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2022 AMC 10A Problem 12

2022 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Logic Puzzles · difficulty 3 of 5 · about 5 min · position P11-15

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?

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

Answer

Key insight

Tabulate each type's yes/no pattern: truth-tellers Y N N, liars Y Y N, alternaters N N N or Y Y Y; then peel the counts from question 3 backward.

Solution

Work out how each kind of child answers the three questions (truth-teller? alternater? liar?).

Truth-teller: yes, no, no.
Liar: the true answers are no, no, yes; a liar negates them: yes, yes, no.
Alternater starting truthful: truthfully "no" to question 1, then lies about being an alternater: "no", then truthfully "no": no, no, no.
Alternater starting with a lie: yes, yes, yes.

Let $T$ , $L$ , $A_1$ , $A_2$ be the numbers of truth-tellers, liars, truth-first alternaters, and lie-first alternaters. Reading the counts from the last question backward:

Question 3 ( $9$ yeses): only lie-first alternaters said yes, so $A_2 = 9$ .
Question 2 ( $15$ yeses): liars and lie-first alternaters, so $L + 9 = 15$ , $L = 6$ .
Question 1 ( $22$ yeses): truth-tellers, liars, lie-first alternaters, so $T + 6 + 9 = 22$ , $T = 7$ .

(Then $A_1 = 31 - 22 = 9$ , consistent.) Each truth-teller received exactly one piece of candy, on question 1, so the truth-tellers received $7$ pieces in all.

The answer is $\boxed{\textbf{(A)}\ 7}$ .

Why this works

Truth-teller/liar problems become bookkeeping once you list each type's response pattern; the "alternater" twist simply creates two sub-types. The three yes-counts form a triangular system, so start with the question answered yes by the fewest types and substitute upward.

The trap

Assuming all alternaters answer alike; those who start truthful say no three times, those who start lying say yes three times.

Common mistakes

Assuming all alternaters answer alike; those who start truthful say no three times, those who start lying say yes three times.
Answering $22$ (everyone who said yes to question 1) or forgetting that truth-tellers get candy only for the first question.

Techniques

Split into exhaustive cases and handle each · Start from the end state / desired conclusion and reverse