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

(Remark (Fake Solve))
(Solution 3)
Line 46: Line 46:
  
 
==Solution 3==
 
==Solution 3==
Suppose that there are <math>T</math> truth-tellers, <math>L</math> liars, <math>A</math> alternaters who answered yes-no-yes, and <math>A'</math> alternaters who answered no-yes-no.
+
Note that:
  
We have the following system of equations:
+
* Truth-tellers would answer yes-no-no to the three questions, in this order.
<cmath>\begin{array}{ccccccccrcccc}
+
 
T &+ &L &+ &A &+ &A' &= &31, & & & & (1) \ [0.5ex]
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* Liars would answer yes-yes-no to the three questions, in this order.
T &+ &L &+ &A & & &= &22, & & & & (2) \ [0.5ex]
+
 
& &L & & &+ &A' &= &15, & & & & (3) \ [0.5ex]
+
* Alternaters who responded truth-lie-truth would answer no-no-no to the three questions, in this order.
& & & & A & & &= &9. & & & & (4)
+
 
\end{array}</cmath>
+
* Alternaters who responded lie-truth-lie would answer yes-yes-yes to the three questions, in this order.
Subtracting <math>(2)</math> from <math>(1)</math> gives <math>A'=9.</math> From <math>(3),</math> it follows that <math>L=15-A'=6.</math> Finally, from <math>(2),</math> we have <math>T=22-A-L=\boxed{\textbf{(A) } 7}.</math>
+
 
 +
Suppose that there are <math>T</math> truth-tellers, <math>L</math> liars, and <math>A</math> alternaters who responded lie-truth-lie.
 +
 
 +
The information of the first two questions implies that
 +
<cmath>\begin{align*}
 +
T+L+A&=22, \
 +
L+A&=15.
 +
\end{align*}</cmath>
 +
Subtracting the second equation from the first, we have <math>T=22-15=\boxed{\textbf{(A) } 7}.</math>
  
 
~[[OrenSH|orenbad]] ~MRENTHUSIASM
 
~[[OrenSH|orenbad]] ~MRENTHUSIASM

Revision as of 08:59, 15 August 2023

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 1

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 Questions 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 answer yes on only the first question, we know that all $7$ of them said yes once, resulting in $\boxed{\textbf{(A) } 7}$ pieces of candy.

~phuang1024

Solution 2

On the first question, truth tellers would say yes. Liars would say yes because they are not truth tellers and thus will say the opposite of no. Some alternaters may lie on this question too, meaning they say yes.

On the second question, Liars would say yes because they are not alternaters and thus will say the opposite of no. Alternaters only say yes to this question if and only if they said yes to the previous question. Thus, the difference between the amount of people that said yes first and that said yes second is the amount of truth tellers, which is $22-15=7$. Because it is obvious that truth tellers only say yes on the first question, our answer is $\boxed{\textbf{(A) } 7}$.

~sigma

Note: This does not even require the information given as the 3rd question asked by the Principal.

(Added by ~SouradipClash_03)

Solution 3

Note that:

  • Truth-tellers would answer yes-no-no to the three questions, in this order.
  • Liars would answer yes-yes-no to the three questions, in this order.
  • Alternaters who responded truth-lie-truth would answer no-no-no to the three questions, in this order.
  • Alternaters who responded lie-truth-lie would answer yes-yes-yes to the three questions, in this order.

Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.

The information of the first two questions implies that \begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*} Subtracting the second equation from the first, we have $T=22-15=\boxed{\textbf{(A) } 7}.$

~orenbad ~MRENTHUSIASM

Remark (Fake Solve)

This problem is broken in an interesting way, that helps the test-taker. Since the true answers alternate for alternaters, you can still get the correct answer if you misinterpret the problem as "alternaters alternate their answer (not their truth value)". And since $A=A'=9$, you can even get the correct answer with the misinterpretation "alternaters are arbitraters of two types: $A$ who answer arbitrarily but all give the same answer as each other, and $A'$ who all answer the opposite of $A$.

It's also notable that the misinterpretation makes the problem harder, so that the solution actually relies on all the information. This suggests that the question-writer may have been mistaken but got lucky.

~oinava, based on demonstration of misinterpretation fakesolve by ~orenbad

Video Solution (One Key Observation)

https://youtu.be/9IQRgWn4NAk

~Education, the Study of Everything

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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