# Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 6"

## Problem

After a $p\%$ price reduction, what increase does it take to restore the original price? $\mathrm{(A) \ }p\% \qquad \mathrm{(B) \ }\frac p{1-p}\% \qquad \mathrm{(C) \ } (100-p)\% \qquad \mathrm{(D) \ } \frac{100p}{100+p}\% \qquad \mathrm{(E) \ } \frac{100p}{100-p}\%$

## Solution

Let the unknown be $x$. Initially, we have something of price $Q$. We reduce the price by $p%$ (Error compiling LaTeX. ! Missing $inserted.) to$Q - Q\cdot p% = Q - Q\frac p{100} = Q\cdot\frac{100 - p}{100}$(Error compiling LaTeX. ! Missing$ inserted.). We now increase this price by $x%$ (Error compiling LaTeX. ! Missing $inserted.) to get$\left(Q\cdot\frac{100 - p}{100}\right) + \left(Q\cdot\frac{100 - p}{100}\right)\cdot x% = \left(Q\cdot\frac{100 - p}{100}\right)\cdot(1 + x%) = Q$(Error compiling LaTeX. ! Missing$ inserted.) We can cancel $Q$ from both sides to get $\frac{100 - p}{100}\cdot\left(1 + x%\right) = 1$ (Error compiling LaTeX. ! Missing $inserted.) so$1 + x% = \frac{100}{100 - p}$(Error compiling LaTeX. ! Missing$ inserted.) and $x% = \frac{p}{100 - p}$ (Error compiling LaTeX. ! Missing $inserted.) and $x = \frac{100p}{100 - p}$, so our answer is $\mathrm{(E) \ }$. Alternatively, select a particular value for $p$ such that the five answer choices all have different values. For instance, let $p=10$. Thus if $100$ dollars was the original price, after the price reduction, we have $90$ dollars. We need $10$ dollars. Thus,$90(1+x%)=100 \Longrightarrow x%=\frac{10}{90}$(Error compiling LaTeX. ! Missing$ inserted.) and $x = \frac{100 \cdot 10}{90}$. This only matches up with answer $\mathrm{(E) \ }$ when we plug in $p = 10$.