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Difference between revisions of "2010 AMC 12A Problems/Problem 12"

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{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #12]] and [[2010 AMC 10A Problems|2010 AMC 10A #15]]}}
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== Problem ==
 
== Problem ==
 
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
 
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
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== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
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We can begin by first looking at Chris and LeRoy.
 
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Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.
 
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Clearly, Chris and LeRoy are different species, and so we have exactly <math>1</math> frog out of the two of them.
 
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Now suppose Mike is a toad. Then what he says is true because we already have <math>2</math> toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.
 
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Therefore, Mike must be a frog. His statement must be false, which means that there is at most <math>1</math> toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.
 
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Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have <math>3</math> frogs total. <math>\boxed{\textbf{(D)}}</math>
 
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=== Solution 2 ===
 
  
 
Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
 
Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
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As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
 
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
  
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Hence we must have one toad and three frogs.
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Hence we must have one toad and <math>\boxed{\textbf{(D)}\ 3}</math> frogs.
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==Video Solution==
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https://youtu.be/kU70k1-ONgM?t=1207
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~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=11|num-a=13|ab=A}}
 
{{AMC12 box|year=2010|num-b=11|num-a=13|ab=A}}
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{{AMC10 box|year=2010|num-b=14|num-a=16|ab=A}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 16:43, 20 September 2020

The following problem is from both the 2010 AMC 12A #12 and 2010 AMC 10A #15, so both problems redirect to this page.

Problem

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.

Brian: "Mike and I are different species."

Chris: "LeRoy is a frog."

LeRoy: "Chris is a frog."

Mike: "Of the four of us, at least two are toads."

How many of these amphibians are frogs?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.

As Mike is a frog, his statement is false, hence there is at most one toad.

As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.

Hence we must have one toad and $\boxed{\textbf{(D)}\ 3}$ frogs.

Video Solution

https://youtu.be/kU70k1-ONgM?t=1207

~IceMatrix

See also

2010 AMC 12A (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 12 Problems and Solutions
2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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

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

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