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 1a

After the first increase by $p$ percent, the shirt price became $(1+p)$ times greater than the original. Upon the decrease in p percent on this price, the shirt price became $(1-p)$ times less than $(1+p)$, or $(1-p)(1+p)$. We know that this price is $84$ percent of the original, so $(1-p)(1+p) = 0.84$.


From here, we can list the factors of $0.84$ and see which are equidistant from $1$. We see that $0.6$ and $1.4$ are both $0.4$ from $1$, so $p = 0.4 = 40 \%$, or choice $\boxed{\textbf{(E)}\ 40}$.


~TaeKim

Solution 2 (Answer options)

We can try out every option and see which one works. 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.

Solution 5 (A Variation of Solution 4)

Let our original cost be $$ 100$, so we are looking for a whole number of $$ 84$. Also, we can see that (A), (C), and (D) give us answers with decimals while we know that (B) and (E) give us whole numbers. Therefore, we only need to try these two: (B) $$100$ increased by 20% = $$120$, and $$120$ decreased by 20% = $$96$, a whole number, and (E) $$100$ increased by 40% = $$140$, and $$140$ decreased by 40% = $$84$, a whole number.

Thus, $40$% or $\boxed{\textbf{(E)}\ 40}$ is the answer.

~ SaxStreak

Video Solution

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

- Happytwin

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

https://youtu.be/gX_l0PGsQao

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

~ MathEx

https://youtu.be/0MoZFiPp8LA

~savannahsolver

https://www.youtube.com/watch?v=DaF8uD8V8u0&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=24

https://www.youtube.com/watch?v=i1x2b3_hmzA

https://youtu.be/2Y294ssDjVQ (The Fastest Way!) OR https://bit.ly/3kDvI9b (The Fastest Way! 2.0)

~ SaxStreak

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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