/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Involve markup, the amount added... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 for the refrigerator is $467.2

Step by step solution

01

Understanding the concept of markup

The markup is the amount added to the cost of a product to determine the selling price. Here, it's given that the markup is 25% of the dealer's cost. So, let dealer's cost be \(x\). Therefore, markup = \(0.25x\)
02

Setting up the equation

Given that the selling price of the refrigerator is $584. From the above step, we know the selling price is summation of the cost price and the markup. Therefore, we can set up the equation as follows: \(584 = x + 0.25x\)
03

Solving the equation

Combine like terms to simplify the equation which becomes \(584 = 1.25x\). Divide both sides of the equation by 1.25 to find the dealer's cost. \(x = 584 / 1.25\)
04

Calculate the dealer's cost

Calculating the cost, the dealer's cost of the refrigerator equals \(467.2\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.