Competition · AMC preparation · step 4 of 4

AMC 8 · 2024 · #19

Grade 4 rate-ratio
Archetype Overlap / Complement Counting └ Sub-archetype Two-Attribute Table: Fill or Optimize a Cell
ratio-proportionfraction-arithmeticcomplementary-counting complementary-countingcasework ↑ Prerequisites: fraction-arithmeticratio-proportion
📏 Medium solution 💡 3 insights 📊 Diagram
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Problem
Jordan owns 15 pairs of sneakers. 35\frac{3}{5} of them are red (the rest white) and 23\frac{2}{3} are high-top (the rest low-top). What is the smallest possible fraction of the 15 pairs that can be both red AND high-top?

Pick an answer.

(A)
0
(B)
$\dfrac{1}{5}$
(C)
$\dfrac{4}{15}$
(D)
$\dfrac{1}{3}$
(E)
$\dfrac{2}{5}$

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Two classifications (color and style) overlap, so the first move is Tool #1 — draw a 2 by 2 table that shows the four cells (red high-top, red low-top, white high-top, white low-top) all at once. The goal "make the red high-top cell as small as possible" is hard to attack directly, but flipping the perspective with Tool #16 turns it into the much easier goal "make the red low-top cell as large as possible." Finally Tool #6 (Guess and Check) lets us try x = 0, 1, 2, 3, 4 in turn and pick the smallest x that keeps every cell ≥ 0. We do not need Tool #13 (algebra) — table + complement + check is enough.

1STEP 1

Count each category

Take a fraction of the whole: 35\frac{3}{5} × 15 = 9 red and 15 - 9 = 6 white; 23\frac{2}{3} × 15 = 10 high-top and 15 - 10 = 5 low-top.

3/5 × 15 = 9, 15 - 9 = 6, 2/3 × 15 = 10, 15 - 10 = 5
2STEP 2

Fill in a two-by-two table

Draw a 2×2 table with x = red high-top; row/column sums force red low-top = 9 - x, white high-top = 10 - x, white low-top = x - 4.

& High-top & Low-top & Total ; Red & x & 9-x & 9 ; White & 10-x & x-4 & 6 ; Total & 10 & 5 & 15
3STEP 3

Flip to the complement

Flip the goal: minimize x by maximizing red low-top = 9 - x. It can't exceed the 5 low-tops, so 9 - x ≤ 5, forcing x ≥ 4.

red low-top = 9 - x ≤ min(5, 9) = 5 ⟹ x ≥ 9 - 5 = 4
4STEP 4

Check the candidate values

Guess and check x = 0,1,2,3: each forces red low-top above 5 — impossible. Only x = 4 gives red low-top 5, white high-top 6, all cells fit.

x = 4: & High-top & Low-top & Total ; Red & 4 & 5 & 9 ; White & 6 & 0 & 6 ; Total & 10 & 5 & 15
5STEP 5

Write the answer as a fraction

Divide the minimum count 4 by 15 to get 415\frac{4}{15}, already in lowest terms — choice (C); the other choices need x = 0 or 3, both impossible.

4/15 → (C)
Answer
4/15
Re-add the x = 4 table by rows and columns: rows give 4 + 5 = 9 (red) and 6 + 0 = 6 (white); columns give 4 + 6 = 10 (high-top) and 5 + 0 = 5 (low-top). All four edge totals match exactly, so the configuration is valid. The fact that the minimum is not 0 also matches a pigeonhole intuition: 9 red pairs cannot all hide among the 5 low-tops, so at least 9 - 5 = 4 red pairs must be high-top. Finally, 415\frac{4}{15} ≈ 0.267 sits comfortably between 15\frac{1}{5} = 0.2 and 13\frac{1}{3} ≈ 0.333 — a reasonable size for "smallest possible."
💡Key takeaway

This AMC 8 problem only needs Grade 4 "fraction of a whole number" and a simple table you already know!

  • Count each category
  • Fill in a two-by-two table
  • Flip to the complement
  • Check the candidate values
  • Write the answer as a fraction

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