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: 10,000100x(x+100)=10910,000(100+x)(100x)=109(100x)10,000(10,000x2)=10009109xx2+109x10009=09x2+10x1000=0(9x+100)(x10)=0 Because $x \geq 0$, $x=10$, and so we choose answer $\boxed{\textbf{(C) }10}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png