Difference between revisions of "2019 AMC 8 Problems/Problem 22"
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==Solution 1== | ==Solution 1== | ||
− | 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>. | + | 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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+ | ==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 | ||
==Solution 2 (Answer options)== | ==Solution 2 (Answer options)== |
Revision as of 23:25, 10 January 2023
Contents
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 of the original price, by what percent was the price increased and decreased
Solution 1
Suppose the fraction of discount is . That means
; so,
, and
, obtaining
. Therefore, the price was increased and decreased by
%, or
.
Solution 1a
After the first increase by percent, the shirt price became
times greater than the original. Upon the decrease in p percent on this price, the shirt price became
times less than
, or
. We know that this price is
percent of the original, so
.
From here, we can list the factors of and see which are equidistant from
. We see that
and
are both
from
, so
, or choice
.
~TaeKim
Solution 2 (Answer options)
We can try out every option and see which one works. By this method, we get .
Solution 3
Let x be the discount. We can also work in reverse such as ()
=
.
Thus, =
. Solving for
gives us
. But
has to be positive. Thus,
=
.
Solution 4 ~ using the answer choices
Let our original cost be We are looking for a result of
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
, and we have the answer; it worked.
Solution 5 (A Variation of Solution 4)
Let our original cost be , so we are looking for a whole number of
. 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)
increased by 20% =
, and
decreased by 20% =
, a whole number, and (E)
increased by 40% =
, and
decreased by 40% =
, a whole number.
Thus, % or
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://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4
~ MathEx
~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 (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.