University of South Carolina High School Math Contest/1993 Exam/Problem 6

Problem

After a price reduction, what increase does it take to restore the original price?

Solution

Let the unknown be . Initially, we have something of price . We reduce the price by $p%$ (Error compiling LaTeX. Unknown error_msg) to $Q - Q\cdot p% = Q - Q\frac p{100} = Q\cdot\frac{100 - p}{100}$ (Error compiling LaTeX. Unknown error_msg). We now increase this price by $x%$ (Error compiling LaTeX. Unknown error_msg) 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. Unknown error_msg) We can cancel from both sides to get $\frac{100 - p}{100}\cdot\left(1 + x%\right) = 1$ (Error compiling LaTeX. Unknown error_msg) so $1 + x% = \frac{100}{100 - p}$ (Error compiling LaTeX. Unknown error_msg) and $x% = \frac{p}{100 - p}$ (Error compiling LaTeX. Unknown error_msg) and , so our answer is .

Alternatively, select a particular value for such that the five answer choices all have different values. For instance, let . Thus if dollars was the original price, after the price reduction, we have dollars. We need dollars. Thus, $90(1+x%)=100 \Longrightarrow x%=\frac{10}{90}$ (Error compiling LaTeX. Unknown error_msg) and . This only matches up with answer when we plug in .

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