|
|
| (8 intermediate revisions by 6 users not shown) |
| Line 1: |
Line 1: |
| − | == Problem ==
| + | #REDIRECT [[2010_AMC_12A_Problems/Problem_12]] |
| − | 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?
| |
| − | | |
| − | <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
| |
| − | | |
| − | == Solution ==
| |
| − | | |
| − | === Solution 1 ===
| |
| − | | |
| − | We can begin by first looking at Chris and LeRoy.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | Clearly, Chris and LeRoy are different species, and so we have at least <math>1</math> frog out of the two of them.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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>
| |
| − | | |
| − | === 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.
| |
| − | | |
| − | 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 three frogs. <math> \boxed{\textbf{(D)}} </math>
| |
| − | | |
| − | == See also ==
| |
| − | {{AMC10 box|year=2010|num-b=14|num-a=16|ab=A}}
| |
| − | | |
| − | [[Category:Introductory Algebra Problems]] | |
