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

## 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 $\boxed{\textbf{(E)}\ 40}$.

We can try out every option and see which one works out. By this method, we get $\boxed{\textbf{(E)}\ 40}$.

## 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 $x$ = $40$.

## Solution 4 ~ using the answer choices

Let our original cost be $100.$ We are looking for a result of $84,$ then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try $\boxed{40\%}$, and we have the answer; it worked.