Difference between revisions of "2023 AMC 12B Problems/Problem 2"
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Let the original price be <math>x</math> dollars. | Let the original price be <math>x</math> dollars. | ||
− | After the discount, the price becomes <math> 80\%x</math> dollars. | + | After the discount, the price becomes <math> 80\%\cdot x</math> dollars. |
− | After tax, the price becomes <math> 80\% \times (1+7.5\%) = 86\% x </math> dollars. | + | After tax, the price becomes <math> 80\% \times (1+7.5\%) = 86\% \cdot x </math> dollars. |
− | So, <math>43=86\%x</math>, <math>x=\boxed{\textbf{(B) }\$50}.</math> | + | So, <math>43=86\% \cdot x</math>, <math>x=\boxed{\textbf{(B) }\$50}.</math> |
~Mintylemon66 | ~Mintylemon66 | ||
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==Solution 5 (Intuition and Guessing)== | ==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 | + | 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>. |
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==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== |
Latest revision as of 12:42, 1 November 2024
- 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.