AMC 8 · 2024 · #19

Grade 4 rate-ratio

ratio-proportionfraction-arithmeticcomplementary-counting complementary-countingcasework ↑ Prerequisites: fraction-arithmeticratio-proportion

📏 Medium solution 💡 3 insights 📊 Diagram

📘 View easy version →

Problem

Jordan owns 15 pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction?

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

Pick an answer.

(A)

(B)

$dfrac{1}{5}$

(C)

$dfrac{4}{15}$

(D)

$dfrac{1}{3}$

(E)

$dfrac{2}{5}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: Jordan owns 15 pairs of sneakers. $\tfrac{3}{5}$ of them are red (the rest white) and $\tfrac{2}{3}$ 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?

Givens: Total pairs of sneakers: $15$; Red pairs: $\tfrac{3}{5}$ of the total, the rest are white; High-top pairs: $\tfrac{2}{3}$ of the total, the rest are low-top; Answer choices: $(A)\ 0,\ (B)\ \tfrac{1}{5},\ (C)\ \tfrac{4}{15},\ (D)\ \tfrac{1}{3},\ (E)\ \tfrac{2}{5}$

Unknowns: The **smallest** possible count of pairs that are both red AND high-top, divided by 15

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #16 Change Focus / Count the Complement, #6 Guess and Check

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 $\ge 0$. We do **not** need Tool #13 (algebra) — table + complement + check is enough.

Execute — Answer: C

#1 Draw a Diagram 4.NF.B.4 Step 1

First find how many pairs are red, white, high-top, and low-top.
"$\tfrac{3}{5}$ of the 15 pairs are red" means split the 15 pairs into 5 equal groups and take 3 of them: $\tfrac{3}{5} \times 15 = 9$ red pairs.
Likewise "$\tfrac{2}{3}$ are high-top" means split into 3 equal groups and take 2: $\tfrac{2}{3} \times 15 = 10$ high-top pairs.
The remainders come from subtraction: white $= 15 - 9 = 6$ and low-top $= 15 - 10 = 5$.

$$\tfrac{3}{5} \times 15 = 9,\ \ 15 - 9 = 6,\ \ \tfrac{2}{3} \times 15 = 10,\ \ 15 - 10 = 5$$

💡 Taking $\tfrac{3}{5}$ or $\tfrac{2}{3}$ of a whole number to find "how many" is exactly Grade 4 multiplication of a fraction by a whole number.

#1 Draw a Diagram 1.MD.C.4 Step 2

Put the four totals (red 9, white 6, high-top 10, low-top 5) on the edges of a 2 by 2 table.
Let $x$ = pairs that are red AND high-top.
Then the other three cells are forced by the row and column totals: red low-top $= 9 - x$, white high-top $= 10 - x$, and white low-top $= x - 4$ (so that the white row sums to 6 and the low-top column sums to 5).
The single picture makes the constraint "every cell must be a non-negative integer" easy to see.

$$\begin{array}{c|cc|c} & \text{High-top} & \text{Low-top} & \text{Total} \\\hline \text{Red} & x & 9-x & 9 \\ \text{White} & 10-x & x-4 & 6 \\\hline \text{Total} & 10 & 5 & 15 \end{array}$$

💡 Sorting items by two categories (color and style) into the cells of a table is the Grade 1 "organize, represent, and interpret data with up to three categories" idea.

#16 Change Focus / Count the Complement 2.NBT.B.5 Step 3

Direct attack on "make $x$ as small as possible" is awkward, so **flip the focus** to its complement in the red row: "make red low-top $9 - x$ as **large** as possible." Red low-top has two ceilings: it cannot exceed the 5 low-tops that exist, and it cannot exceed the 9 reds that exist.
The tighter ceiling is the smaller one, $\min(5, 9) = 5$.
So red low-top $\le 5$, which means $9 - x \le 5$, i.e.
$x \ge 9 - 5 = 4$.

$$\text{red low-top} = 9 - x \le \min(5,\ 9) = 5\ \Longrightarrow\ x \ge 9 - 5 = 4$$

💡 Swapping a hard "minimize" question for an easier "maximize the leftover" question, then doing $9 - 5 = 4$, is just subtraction within 100 — a Grade 2 skill.

