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

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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>.
 
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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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 12:28, 23 July 2006

Problem

After a $p%$ (Error compiling LaTeX. Unknown error_msg) price reduction, what increase does it take to restore the original price?

$\mathrm{(A) \ }p% \qquad \mathrm{(B) \ }\frac p{1-p}% \qquad \mathrm{(C) \ } (100-p)% \qquad \mathrm{(D) \ } \frac{100p}{100+p}% \qquad \mathrm{(E) \ } \frac{100p}{100-p}%$ (Error compiling LaTeX. Unknown error_msg)

Solution

Let $p=10$. Thus if $100$ dollars was the original price, after the price reduction, we have $90$ dollars. We need $10$ dollars. Thus, $90(1+x)=100 \Longrightarrow x=\frac{10}{90}$. The percentage thus is choice $(e)$.