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

Revision as of 21:06, 5 July 2017 by Mathematical-phi (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $p\%$ 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 $\left(Q\cdot\frac{100 - p}{100}\right) + \left(Q\cdot\frac{100 - p}{100}\right)\cdot x\% = \left(Q\cdot\frac{100 - p}{100}\right)\cdot(1 + x\%) = Q$ 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 $p = 10$.


Invalid username
Login to AoPS