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?
Discussion on Bangladesh National Math Camp
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- Swapnil Barua
- Joined:Tue Apr 13, 2021 2:55 pm
- asif e elahi
- Joined:Mon Aug 05, 2013 12:36 pm
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.
3 posts •Page 1 of 1