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.

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 (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 2

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

~Mintylemon66

~ Minor tweak:Multpi12

Solution 3

We can assign a variable $c$ to represent the original cost of the running 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 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 $\textbf{(B) }$ , see that the discoutn price will be 40, and with sales tax applied it will be 43, so the answer choice is $\boxed{\textbf{(B) }$50}$ .

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.

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