Difference between revisions of "2019 AMC 8 Problems/Problem 22"
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Let the price be 100. Then multiply by 40/100 to get 140. Then, when you subtract by 40 percent, you will end up at <math>\boxed{\textbf{(E)}\ 40}</math>. | Let the price be 100. Then multiply by 40/100 to get 140. Then, when you subtract by 40 percent, you will end up at <math>\boxed{\textbf{(E)}\ 40}</math>. | ||
| − | + | ~Gr8 | |
==Solution 3== | ==Solution 3== | ||
Revision as of 22:19, 3 November 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. Then multiply by 40/100 to get 140. Then, when you subtract by 40 percent, you will end up at .
~Gr8
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
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
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. 
