AMC 10 · 2022 · #12

Grade 6 arithmetic

logical-deductionsystems-of-equationscasework caseworkconvert-to-algebra ↑ Prerequisites: logical-deduction

📏 Medium solution 💡 3 insights

Problem

On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.

"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$
children who answered yes.

"Are you an alternater?" The principal gave a piece of candy to each of the $15$
children who answered yes.

"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.

How many pieces of candy in all did the principal give to the children who always
tell the truth?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 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: $31$ children come in three flavors — truth-tellers (always say true things), liars (always say false things), and alternaters (flip between truth and lie, starting with either). The principal asks each child the same three Yes/No questions in order: (1) Are you a truth-teller? (2) Are you an alternater? (3) Are you a liar? After each question, every child who said Yes gets one candy. The three Yes-counts are $22$, $15$, $9$. How much total candy went to the truth-tellers?

Givens: Total children: $31$; Yes-counts: question 1 -> $22$, question 2 -> $15$, question 3 -> $9$; Truth-teller always tells the truth; liar always lies; alternater alternates and may start either way

Unknowns: Total candies received by the truth-tellers

Understand

Plan

Primary tool: #4 Use Matrix Logic

Secondary: #2 Make a Systematic List, #7 Identify Subproblems, #13 Convert to Algebra, #3 Eliminate Possibilities

Tool #4 (Matrix Logic) is the natural fit: build a small grid whose rows are the four kinds of children (truth-teller, liar, alternater-starts-truth, alternater-starts-lie) and whose columns are the three questions. Tool #2 (Systematic List) drives row-by-row Yes/No filling — for each row we just apply the rule mechanically. Tool #7 (Subproblems) breaks the work into (a) build the grid, (b) translate Yes-counts into equations, (c) solve. Tool #13 (Algebra) lets us subtract equations to isolate the truth-teller count without solving the whole system. The final candy count is just (number of truth-tellers) x (Yes answers each gives), and the row for truth-tellers in the grid will read Yes-No-No, so each contributes exactly one candy.

Execute — Answer: A

#4 Use Matrix Logic K.MD.B.3 Step 1

Build the grid (Tool #4).
Rows: T = truth-teller, L = liar, $A_t$ = alternater starting truth, $A_\ell$ = alternater starting lie.
Columns: Q1, Q2, Q3.
For each cell, ask "Is the child's spoken answer Yes or No?".
A truth-teller says the true statement; a liar says the negation.
Filling row by row gives the response table.

$$\begin{array}{c|ccc} & Q1 & Q2 & Q3 \\ \hline T & \text{Y} & \text{N} & \text{N} \\ L & \text{Y} & \text{Y} & \text{N} \\ A_t & \text{N} & \text{N} & \text{N} \\ A_\ell & \text{Y} & \text{Y} & \text{Y} \end{array}$$

💡 Kindergarten-level classification: sort children into categories and count Yes per row — the only "hard" part is reading the rules out loud one row at a time.

#2 Make a Systematic List K.MD.B.3 Step 2

Spot-check the trickiest row, $A_\ell$ (alternater starting with a lie).
Q1 "Are you a truth-teller?" — truth would be No, this child lies first, so says Yes.
Q2 "Are you an alternater?" — truth, this child now tells the truth, says Yes.
Q3 "Are you a liar?" — truth would be No, child lies again, says Yes.
All three Yes.

$$A_\ell: \;\text{Y}, \text{Y}, \text{Y}$$

💡 Just walk through Yes/No one question at a time using each child's rule — Kindergarten classification, no arithmetic yet.

#7 Identify Subproblems 6.EE.B.6 Step 3

Read the Yes columns.
Q1 "Yes" comes from T, L, $A_\ell$ -> $T + L + A_\ell = 22$.
Q2 "Yes" comes from L, $A_\ell$ -> $L + A_\ell = 15$.
Q3 "Yes" comes only from $A_\ell$ -> $A_\ell = 9$.

$$T + L + A_\ell = 22, \quad L + A_\ell = 15, \quad A_\ell = 9$$

💡 Each Yes count is just "add up the rows that said Yes in that column" — Grade 6 expression-writing turns the grid into three equations.

