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

Solution 2 (Answer options)

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.

Video explaining solution

Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc

https://youtu.be/gX_l0PGsQao

https://www.youtube.com/watch?v=_TheVi-6LWE

https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 ~ MathEx

https://youtu.be/h2GlK7itc2g

~savannahsolver

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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