numerical terms on the other by subtracting 0.7x from both sides, and subtracting 40 from both sides. We get: \( 0.2x = 260 \). Now, solve for \( x \) by dividing both sides by 0.2. The solution is \( x = 260 / 0.2 = 1300 \) dollars worth of merchandise.

Step 4: Calculate the Total Cost under each Plan

For the amount of merchandise \( x = 1300 \) dollars, calculate the total cost under each plan. So, under Plan A the total cost is \( 300 + 0.7*1300 = 1210 \) dollars, and under Plan B the total cost is \( 40 + 0.9*1300 = 1210 \) dollars. Thus, the cost is indeed the same for the amount of 1300 dollars worth of merchandise.

Key Concepts

Understanding Cost Equations

In algebraic problem solving, cost equations are essential when comparing different pricing plans or financial options. They help us understand how different variables affect the overall cost. In this exercise, we have two plans with different membership fees and discounts applied to the merchandise price. To create cost equations from each plan:

• Plan A: The equation is set up as the membership fee plus a percentage of the merchandise price: \(300 + 0.7x\). Here, 300 represents the fixed annual fee, while \(0.7x\) accounts for paying 70% of the merchandise price.
• Plan B: Similar to Plan A, the total cost here is represented as \(40 + 0.9x\). The fixed amount is 40 dollars, and the merchandise is priced at 90% of its original cost, hence \(0.9x\).

These equations simplify complex pricing structures into manageable calculations allowing for direct comparison and problem-solving.

Exploring System of Equations

A system of equations arises when we have two or more equations that we want to solve at the same time. In this scenario, the goal is to find the value of \(x\) where the total costs of both plans are equal. By doing this, it allows us to determine the point of parity between Plan A and Plan B.Here, the system comprises:

• The cost equation for Plan A: \(300 + 0.7x\)
• The cost equation for Plan B: \(40 + 0.9x\)

The task is to find where these two equations have the same result, meaning where both costs are ultimately equal. By setting the equations equal (i.e., \(300 + 0.7x = 40 + 0.9x\)), we are able to find the crossover point of \(x\), which is the merchandise purchase amount required to equate both plans in terms of cost.

Techniques for Solving Equations

Solving equations is all about manipulating mathematical expressions to find unknown values, such as \(x\) in this exercise. The equation \(300 + 0.7x = 40 + 0.9x\) is a good example that demonstrates the step-by-step process of solving.To solve:

• First, we need to isolate the term with \(x\). Subtract \(0.7x\) from both sides: \(300 = 40 + 0.9x - 0.7x\).
• This simplifies to \(300 = 40 + 0.2x\).
• Next, eliminate the constant on the right by subtracting 40 from both sides: \(260 = 0.2x\).
• Finally, divide both sides by 0.2 to solve for \(x\): \(x = \frac{260}{0.2} = 1300\).

Through these steps, we uncover that 1300 dollars' worth of merchandise results in equal costs under both plans, illustrating the power and utility of algebraic problem solving in real-world scenarios.