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)=0.84x</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>.
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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>.
  
 
==Solution 2(Answer options)==
 
==Solution 2(Answer options)==

Revision as of 23:31, 28 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 $84\%$ of the original price, by what percent was the price increased and decreased?

$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

Solution 1

Suppose the fraction of discount is $x$. That means $(1-x)(1+x)=0.84$; so $1-x^{2}=0.84$, and $(x^{2})=0.16$, obtaining $x=0.4$. Therefore, the price was increased and decreased by $40$%, or $\boxed{\textbf{(E)}\ 40}$.

Solution 2(Answer options)

Let the price be $100$ and then trying for each option leads to $\boxed{\textbf{(E)}\ 40}$.

~phoenixfire

Solution 3

Let x be the discount. We can also work in reverse such as ($84$)$(\frac{100}{100-x})$$(\frac{100}{100+x})$ = $100$.

Thus $8400$ = $(100+x)(100-x)$. Solving for $x$ gives us $x = 40, -40$. But $x$ has to be positive. Thus $x$ = $40$.

~phoenixfire

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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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