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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 6"
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== Problem ==
 
== Problem ==
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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?
  
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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>
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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 ==
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Let <math>p=10</math>. Thus if <math>100</math> dollars was the original price, after the price reduction, we have <math>90</math> dollars. We need <math>10</math> dollars. Thus, <math>90(1+x)=100 \Longrightarrow x=\frac{10}{90}</math>. The percentage thus is choice <math>(e)</math>.
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Let the unknown be <math>x</math>.  Initially, we have something of price <math>Q</math>.  We reduce the price by <math>p\%</math> to <math>Q - Q\cdot p\% = Q - Q\frac p{100} = Q\cdot\frac{100 - p}{100}</math>.  We now increase this price by <math>x\%</math> to get <math>\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</math>  We can cancel <math>Q</math> from both sides to get <math>\frac{100 - p}{100}\cdot\left(1 + x\%\right) = 1</math> so <math>1 + x\% = \frac{100}{100 - p}</math> and <math>x\% = \frac{p}{100 - p}</math> and <math>x = \frac{100p}{100 - p}</math>, so our answer is <math>\mathrm{(E) \ } </math>.
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Alternatively, select a particular value for <math>p</math> such that the five answer choices all have different values.  For instance, let <math>p=10</math>. Thus if <math>100</math> dollars was the original price, after the price reduction, we have <math>90</math> dollars. We need <math>10</math> dollars. Thus, <math>90(1+x\%)=100 \Longrightarrow x\%=\frac{10}{90}</math> and <math>x = \frac{100 \cdot 10}{90}</math>. This only matches up with answer <math>\mathrm{(E) \ } </math> when we plug in <math>p = 10</math>.
 
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[[University of South Carolina High School Math Contest/1993 Exam/Problem 7|Next Problem]]
 
  
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[[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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*[[University of South Carolina High School Math Contest/1993 Exam/Problem 5|Previous Problem]]
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 21:06, 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 $p\%$ 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 $p = 10$.

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