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Area Man Who Talks a Lot About Teaching Teaches His First Full Day in >10
Years
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I have taught demo and observational classes regularly since I left
full-time teaching but yesterday was the first time I taught every class
for the day. L...
4 years ago
Answer: There must have been a point where the winning percentage was exactly 80%, but there need not have been a point where it was exactly 70%.
ReplyDeleteConsider the first time that the team reached or exceeded 80%. Say that they have played N games, and won k of them. Thus k/N >= 4/5.
Since this was the first time they reached 80%, the team must have won the Nth game. So, they had only k-1 wins after N-1 games, and (k-1)/(N-1) < 4/5.
Now, (k-1)/(N-1) < 4/5 is equivalent to 4N > 5k-1, and k/N >= 4/5 is equivalent to 4N <= 5k.
Since 4N is an integer, and there are no integers between 5k-1 and 5k, we conclude that 4N = 5k, or k = 4/5 N. Thus, the winning percentage was exactly 80% after N games.
For the other part of the question, note that if the team loses the first game and wins the next nine games, then the winning percentage will never be exactly 70%.