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

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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+...

Solution

Let the unknown be x. Initially, we have something of price Q. We reduce the price by p% to Q - Q\cdot p% = Q - Q\frac p{100} = Q\cdot\frac{100 - p}{100}. We now increase this price by x% 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}\rig... We can cancel Q from both sides to get \frac{100 - p}{100}\cdot\left(1 + x%\right) = 1 so 1 + x% = \frac{100}{100 - p} and x% = \frac{p}{100 - p} 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} and x = \frac{100 \cdot 10}{90}. This only matches up with answer \mathrm{(E) \ } when we plug in p = 10.


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