AMC 10 · 2022 · #12

Grade 6 arithmetic
logical-deductionsystems-of-equationscasework caseworkconvert-to-algebra ↑ Prerequisites: logical-deduction
📏 Medium solution 💡 3 insights
Problem
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?

Pick an answer.

(A)
7
(B)
12
(C)
21
(D)
27
(E)
31

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Use Matrix Logic

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.

1STEP 1

Build the Yes/No grid (Tool #4): rows T, L, A_t, A_ℓ; columns Q1–Q3. Apply each child's rule cell by cell to fill the table.

& Q1 & Q2 & Q3 ; T & Y & N & N ; L & Y & Y & N ; A_t & N & N & N ; A_ℓ & Y & Y & Y
2STEP 2

Spot-check the trickiest row A_ℓ (starts with a lie): applying its lie-truth-lie rule across Q1–Q3 gives Yes, Yes, Yes.

A_ℓ: Y, Y, Y
3STEP 3

Read each Yes column into an equation: Q1 gives T + L + A_ℓ = 22, Q2 gives L + A_ℓ = 15, Q3 gives A_ℓ = 9.

T + L + A_ℓ = 22, L + A_ℓ = 15, A_ℓ = 9
4STEP 4

Subtract equation 2 from equation 1 (Tool #13): L and A_ℓ cancel at once, leaving the truth-teller count T = 7.

(T + L + A_ℓ) - (L + A_ℓ) = 22 - 15 → T = 7
5STEP 5

The T-row reads Yes-No-No, so each truth-teller earns just 1 candy (from Q1); with 7 of them the candy total is 7 × 1 = 7.

candy to truth-tellers = T × 1 = 7 × 1 = 7
6STEP 6

Matching the total to the choices gives option (A).

answer = 7 → (A)
Answer
7
Sanity check the whole population. From the equations A_ℓ = 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_ℓ = 7 + 6 + 9 = 22 ✓. Q2 Yes from L + A_ℓ = 6 + 9 = 15 ✓. Q3 Yes from A_ℓ = 9 ✓. All three counts match the given data, so the model is consistent.
💡Key takeaway

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.