The selling price of a refrigerator is $528.15. If the markup is 5% of the dealer's cost, what is the dealer's cost of the refrigerator?

Let the dealer's cost of the refrigerator be denoted by $ C $. The selling price is calculated by adding a 5% markup to the dealer's cost, which can be expressed as:

\[
\text{Selling Price} = C + 0.05C = 1.05C
\]

Given that the selling price is $528.15, we can set up the equation:

\[
1.05C = 528.15
\]

Now, solve for $ C $ by dividing both sides of the equation by 1.05:

\[
C = \frac{528.15}{1.05}
\]

Let me calculate that for you.The dealer's cost of the refrigerator is approximately **$503.00**.

Would you like further details or have any questions about this process? Here are 5 related questions to expand your understanding:

1. How would the dealer's cost change if the markup were 10% instead of 5%?
2. What is the selling price if the dealer's cost is $600 and the markup is 5%?
3. How do you calculate markup as a percentage of the selling price rather than the cost?
4. What is the dealer's cost if the selling price is $528.15 and the markup is 8%?
5. Can the markup percentage ever exceed 100%, and if so, how would that affect the dealer's cost?

**Tip:** When dealing with percentages, always express the markup as a fraction of the original value for clearer calculations.

Learn how to calculate the dealer's cost of a refrigerator with a selling price of $528.15 and a 5% markup. This problem involves algebra and percentage-based markup calculations, suitable for students in Grades 8-10.

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