Difference between revisions of "2019 AMC 8 Problems/Problem 22"

(Solution 1)
(Solution 3)
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==Solution 3==
 
==Solution 3==
Let x be the discount. We can also work in reverse such as <math>84</math>*<math>\frac{100}{100-x}</math>*<math>\frac{100}{100+x}</math> = <math>100</math>.
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Let x be the discount. We can also work in reverse such as (<math>84</math>)(<math>\frac{100}{100-x}</math>)(<math>\frac{100}{100+x}</math>) = <math>100</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>.
 
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>.

Revision as of 01:04, 25 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 $(1-x)(1+x)=0.84x$; 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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