2023 AMC 12B Problems/Problem 2: Difference between revisions
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<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== | ==Solution 1 (easy)== | ||
We can create the equation: | We can create the equation: | ||
<cmath>0.8x \cdot 1.075 = 43</cmath> | <cmath>0.8x \cdot 1.075 = 43</cmath> | ||
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~lprado | ~lprado | ||
==Solution 2 | ==Solution 2== | ||
Let the original price be <math>x</math> dollars. | Let the original price be <math>x</math> dollars. | ||
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~ Minor tweak:Multpi12 | ~ Minor tweak:Multpi12 | ||
==Solution | ==Solution 3== | ||
We can assign a variable <math>c</math> to represent the original cost of the running 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>. | We can assign a variable <math>c</math> to represent the original cost of the running 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 | ~vsinghminhas | ||
==Solution | ==Solution 4 (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 discoutn price will be 40, and with sales tax applied it will be 43, so the answer choice is <math>\boxed{\textbf{(B) }\$50}</math>. | 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 discoutn price will be 40, and with sales tax applied it will be 43, so the answer choice is <math>\boxed{\textbf{(B) }\$50}</math>. | ||
Revision as of 20:43, 15 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.
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:
![\[0.8x \cdot 1.075 = 43\]](http://latex.artofproblemsolving.com/a/9/9/a99ff0797faf6d76f376fba7bc0a084a8d4ea5c9.png)
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
![\[\frac{4}{5} \cdot x \cdot \frac{43}{40} = 43\]](http://latex.artofproblemsolving.com/c/9/5/c95c2d267bbcce7ba7040eb5af23664cb7a18a8a.png)
![\[\frac{4}{5} \cdot x \cdot \frac{1}{40} = 1\]](http://latex.artofproblemsolving.com/3/d/a/3da7b3c51c8e58b7d1253603a69bfb4d196976f5.png)
![\[\frac{1}{5} \cdot x \cdot \frac{1}{10} = 1\]](http://latex.artofproblemsolving.com/5/0/4/50451ecbdc9c61b1b4221ebac9baf0dce18704ea.png)
![\[x = \boxed{50}\]](http://latex.artofproblemsolving.com/8/5/c/85c6e9e3aed031fa6a0577bb820b7ede184c97af.png)
~lprado
Solution 2
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 3
We can assign a variable 
to represent the original cost of the running shoes. Next, we set up the equation 
. We can solve this equation for and get 
.
~vsinghminhas
Solution 4 (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 .
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
