# 1965 AHSME Problems/Problem 32

## Problem

An article costing $C$ dollars is sold for \$100 at a loss of $x$ percent of the selling price. It is then resold at a profit of $x$ percent of the new selling price $S'$. If the difference between $S'$ and $C$ is $1\frac {1}{9}$ dollars, then $x$ is:

$\textbf{(A)}\ \text{undetermined} \qquad \textbf{(B) }\ \frac {80}{9} \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ \frac{95}{9}\qquad \textbf{(E) }\ \frac{100}{9}$

## Solution

The magnitude of the loss after the first sale is $C-100$, which equals $x$% of the selling price, $100$. Thus, $C-100=100*\frac{x}{100}$, and so $C=x+100$. The profit made after the second sale, $S'-100$, is $x$% of the new selling price, and this quantity is represented by $S'*\frac{x}{100}$. Equating these two expressions, we see that $S'-100=S'*\frac{x}{100}$, and so $S'=\frac{10,000}{100-x}$. Because we know that the difference between $S'$ and $C$ is $\frac{10}{9}$, we can solve the following equation: \begin{align*} \\ \frac{10,000}{100-x}-(x+100)&=\frac{10}{9} \\ 10,000-(100+x)(100-x)&=\frac{10}{9}(100-x) \\ 10,000-(10,000-x^2)&=\frac{1000}{9}-\frac{10}{9}x \\ x^2+\frac{10}{9}x-\frac{1000}{9}&=0 \\ 9x^2+10x-1000&=0 \\ (9x+100)(x-10)&=0 \\ \end{align*} Because $x \geq 0$, $x=10$, and so we choose answer $\boxed{\textbf{(C) }10}$.

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