Difference between revisions of "2023 AMC 12B Problems/Problem 2"

(Solution 5 (Intuition and Guessing))
(Solution 1)
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==Solution 1==
 
==Solution 1==
  
Let the price originally be \( x \). Then, after a 20\% discount, the price is now \( x - \frac{1}{5}x = \frac{4}{5}x \).  
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Let the price originally be \( x \). Then, after a \(20%\) discount, the price is now \( x - \frac{1}{5}x = \frac{4}{5}x \).  
  
 
From the discounted price \( \frac{4}{5}x \), we now take \( \frac{7.5}{100} \) of \( \frac{4}{5}x \) and add it to \( \frac{4}{5}x \), giving us \( \frac{4}{5}x + \left(\frac{7.5}{100}\right)\left(\frac{4}{5}x\right) = \frac{4}{5}x + \frac{30}{500}x = \frac{4}{5}x + \frac{3}{50}x = \frac{43}{50}x \).  
 
From the discounted price \( \frac{4}{5}x \), we now take \( \frac{7.5}{100} \) of \( \frac{4}{5}x \) and add it to \( \frac{4}{5}x \), giving us \( \frac{4}{5}x + \left(\frac{7.5}{100}\right)\left(\frac{4}{5}x\right) = \frac{4}{5}x + \frac{30}{500}x = \frac{4}{5}x + \frac{3}{50}x = \frac{43}{50}x \).  

Revision as of 11:07, 22 August 2025

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 $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?


$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$

Solution 1

Let the price originally be \( x \). Then, after a \(20%\) discount, the price is now \( x - \frac{1}{5}x = \frac{4}{5}x \).

From the discounted price \( \frac{4}{5}x \), we now take \( \frac{7.5}{100} \) of \( \frac{4}{5}x \) and add it to \( \frac{4}{5}x \), giving us \( \frac{4}{5}x + \left(\frac{7.5}{100}\right)\left(\frac{4}{5}x\right) = \frac{4}{5}x + \frac{30}{500}x = \frac{4}{5}x + \frac{3}{50}x = \frac{43}{50}x \).

Now we write the inequality \( \frac{43}{50}x \leq $43 \) and multiply by \( \frac{50}{43} \) on both sides to get \( x \leq $50 \).

We want the greatest original price, which would be \($50\) or option choice \(B\).

~Pinotation

Solution 2 (easy)

We can create the equation: \[0.8x \cdot 1.075 = 43\] 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 \[\frac{4}{5} \cdot x \cdot \frac{43}{40} = 43\] \[\frac{4}{5} \cdot x \cdot \frac{1}{40} = 1\] \[\frac{1}{5} \cdot x \cdot \frac{1}{10} = 1\] \[x  = \boxed{50}\]

~lprado

Solution 3 (One-Step Equation)

The discounted shoe is $20\%$ off the original price. So that means $1 - 0.2 = 0.8$. There is also a $7.5\%$ sales tax charge, so $0.8 * 1.075 = 0.86$. Now we can set up the equation $0.86x = 43$, and solving that we get $x=\boxed{\textbf{(B) }50}$ ~ kabbybear

Solution 4

Let the original price be $x$ dollars. After the discount, the price becomes $80\%\cdot x$ dollars. After tax, the price becomes $80\% \times (1+7.5\%) = 86\% \cdot x$ dollars. So, $43=86\% \cdot x$, $x=\boxed{\textbf{(B) }$50}.$

~Mintylemon66

~ Minor tweak:Multpi12

Solution 5

We can assign a variable $c$ to represent the original cost of the shoes. Next, we set up the equation $80\%\cdot107.5\%\cdot c=43$. We can solve this equation for $c$ and get $\boxed{\textbf{(B) }$50}$.

~vsinghminhas

Solution 6 (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 $\textbf{(B) }$, see that the discount price will be 40, and with sales tax applied it will be 43, so the answer choice is $\boxed{\textbf{(B) }$50}$.

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

2023 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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