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? | + | 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}% | + | <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 == | ||
− | Let <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>. |
− | [[University of South Carolina High School Math Contest/1993 Exam/Problem 7|Next Problem]] | + | 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>. |
− | [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | + | ---- |
+ | |||
+ | *[[University of South Carolina High School Math Contest/1993 Exam/Problem 5|Previous Problem]] | ||
+ | *[[University of South Carolina High School Math Contest/1993 Exam/Problem 7|Next Problem]] | ||
+ | *[[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Latest revision as of 21:06, 5 July 2017
Problem
After a price reduction, what increase does it take to restore the original price?
![$\mathrm{(A) \ }p\% \qquad \mathrm{(B) \ }\frac p{1-p}\% \qquad \mathrm{(C) \ } (100-p)\% \qquad \mathrm{(D) \ } \frac{100p}{100+p}\% \qquad \mathrm{(E) \ } \frac{100p}{100-p}\%$](http://latex.artofproblemsolving.com/b/3/9/b3913d5fd3abbcbc40bdcd74cd4b700fa906e2c9.png)
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
.