Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - 6.3 Applications of Linear Equations - Exercise Set 6.3 - Page 375: 41



Work Step by Step

Let dealer’s cost be \[p\] and the selling price be \[s\] \[\therefore s=584\] And, Markup price \[\begin{align} & =p\times \frac{25}{100} \\ & =\frac{p}{4} \\ \end{align}\] Since markup is the amount added to the dealer’s cost of an item to arrive at the selling price of that item.So, we can write: Selling price \[=\]dealer’s cost +markup Putting corresponding values obtained, we get: \[\begin{align} & s=p+\frac{p}{4} \\ & =\frac{4p+p}{4} \\ & =\frac{5p}{4} \end{align}\] After cross multiplying, we get: \[p=\frac{4s}{5}\] Now on putting here the given value of \[s\],we get: \[\begin{align} & p=\frac{4\times 584}{5} \\ & =\frac{2336}{5} \\ & =467.2 \end{align}\] Hence dealer’s cost is \[\$467.2\]
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