Peter, Paul, David join a table tennis tournament. On the first day, two of them were
randomly chosen to play a game against each other. On each subsequent day, the loser of the game on
the previous day would be benched and the other two would play a game. After a certain number of
days, it was found that Peter had won 22 games, Paul had won 20 and David had won 32. How many
games had Peter played?
AIME
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: AIME
Every time Peter loses, he doesn't get to play in the next game. So we can (almost) make an one-to-one map between Peter's losing and not playing days. i,e if $x=$ number of games Peter lost, then $x = (22 + 20 + 32) - 22 - x \pm 1 \Rightarrow x=26$. So Peter played $26 + 22 = 48$ games.