AMC 8 · 2012 · #19

Grade 6 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

systems-of-equationslinear-equations-two-varlogical-deduction convert-to-algebraidentify-subproblems ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?

Pick an answer.

(A)

$hspace{.05in}6$

(B)

$hspace{.05in}8$

(C)

$hspace{.05in}9$

(D)

$hspace{.05in}10$

(E)

$hspace{.05in}12$

AMC 8 2012 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.

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Toolkit + CCSS Solution

Understand

Restated: A jar holds only red, green, and blue marbles. Three clues are given: "all but $6$ are red", "all but $8$ are green", and "all but $4$ are blue." Find the total number of marbles in the jar.

Givens: All marbles are red, green, or blue (no other colors); All but $6$ are red $\;\Rightarrow\;$ the non-red marbles number $6$; All but $8$ are green $\;\Rightarrow\;$ the non-green marbles number $8$; All but $4$ are blue $\;\Rightarrow\;$ the non-blue marbles number $4$; Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $10$, (E) $12$

Unknowns: The total number of marbles $T$ in the jar

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #13 Convert to Algebra

Each clue is phrased as a complement: "all but $6$ are red" tells us $T - r = 6$, not $r$ directly. Tool #16 (Count the Complement) is exactly that move — read "all but $k$" as the size of the non-that-color set. Tool #13 (Convert to Algebra) then lets us name $T$, $r$, $g$, $b$ and add the three complement equations. The three left sides add to $3T - (r+g+b) = 3T - T = 2T$, and the right sides add to $6 + 8 + 4 = 18$, so $T$ falls out in one line.

Execute — Answer: C

#16 Change Focus / Count the Complement 6.EE.A.2 Step 1

Reframe each clue as a complement count.
"All but $6$ are red" means the marbles that are not red total $6$.
The non-red marbles are exactly the green and blue ones, so $g + b = 6$, i.e.
$T - r = 6$.
The other two clues work the same way.

$$T - r = 6, \quad T - g = 8, \quad T - b = 4$$

💡 "All but $k$ are X" is a complement statement: it counts everything that is NOT X.

#13 Convert to Algebra 4.OA.A.3 Step 2

Add the three complement equations side by side.
The left side becomes $3T - (r + g + b)$, and the right side becomes $6 + 8 + 4 = 18$.

$$(T - r) + (T - g) + (T - b) = 6 + 8 + 4$$

💡 Adding three clean equations packages all three clues into one statement about $T$.

#13 Convert to Algebra 6.EE.B.7 Step 3

Use $r + g + b = T$ to simplify the left side: $3T - (r + g + b) = 3T - T = 2T$.
So $2T = 18$, giving $T = 9$.

$$3T - T = 18 \;\Rightarrow\; 2T = 18 \;\Rightarrow\; T = 9 \;\Rightarrow\; \textbf{(C)}$$

💡 Replacing $r + g + b$ with $T$ collapses three unknowns into one, and a one-step linear equation finishes the job.

[1] #16 6.EE.A.2 Reframe each clue as a complement count. "All but $6$ are red" means the marbles

[2] #13 4.OA.A.3 Add the three complement equations side by side. The left side becomes $3T - (r

[3] #13 6.EE.B.7 Use $r + g + b = T$ to simplify the left side: $3T - (r + g + b) = 3T - T = 2T$.

Review

Reasonableness: Recover each color count: $r = T - 6 = 3$, $g = T - 8 = 1$, $b = T - 4 = 5$. Total: $3 + 1 + 5 = 9$. Check the clues: non-red $= 1 + 5 = 6$ ✓, non-green $= 3 + 5 = 8$ ✓, non-blue $= 3 + 1 = 4$ ✓. All three statements hold, and every count is a non-negative integer, so $T = 9$ is consistent.

Alternative: Tool #6 (Guess and Check) on the answer choices. For each candidate $T$, compute $r = T - 6$, $g = T - 8$, $b = T - 4$ and test $r + g + b = T$. $T = 6$: $0 + (-2) + 2 = 0 \neq 6$ (and $g < 0$). $T = 8$: $2 + 0 + 4 = 6 \neq 8$. $T = 9$: $3 + 1 + 5 = 9$ ✓. $T = 10$: $4 + 2 + 6 = 12 \neq 10$. $T = 12$: $6 + 4 + 8 = 18 \neq 12$. Only (C) works.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Adding the three right-hand sides $6 + 8 + 4 = 18$ and combining whole-number clues into a single arithmetic statement.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming $T$, $r$, $g$, $b$ and translating each "all but $k$" clue into the expression $T - (\text{color count}) = k$.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Solving the one-variable linear equation $2T = 18$ to get $T = 9$.)

⭐ When a problem says "all but $k$", it's counting the OPPOSITE — write that down, add the clues, and the total appears in one step.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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