Difference between revisions of "2019 AMC 8 Problems/Problem 22"
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Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0.84</math>; so <math>1-x^{2}=0.84</math>, and <math>(x^{2})=0.16</math>, obtaining <math>x=0.4</math>. Therefore, the price was increased and decreased by <math>40</math>%, or <math>\boxed{\textbf{(E)}\ 40}</math>. | Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0.84</math>; so <math>1-x^{2}=0.84</math>, and <math>(x^{2})=0.16</math>, obtaining <math>x=0.4</math>. Therefore, the price was increased and decreased by <math>40</math>%, or <math>\boxed{\textbf{(E)}\ 40}</math>. | ||
| − | ==Solution 2(Answer options)== | + | ==Solution 2 (Answer options)== |
Let the price be 100 and then trying for each option leads to <math>\boxed{\textbf{(E)}\ 40}</math>. | Let the price be 100 and then trying for each option leads to <math>\boxed{\textbf{(E)}\ 40}</math>. | ||
Revision as of 16:53, 13 August 2020
Contents
Problem 22
A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was 
of the original price, by what percent was the price increased and decreased?
Solution 1
Suppose the fraction of discount is 
. That means 
; so 
, and 
, obtaining 
. Therefore, the price was increased and decreased by 
%, or 
.
Solution 2 (Answer options)
Let the price be 100 and then trying for each option leads to .
-phoenixfire
Solution 3
Let x be the discount. We can also work in reverse such as (
)
= 
.
Thus 
= 
. Solving for gives us 
. But has to be positive. Thus = .
~phoenixfire
Video explaining solution
https://www.youtube.com/watch?v=_TheVi-6LWE
https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 ~ MathEx
See Also
| 2019 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
