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

Revision as of 14:42, 29 April 2020

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 $84\%$ 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

Video explaining solution: https://youtu.be/gX_l0PGsQao

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.

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