AMC 10 · 2023 · #2

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

percentageratedecimal-arithmeticlinear-equations-one-var dimensional-analysisguess-and-checkidentify-subproblems ↑ Prerequisites: percentagedecimal-arithmetic

📏 Short solution 💡 2 insights

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?

Pick an answer.

(A)

$46

(B)

$50

(C)

$48

(D)

$47

(E)

$49

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Shoes are listed at $\$x$, marked down $20\%$, then taxed $7.5\%$ at the register. Carlos has exactly $\$43$. Find the largest original price $x$ he can afford.

Givens: Original price $x$ dollars; Sale discount: $20\%$ off — pays $80\%$ of $x$; Sales tax: $7.5\%$ on the discounted price; Carlos has $\$43$; Answer choices: (A) $\$46$, (B) $\$50$, (C) $\$48$, (D) $\$47$, (E) $\$49$

Unknowns: Largest $x$ such that the final price stays at or below $\$43$

Understand

Restated: Shoes are listed at $\$x$, marked down $20\%$, then taxed $7.5\%$ at the register. Carlos has exactly $\$43$. Find the largest original price $x$ he can afford.

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #6 Guess and Check, #8 Analyze the Units

There are only five candidate prices and each one is easy to push through the "discount then tax" pipeline. Tool #3 (Eliminate) frames the work: plug each choice in and keep only the ones that produce $\le \$43$. Tool #6 (Guess and Check) is the engine for each test, and the largest surviving candidate is the answer. Tool #8 (Units) keeps the percentages honest — $\$$ in, $\times 0.8$, $\times 1.075$, $\$$ out. Algebra (#13) also works but is heavier than just multiplying $5$ numbers.

Execute — Answer: B

#8 Analyze the Units 6.RP.A.3 Step 1

Translate the two percent steps into multipliers.
A $20\%$ discount means paying $80\% = 0.8$ of the sticker price.
A $7.5\%$ tax on top of that means multiplying by $107.5\% = 1.075$.
Composing the two gives the single "sticker $\to$ register" rate.

$$0.8 \times 1.075 = 0.86$$

💡 Each percent is a Grade 6 rate; chaining two of them is the same as multiplying the rates.

#3 Eliminate Possibilities 6.RP.A.3 Step 2

So the register charge for a sticker price of $\$x$ is $0.86x$. The budget condition becomes $0.86x \le 43$.
Now plug in each choice from largest to smallest and stop at the first one that fits.

$$\text{register} = 0.86 \cdot x \quad ; \quad 0.86 \cdot x \le 43$$

💡 Once two percent moves collapse into one multiplier, plugging in answer choices is a single multiplication each.

#6 Guess and Check 5.NBT.B.7 Step 3

Test the largest choice $x = 50$: $0.86 \cdot 50 = 43$.
That hits the budget exactly — $\$43$ pays for it. So $\$50$ is affordable, and nothing larger appears in the list.

$$0.86 \cdot 50 = 43.00 \le 43 \;\checkmark$$

💡 Multiplying $0.86 \cdot 50$ as $86 \cdot 50 \div 100 = 4300 \div 100 = 43$ is direct Grade 5 decimal arithmetic.

#3 Eliminate Possibilities 6.EE.B.5 Step 4

Cross-check by ruling out the only larger imaginable case.
The choice list tops out at $\$50$, so any $x > 50$ would have to give $0.86x > 43$ — too expensive. So $\$50$ is the maximum allowed and the answer is (B).

$$x = 50 \Rightarrow \textbf{(B)}$$

💡 Among five candidates, the largest one that still satisfies $0.86x \le 43$ is the answer — pure Grade 6 substitute-and-check.

[1] #8 6.RP.A.3 Translate the two percent steps into multipliers. A $20\%$ discount means paying

[2] #3 6.RP.A.3 So the register charge for a sticker price of $\$x$ is $0.86x$. The budget condi

[3] #6 5.NBT.B.7 Test the largest choice $x = 50$: $0.86 \cdot 50 = 43$. That hits the budget exa

[4] #3 6.EE.B.5 Cross-check by ruling out the only larger imaginable case. The choice list tops

Review

Reasonableness: Magnitude check: a $20\%$ cut on $\$50$ gives $\$40$, and a $7.5\%$ tax on $\$40$ is $\$3$, total $\$43$ — matches Carlos's wallet exactly. Smaller choices ($\$46, \$47, \$48, \$49$) would also fit but aren't the *most expensive*, and the problem asks for the maximum, so (B) is correct.

Alternative: Tool #13 (Convert to Algebra). Solve $0.86x = 43$ directly: $x = \frac{43}{0.86} = \frac{4300}{86} = 50$. Same answer, but requires either a calculator-style long division or the insight to multiply numerator and denominator by $100$ first.

CCSS standards used (min grade 6)

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Computing $0.86 \cdot 50 = 43$ to test the largest choice.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Turning $20\%$ off and $7.5\%$ tax into the single multiplier $0.8 \cdot 1.075 = 0.86$.)
6.EE.B.5 Understand solving an equation or inequality as finding values that make it true (Choosing the largest $x$ from the choice list that keeps $0.86x \le 43$.)

⭐ This AMC 10 problem only needs the Grade 6 idea that "$20\%$ off then $7.5\%$ tax" is the same as multiplying by $0.86$ — once the rate is collapsed, just plug in each choice.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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