Difference between revisions of "2019 AMC 8 Problems/Problem 22"
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==Solution 1== | ==Solution 1== | ||
− | Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0. | + | 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)== |
Revision as of 00:31, 29 November 2019
Problem 22
A store increased the original price of a shirt by a certain percent and then decreased 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 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
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 |
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