#6 Guess and Check 2.NBT.B.5 Step 4

Now **guess and check** the candidate values $x = 0, 1, 2, 3, 4$ in the table.
For $x = 0$: red low-top would be $9$, but only $5$ low-tops exist — fail.
For $x = 1, 2, 3$: red low-top would be $8, 7, 6$ respectively — each still exceeds $5$, so all fail.
For $x = 4$: red low-top $= 5$, white high-top $= 6$, white low-top $= 0$, and every cell is a non-negative integer.
So $x = 4$ is the smallest feasible value.

$$x = 4:\ \begin{array}{c|cc|c} & \text{High-top} & \text{Low-top} & \text{Total}\\\hline \text{Red} & 4 & 5 & 9\\ \text{White} & 6 & 0 & 6\\\hline \text{Total} & 10 & 5 & 15 \end{array}$$

💡 Plugging in candidate values and checking whether subtractions stay non-negative is Grade 2 subtraction-within-100 work.

#1 Draw a Diagram 3.NF.A.1 Step 5

Divide the minimum count of red high-tops, $4$, by the total $15$ to get the fraction: $\tfrac{4}{15}$.
The denominator $15 = 3 \times 5$ shares no common factor with $4$, so $\tfrac{4}{15}$ is already in lowest terms.
Matching to the choices, $\tfrac{4}{15}$ is exactly (C).
(A) $0$ would mean $x = 0$ (impossible), (B) $\tfrac{1}{5} = \tfrac{3}{15}$ would mean $x = 3$ (impossible), and (D) $\tfrac{1}{3}$, (E) $\tfrac{2}{5}$ are both bigger than the minimum, so they cannot be the least value.

$$\dfrac{4}{15}\ \Rightarrow\ \textbf{(C)}$$

💡 Writing "$4$ out of $15$ equal pairs" as the fraction $\tfrac{4}{15}$ is the Grade 3 idea of a fraction as a part of a whole split into equal parts.

[1] #1 4.NF.B.4 First find how many pairs are red, white, high-top, and low-top. "$\tfrac{3}{5}$

[2] #1 1.MD.C.4 Put the four totals (red 9, white 6, high-top 10, low-top 5) on the edges of a 2

[3] #16 2.NBT.B.5 Direct attack on "make $x$ as small as possible" is awkward, so **flip the focus

[4] #6 2.NBT.B.5 Now **guess and check** the candidate values $x = 0, 1, 2, 3, 4$ in the table. F

[5] #1 3.NF.A.1 Divide the minimum count of red high-tops, $4$, by the total $15$ to get the fra

Review

Reasonableness: 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, $\tfrac{4}{15} \approx 0.267$ sits comfortably between $\tfrac{1}{5} = 0.2$ and $\tfrac{1}{3} \approx 0.333$ — a reasonable size for "smallest possible."

Alternative: Tool #12 (Venn diagram) works too: draw circles "Red" and "High-top" inside a universe of $15$. With intersection $x$, the "red only" region is $9 - x$, the "high-top only" region is $10 - x$, and the "outside both" region (white low-top) is $15 - (9 + 10 - x) = x - 4 \ge 0$, which immediately gives $x \ge 4$. Same answer $\tfrac{4}{15}$, but with the inclusion-exclusion picture instead of the 2 by 2 table.

CCSS standards used (min grade 4)

1.MD.C.4 Organize, represent, and interpret data with up to three categories (Sorting sneakers by the two categories (color and style) into the four cells of a 2 by 2 table.)
2.NBT.B.5 Fluently add and subtract within 100 (Filling the table with $15 - 9 = 6,\ 15 - 10 = 5,\ 9 - 5 = 4$ and checking each candidate $x$ for non-negative cells.)
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole (Writing "$4$ out of $15$ equal pairs" as the fraction $\tfrac{4}{15}$ and matching it to a multiple-choice option.)
4.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a whole number (Computing $\tfrac{3}{5} \times 15 = 9$ red pairs and $\tfrac{2}{3} \times 15 = 10$ high-top pairs.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

More like this