Difference between revisions of "2023 AMC 12B Problems/Problem 2"
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+ | {{duplicate|[[2023 AMC 10B Problems/Problem 2|2023 AMC 10B #2]] and [[2023 AMC 12B Problems/Problem 2|2023 AMC 12B #2]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | |||
Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by <math>20\%</math> on every pair of shoes. Carlos also knew that he had to pay a <math>7.5\%</math> sales tax on the discounted price. He had <math>\$43</math> dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy? | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by <math>20\%</math> on every pair of shoes. Carlos also knew that he had to pay a <math>7.5\%</math> sales tax on the discounted price. He had <math>\$43</math> dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy? | ||
<math>\textbf{(A) }\$46\qquad\textbf{(B) }\$50\qquad\textbf{(C) }\$48\qquad\textbf{(D) }\$47\qquad\textbf{(E) }\$49 </math> | <math>\textbf{(A) }\$46\qquad\textbf{(B) }\$50\qquad\textbf{(C) }\$48\qquad\textbf{(D) }\$47\qquad\textbf{(E) }\$49 </math> | ||
+ | |||
+ | ==Solution 1 (easy)== | ||
+ | We can create the equation: | ||
+ | <cmath>0.8x \cdot 1.075 = 43</cmath> | ||
+ | using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get | ||
+ | <cmath>\frac{4}{5} \cdot x \cdot \frac{43}{40} = 43</cmath> | ||
+ | <cmath>\frac{4}{5} \cdot x \cdot \frac{1}{40} = 1</cmath> | ||
+ | <cmath>\frac{1}{5} \cdot x \cdot \frac{1}{10} = 1</cmath> | ||
+ | <cmath>x = \boxed{50}</cmath> | ||
+ | |||
+ | ~lprado | ||
+ | |||
+ | ==Solution 2 (One-Step Equation)== | ||
+ | The discounted shoe is <math>20\%</math> off the original price. So that means <math>1 - 0.2 = 0.8</math>. There is also a <math>7.5\%</math> sales tax charge, so <math>0.8 * 1.075 = 0.86</math>. Now we can set up the equation <math>0.86x = 43</math>, and solving that we get <math>x=\boxed{\textbf{(B) }50}</math> ~ kabbybear | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Let the original price be <math>x</math> dollars. | ||
+ | After the discount, the price becomes <math> 80\%x</math> dollars. | ||
+ | After tax, the price becomes <math> 80\% \times (1+7.5\%) = 86\% x </math> dollars. | ||
+ | So, <math>43=86\%x</math>, <math>x=\boxed{\textbf{(B) }\$50}.</math> | ||
+ | |||
+ | ~Mintylemon66 | ||
+ | ~ Minor tweak:Multpi12 | ||
+ | |||
+ | ==Solution 4== | ||
+ | We can assign a variable <math>c</math> to represent the original cost of the shoes. Next, we set up the equation <math>80\%\cdot107.5\%\cdot c=43</math>. We can solve this equation for <math>c</math> and get <math>\boxed{\textbf{(B) }\$50}</math>. | ||
+ | |||
+ | ~vsinghminhas | ||
+ | |||
+ | ==Solution 5 (Intuition and Guessing)== | ||
+ | We know the discount price will be 5/4, and 0.075 is equal to 3/40. So we look at answer choice <math>\textbf{(B) }</math>, see that the discount price will be 40, and with sales tax applied it will be 43, so the answer choice is <math>\boxed{\textbf{(B) }\$50}</math>. | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/EuLkw8HFdk4?si=IheDCDn6eMjae8SD&t=285 | ||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (Quick and Easy!)== | ||
+ | https://youtu.be/Li3znsdPT1s | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | |||
+ | https://www.youtube.com/watch?v=SUnhwbA5_So | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/KxLx1gSyESU | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2023|ab=B|num-b=1|num-a=3}} | ||
+ | {{AMC12 box|year=2023|ab=B|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:58, 24 November 2023
- The following problem is from both the 2023 AMC 10B #2 and 2023 AMC 12B #2, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (easy)
- 3 Solution 2 (One-Step Equation)
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (Intuition and Guessing)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution (Quick and Easy!)
- 9 Video Solution by SpreadTheMathLove
- 10 Video Solution
- 11 Video Solution by Interstigation
- 12 See also
Problem
Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by on every pair of shoes. Carlos also knew that he had to pay a
sales tax on the discounted price. He had
dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
Solution 1 (easy)
We can create the equation:
using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get
~lprado
Solution 2 (One-Step Equation)
The discounted shoe is off the original price. So that means
. There is also a
sales tax charge, so
. Now we can set up the equation
, and solving that we get
~ kabbybear
Solution 3
Let the original price be dollars.
After the discount, the price becomes
dollars.
After tax, the price becomes
dollars.
So,
,
~Mintylemon66
~ Minor tweak:Multpi12
Solution 4
We can assign a variable to represent the original cost of the shoes. Next, we set up the equation
. We can solve this equation for
and get
.
~vsinghminhas
Solution 5 (Intuition and Guessing)
We know the discount price will be 5/4, and 0.075 is equal to 3/40. So we look at answer choice , see that the discount price will be 40, and with sales tax applied it will be 43, so the answer choice is
.
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/EuLkw8HFdk4?si=IheDCDn6eMjae8SD&t=285 ~Math-X
Video Solution (Quick and Easy!)
~Education, the Study of Everything
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 AMC 10 Problems and Solutions |
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.