2011 AMC 12A Problem 9
Attempt Problem 9 of the 2011 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 2011 AMC 12A solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
9.
At a twins and triplets convention, there were sets of twins and sets of triplets, all from different families. Each twin shook hands with all the twins except his/her sibling and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?
Answer: B
Small hint:
There are twins and triplets; count twin-twin, triplet-triplet, and twin-triplet handshakes separately
Big hint:
Each twin greets twins and each triplet greets triplets; divide those two totals by
Solution:
There are twins and triplets.
Twin-twin handshakes: each twin shakes other twins, giving
Triplet-triplet handshakes: each triplet shakes other triplets, giving
Twin-triplet handshakes: each twin shakes half the triplets, giving (each such handshake counted once).
The total is
Thus, the correct answer is B.
Problem 9 in other years
- 1950 AMC 12
- 1951 AMC 12
- 1952 AMC 12
- 1953 AMC 12
- 1954 AMC 12
- 1955 AMC 12
- 1956 AMC 12
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