#13 Convert to Algebra 6.EE.B.7 Step 4

Tool #13 (algebra) on the cleanest move: subtract equation 2 from equation 1 to kill $L$ and $A_\ell$ at once.
The number of truth-tellers $T$ falls out directly.

$$(T + L + A_\ell) - (L + A_\ell) = 22 - 15 \;\Rightarrow\; T = 7$$

💡 One subtraction isolates $T$ — Grade 6 equation-solving avoids touching the other unknowns.

#4 Use Matrix Logic 3.OA.A.3 Step 5

Count the candy.
Look at the T-row of the grid: Yes-No-No, so every truth-teller earns exactly $1$ candy (only from Q1).
With $T = 7$ truth-tellers, total candy to truth-tellers $= 7 \times 1 = 7$.

$$\text{candy to truth-tellers} = T \times 1 = 7 \times 1 = 7$$

💡 Multiply candies-per-child by number of truth-tellers — Grade 3 multiplication.

#3 Eliminate Possibilities K.MD.B.3 Step 6

Match $7$ to the choices: it's option (A).

$$\text{answer} = 7 \;\Rightarrow\; \textbf{(A)}$$

💡 Final compare-to-options — the smallest, cleanest number on the list.

[1] #4 K.MD.B.3 Build the grid (Tool #4). Rows: T = truth-teller, L = liar, $A_t$ = alternater s

[2] #2 K.MD.B.3 Spot-check the trickiest row, $A_\ell$ (alternater starting with a lie). Q1 "Are

[3] #7 6.EE.B.6 Read the Yes columns. Q1 "Yes" comes from T, L, $A_\ell$ -> $T + L + A_\ell = 22

[4] #13 6.EE.B.7 Tool #13 (algebra) on the cleanest move: subtract equation 2 from equation 1 to

[5] #4 3.OA.A.3 Count the candy. Look at the T-row of the grid: Yes-No-No, so every truth-teller

[6] #3 K.MD.B.3 Match $7$ to the choices: it's option (A).

Review

Reasonableness: Sanity check the whole population. From the equations $A_\ell = 9$, $L = 15 - 9 = 6$, $T = 7$, so $A_t = 31 - 7 - 6 - 9 = 9$. That's $9$ alternaters of each starting type — perfectly reasonable. Re-totalling the Yes columns: Q1 Yes from $T + L + A_\ell = 7 + 6 + 9 = 22$ ✓. Q2 Yes from $L + A_\ell = 6 + 9 = 15$ ✓. Q3 Yes from $A_\ell = 9$ ✓. All three counts match the given data, so the model is consistent.

Alternative: Tool #16 (Change Focus / Complement): instead of finding $T$ directly, notice that the difference $Q1 - Q2 = 22 - 15 = 7$ exactly counts the rows that said Yes to Q1 but No to Q2. Looking at the grid, only the truth-tellers fit that signature, so $T = 7$ and the truth-teller candy is also $7$. Same answer (A), reached by complementary subtraction in one step.

CCSS standards used (min grade 6)

K.MD.B.3 Classify objects into given categories and count the numbers in each (Sorting the $31$ children into the four behavior categories (T, L, $A_t$, $A_\ell$) and filling a Yes/No grid row by row.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Multiplying the count of truth-tellers ($7$) by the candies each one earns ($1$) to get the total candy ($7$).)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $T, L, A_\ell$ stand for the populations of each behavior and translating each Yes column into a sum expression.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Subtracting $L + A_\ell = 15$ from $T + L + A_\ell = 22$ to isolate $T = 7$ in one move.)

⭐ This AMC 10 problem only needs Grade 6 expression-writing you already know — build a tiny Yes/No grid for the four kinds of children, read off the three Yes-counts as equations, and one subtraction ($22 - 15$) gives the truth-teller count directly. They say Yes only once, so the candy total is exactly $7$.