2019 AMC 8 Problems/Problem 22

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

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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2019 AMC 8 Problems/Problem 22

Contents

Problem 22

Solution 1

Solution 2(Answer options)

Solution 3

See Also