Difference between revisions of "2019 AMC 8 Problems/Problem 22"
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Thus <math>8400</math> = <math>(100+x)(100-x)</math>. Solving for <math>x</math> gives us <math>x = 40, -40</math>. But <math>x</math> has to be positive. Thus <math>x</math> = <math>40</math>. | Thus <math>8400</math> = <math>(100+x)(100-x)</math>. Solving for <math>x</math> gives us <math>x = 40, -40</math>. But <math>x</math> has to be positive. Thus <math>x</math> = <math>40</math>. | ||
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==Solution 4 ~ using the answer choices== | ==Solution 4 ~ using the answer choices== | ||
Revision as of 22:35, 26 December 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)
We can try out every option and see which one works out. By this method, we get . ~avamarora
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 = .
Solution 4 ~ using the answer choices
Let our original cost be 
We are looking for a result of 
then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try 
, and we have the answer; it worked.
Video explaining solution
Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc
https://www.youtube.com/watch?v=_TheVi-6LWE
https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 ~ MathEx
~savannahsolver
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. 
