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Difference between revisions of "2019 AMC 8 Problems/Problem 22"
 
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==Problem==
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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 <math>84\%</math> of the original price, by what percent was the price increased and decreased<math>?</math>
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<math>\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math>
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==Solution 1==
 
==Solution 1==
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Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0.84</math>; so <math>1-x^{2}=0.84</math>, and <math>(x^{2})=0.16</math>, obtaining <math>x=0.4</math>. Therefore, the price was increased and decreased by <math>40</math>%, or <math>\boxed{\textbf{(E)}\ 40}</math>.
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Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0.84</math>; so, <math>1-x^{2}=0.84</math>, and <math>(x^{2})=0.16</math>, procuring <math>x=0.4</math>. Therefore, the price was increased and decreased by <math>40</math>%, or <math>\boxed{\textbf{(E)}\ 40}</math>.
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==Solution 1a ==
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After the first increase by <math>p</math> percent, the shirt price became <math>(1+p)</math> times greater than the original. Upon the decrease in p percent on this price, the shirt price became <math>(1-p)</math> times less than <math>(1+p)</math>, or <math>(1-p)(1+p)</math>. We know that this price is <math>84</math> percent of the original, so <math>(1-p)(1+p) = 0.84</math>.
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From here, we can list the factors of <math>0.84</math> and see which are equidistant from <math>1</math>. We see that <math>0.6</math> and <math>1.4</math> are both <math>0.4</math> from <math>1</math>, so <math>p = 0.4 = 40 \%</math>, or choice <math>\boxed{\textbf{(E)}\ 40}</math>.
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~TaeKim
  
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==Solution 2 (Answer options)==
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==Solution 2 (Time-Consuming)==
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We can try out every option and see which one works out. By this method, we get <math>\boxed{\textbf{(E)}\ 40}</math>.
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We can try out every option and see which one works. By this method, we get <math>\boxed{\textbf{(E)}\ 40}</math>.
  
 
==Solution 3==
 
==Solution 3==
 
Let x be the discount. We can also work in reverse such as (<math>84</math>)<math>(\frac{100}{100-x})</math><math>(\frac{100}{100+x})</math> = <math>100</math>.
 
Let x be the discount. We can also work in reverse such as (<math>84</math>)<math>(\frac{100}{100-x})</math><math>(\frac{100}{100+x})</math> = <math>100</math>.
  
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Thus <math>8400</math> = <math>(100+x)(100-x)</math>. Solving for <math>x</math> gives us <math>x = 40, -40</math>. But <math>x</math> has to be positive. Thus <math>x</math> = <math>40</math>.
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Thus, <math>8400</math> = <math>(100+x)(100-x)</math>. Solving for <math>x</math> gives us <math>x = 40, -40</math>. But <math>x</math> has to be positive. Thus, <math>x</math> = <math>40</math>.
  
 
==Solution 4 ~ using the answer choices==
 
==Solution 4 ~ using the answer choices==
  
 
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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(OR: try (C) first to eliminate 2 answer choices)
  
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==Video explaining solution==  
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==Solution 5 (A Variation of Solution 4)==
  
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https://www.youtube.com/watch?v=_TheVi-6LWE -  Happytwin
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Let our original cost be <math>\$ 100</math>, so we are looking for a whole number of <math>\$ 84</math>. 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) <math>\$100</math> increased by 20% = <math>\$120</math>, and <math>\$120</math> decreased by 20% = <math>\$96</math>, a whole number, and (E) <math>\$100</math> increased by 40% = <math>\$140</math>, and <math>\$140</math> decreased by 40% = <math>\$84</math>, a whole number.
  
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Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc
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Thus, <math>40</math>% or <math>\boxed{\textbf{(E)}\ 40}</math> is the answer.
  
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https://youtu.be/gX_l0PGsQao
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~ SaxStreak
  
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https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 ~ MathEx
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==Video Solution==
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==Video Solution by Marshmallow== (Video Unavailable)
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https://youtu.be/si0qSZhYeho
  
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https://youtu.be/h2GlK7itc2g
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==Video Solution by Marshmallow  (Video Unavailable)
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==
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https://youtu.be/si0qSZhYeho
  
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~savannahsolver
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==Video Solution by The Power of Logic(1 to 25 Full Solution)==
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https://youtu.be/Xm4ZGND9WoY
  
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https://www.youtube.com/watch?v=DaF8uD8V8u0&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=24
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~Hayabusa1
  
 
==See Also==
 
==See Also==

Latest revision as of 09:33, 9 November 2024

Problem

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 $(x^{2})=0.16$, procuring $x=0.4$. 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 (Time-Consuming)

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. (OR: try (C) first to eliminate 2 answer choices)

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

==Video Solution by Marshmallow== (Video Unavailable) https://youtu.be/si0qSZhYeho

==Video Solution by Marshmallow (Video Unavailable) == https://youtu.be/si0qSZhYeho

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

See Also

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. AMC logo.png

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