Difference between revisions of "2022 AMC 10A Problems/Problem 12"

Revision as of 00:29, 12 November 2022

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?

$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$

Solution

Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all 9 children that answered yes are alternaters that falsely answer question 1 and 3, and truthfully answer question 2. The rest of the alternaters, however many there are, have the opposite behavior.

Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that 9 alternaters truthfully answer here. Because only liars and 9 alternaters answer yes, we can deduce that there are $15-9=6$ liars.

Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that 9 alternaters lie here. From the previous part, we know that there are 6 liars. Because only the number of truth tellers is unknown here, we can deduce that there are $22-9-6=7$ truth tellers.

The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers only answer yes on the first question, we know that all 7 of them said yes once, resulting in $\boxed{\textbf{(A) } 7}$ pieces of candy.

- phuang1024

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-#redirect [[2022 AMC 12A Problems/Problem 9]]
+==Problem==
+On Halloween <math>31</math> 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 <math>22</math>
+children who answered yes.
+"Are you an alternater?" The principal gave a piece of candy to each of the <math>15</math>
+children who answered yes.
+"Are you a liar?" The principal gave a piece of candy to each of the <math>9</math> children who
+answered yes.
+How many pieces of candy in all did the principal give to the children who always
+tell the truth?
+<math>\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31</math>
+==Solution==
+Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all 9 children that answered yes are alternaters that falsely answer question 1 and 3, and truthfully answer question 2. The rest of the alternaters, however many there are, have the opposite behavior.
+Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that 9 alternaters truthfully answer here. Because only liars and 9 alternaters answer yes, we can deduce that there are <math>15-9=6</math> liars.
+Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that 9 alternaters lie here. From the previous part, we know that there are 6 liars. Because only the number of truth tellers is unknown here, we can deduce that there are <math>22-9-6=7</math> truth tellers.
+The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers only answer yes on the first question, we know that all 7 of them said yes once, resulting in <math>\boxed{\textbf{(A) } 7}</math> pieces of candy.
+- phuang1024
+== See Also ==
+{{AMC10 box|year=2022|ab=A|num-b=11|num-a=13}}
+{{AMC12 box|year=2022|ab=A|num-b=8|num-a=10}}
+{{MAA Notice}}

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Difference between revisions of "2022 AMC 10A Problems/Problem 12"

Revision as of 00:29, 12 November 2022

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