## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 6 - Algebra: Equations and Inequalities - 6.3 Applications of Linear Equations - Exercise Set 6.3 - Page 375: 42

#### Answer

$\12$.

#### Work Step by Step

Let dealer’s cost be $p$ and the selling price be $s$ $\therefore s=15$and Markup \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. By using formula 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 $p=\frac{4s}{5}$ On putting the given value of $s$in the above equation, we get: \begin{align} & p=\frac{4\times 15}{5} \\ & =\frac{60}{5} \\ & =12 \end{align} Hence dealer’s cost is $\12$.

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