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Chapter 1: Problem 37

Exercises \(37-38\) involve markup, the amount added to the dealer's cost of an item to arrive at the selling price of that item. The selling price of a refrigerator is \(\$ 584 .\) If the markup is \(25 \%\) of the dealer's cost, what is the dealer's cost of the refrigerator?

Short Answer

Expert verified
The dealer's cost of the refrigerator is approximately \$467.20.

Step by step solution

01

Express markup as a fraction of the dealer's cost

Since the markup is given as 25%, this can be expressed as a fraction of the dealer's cost, where 0.25 represents 25%.
02

Set up equation

We know that the selling price is the sum of the dealer's cost and the markup. We can express this in an equation, like so: Selling price = Dealer's cost + Markup. Given that the markup is 25% of the dealer's cost, we substitute this into the equation: \(\$584 = Dealer's cost + 0.25 * Dealer's cost\).
03

Solve for the dealer's cost

Combine like terms in the equation to give \(\$584 = 1.25 * Dealer's cost\). Now solve for Dealer's cost by dividing the selling price (\$584) by the total percentage (1.25). Thus, Dealer's cost = \(\$584/1.25\)
04

Calculate the value

Perform the division to give the dealer's cost. Hence, the dealer's cost of the refrigerator amounts to approximately \$467.20

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Key Concepts

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

Dealer's Cost
The dealer's cost is essentially what a retailer or a business owner pays to acquire a product, before adding any markup for profit. This is a fundamental concept because businesses need to recoup this cost through the selling price to ensure profitability.

In our example, a business wants to determine the original cost of a refrigerator, given a certain selling price and percentage markup. Understanding the dealer's cost allows for better pricing strategies and financial planning.
  • It's crucial to differentiate between the dealer's cost and other factors like markup and selling price.
  • In exercises like the one provided, identifying the dealer's cost is the first step in many pricing analyses.
Selling Price
The selling price is the final price at which a product is sold to a consumer. It includes both the dealer's cost and the markup added by the retailer to cover various expenses and ensure profit.

In the scenario with the refrigerator, the selling price is given as $584.
  • This price represents what a customer would pay.
  • It's critical to cover all costs, including the dealer's cost and any desired profits, through the selling price.
Thus, calculating the selling price accurately is crucial for successful business operations and satisfying budgeting needs.
Percentage Markup
Percentage markup refers to the additional amount or percentage added to the dealer's cost to create the selling price. It's expressed as a percentage of the dealer's cost, making it a flexible tool for businesses to adjust prices based on market factors or desired profit margins.

In the example provided, the markup is 25% of the dealer's cost. This means if the dealer's cost were $100, the markup would be $25, making the selling price $125.
  • Adjusting the percentage markup can help businesses remain competitive and profitable.
  • Understanding percentage markup can lead to more precise pricing strategies.
In different markets or industries, the percentage of markup can vary widely, so knowing how to calculate it is a valuable skill.
Algebraic Equation
An algebraic equation is a mathematical statement that shows the equality between two expressions. In this context, algebraic equations help to break down and calculate unknown components like the dealer's cost when you know the selling price and markup.

For our refrigerator problem, the equation set up is: \[\text{Selling price} = \text{Dealer's cost} + 0.25 \times \text{Dealer's cost}\]
  • This simplifies to \( 584 = 1.25 \times \text{Dealer's cost} \).
  • Solving this equation by dividing both sides by 1.25 reveals the dealer's cost.
This process highlights the power of algebra in solving real-world business issues by isolating variables and finding precise solutions.

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Most popular questions from this chapter

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. To earn an A in a course, you must have a final average of at least \(90 \% .\) On the first four examinations, you have grades of \(86 \% .88 \% .92 \%,\) and \(84 \% .\) If the final examination counts as two grades, what must you get on the final to earn an A in the course?

This will help you prepare for the material covered in the next section. Is \(-1\) a solution of \(3-2 x \leq 11 ?\)

The formula $$1-\frac{1}{4^{x}+26}$$ models the percentage of U.S. households with an interfaith marriage, \(I, x\) years after \(1988 .\) The formula $$N-\frac{1}{4} x+6$$ models the percentage of U.S households in which a person of faith is married to someone with no religion, \(N, x\) years after \(\overline{l 9} 88\). Use these models to solve. a. In which years will more than \(34 \%\) of U.S. households. have an interfaith marriage? b. In which years will more than \(15 \%\) of U.S. households have a person of faith married to someone with no religion? c. Based on your answers to parts (a) and (b), in which years will more than \(34 \%\) of households have an interfaith marriage and more than \(15 \%\) have a faith/no religion marriage? d. Based on your answers to parts (a) and (b), in which years will more than \(34 \%\) of households have an interfaith marriage or more than \(15 \%\) have a faith/no religion marriage?

Solve absolute value inequality. \(4<|2-x|\)

This will help you prepare for the material covered in the next section. $$\text { Solve: } \frac{x+3}{4}-\frac{x-2}{3}+\frac{1}{4}$$
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