2010 AMC 12A Problem 12
Attempt Problem 12 of the 2010 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12A solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
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 four amphibians are frogs?
Answer: D
Small hint:
Chris and LeRoy cannot have the same species
Big hint:
If Brian were a toad, Mike’s statement would create a contradiction
Solution:
Chris and LeRoy are of opposite species: if Chris is a toad, his statement makes LeRoy a frog; if Chris is a frog, his false statement makes LeRoy a toad. Thus exactly one of them is a toad.
If Brian is a toad, his true statement makes Mike a frog. If Brian is a frog, his false statement says that he and Mike are the same species, again making Mike a frog. Therefore Mike is always a frog.
Mike’s statement is therefore false, so there are fewer than two toads. Chris and LeRoy already supply exactly one toad, forcing Brian to be a frog. Hence there is one toad and frogs.
Thus, D is the correct answer.
Problem 12 in other years
- 1950 AMC 12
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