2023 AMC 12B Problems/Problem 2
- The following problem is from both the 2023 AMC 10B #2 and 2023 AMC 12B #2, so both problems redirect to this page.
Contents
[hide]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 discoutn price will be 40, and with sales tax applied it will be 43, so the answer choice is .
Solution 6 (Not really a solution, DON'T DO THIS ON A REAL TEST)
Open up a coding IDE and use Python to solve this problem. Python code:
budget = 43.0 discount_percentage = 20.0 sales_tax_percentage = 7.5 discounted_price = budget / 1.075 / 0.8 print(f"${discounted_price:.2f}")
~Ishaan Garg
Video Solution (Quick and Easy!)
~Education, the Study of Everything
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
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.