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

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== Problem ==
 
== Problem ==
After a <math>p%</math> price reduction, what increase does it take to restore the original price?
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After a <math>p\%</math> price reduction, what increase does it take to restore the original price?
  
<center><math> \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}% </math></center>
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<center><math> \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}\%   </math></center>
  
 
== Solution ==
 
== Solution ==

Revision as of 21:04, 5 July 2017

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 $p = 10$.


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