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

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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>. | + | 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>. |

− | 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>. | + | 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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## Latest revision as of 21:06, 5 July 2017

## Problem

After a price reduction, what increase does it take to restore the original price?

## Solution

Let the unknown be . Initially, we have something of price . We reduce the price by to . We now increase this price by to get We can cancel from both sides to get so and and , so our answer is .

Alternatively, select a particular value for such that the five answer choices all have different values. For instance, let . Thus if dollars was the original price, after the price reduction, we have dollars. We need dollars. Thus, and . This only matches up with answer when we plug in .