Earlier today I set you this problem from last year’s British Mathematical Olympiad (BMO), the UK’s top maths competition for pre-university students, which is taken by almost 2,000 teenagers a year.
The question was attempted by 90 per cent of the contestants, and about 1 in 3 got full marks. How did you get on?
Ping pong percentages
Arun and Disha played several games of table tennis. At five points during the day, Arun calculated the percentage of the games played so far that he had won. The results of these calculations were exactly 30 per cent, exactly 40 per cent, exactly 50 per cent, exactly 60 per cent and exactly 70 per cent, but not necessarily in that order.
What is the smallest possible number of games they played?
Here’s a hint. Arun wins one of the first two games.
Here’s how you might have gone about it. First, let G be the number of games played so far.
If at any stage Arun has won 30 per cent of the games he played, then we can write the equation:
W/G = 3/10
where W are the games he has won so far. We can rearrange this equation as 10W = 3G, from which we can deduce that G must be a multiple of 10.
However we also know that there is a point at which he has won 70 per cent of games. In other words there is a w such that 10w = 7g.
We can eliminate the chance that the total number of games is 10, since if it was then this would mean that after ten games Arun would be finishing having won both 70 and 30 per cent of games, which is impossible.
Let’s say the total number of games is 20. Then either (i) after 10 games he has won 3 of them, and after 20 games he has won 14 of them, or (ii) after 10 games he has won 7 of them and after 20 games he has won 6 of them. [This is because we are only going to achieve the 30 per cent, and 70 per cent, of wins when the games played is divisible by 10]
Both (i) and (ii) are impossible. (i) because there are not enough games to go from 3 to 14 wins in 10 games, and (ii) because you can’t go from 7 wins to 6.
Now let’s try with G = 30. I told you that Arun wins now of the first two games. So, after two games his running total of wins/games is 1/2, which is 50 per cent.
Let’s now rewrite the other percentages required as fractions: 30, 40, 60 and 70 per cent are 3/10, 2/5, 3/5 and 7/10.
Let’s say that after winning 1/2, Arun wins one of the next three. That get’s him to 2/5, when he will have won 40 per cent. If he wins one of the next five, that get’s him to 3/10, which is 30 per cent. If he wins nine of the next ten, that get’s him to 12/20, which is 60 per cent. If he wins another nine of the next ten, he finishes with 21/30, which is 70 per cent.
We’re done! G = 30.
I hope you enjoyed today’s puzzle. I’ll be back in two weeks.
Thanks to the UK Mathematics Trust for today’s puzzle. The trust organizations the BMO, as well as many other school maths competitions. For more information, please go to ukmt.org.uk.
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
I’m the author of several books of puzzles, most recently the Language Lover’s Puzzle Book. I also give school talks about maths and puzzles (online and in person). If your school is interested please get in touch.