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

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Let our original cost be <math>\$ 100.</math> We are looking for a result of <math>\$ 84,</math> 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 <math>\boxed{40\%}</math>, and we have the answer; it worked.
 
Let our original cost be <math>\$ 100.</math> We are looking for a result of <math>\$ 84,</math> 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 <math>\boxed{40\%}</math>, and we have the answer; it worked.
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==Solution 5 (A Variation of Solution 4)==
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Let our original cost be <math>\$ 100.</math> We are looking for a whole number of <math>\$ 84</math> and can see that (A), (C), and (D) give us answers with decimals. We know that (B) and (E) give us whole numbers, so 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.
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Thus, <math>40</math>% or <math>\boxed{\textbf{(E)}\ 40}</math> is the answer.
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~ SaxStreak
  
 
==Video Explaining Solution==  
 
==Video Explaining Solution==  
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https://youtu.be/2Y294ssDjVQ (The Fastest Way!)
 
https://youtu.be/2Y294ssDjVQ (The Fastest Way!)
  
~ Saxstreak
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~ SaxStreak
  
 
==See Also==
 
==See Also==

Revision as of 06:16, 22 December 2022

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. 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 $$ 100.$ We are looking for a whole number of $$ 84$ and can see that (A), (C), and (D) give us answers with decimals. We know that (B) and (E) give us whole numbers, so 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 Explaining 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!)

~ 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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