This problem solving activity has a number focus.
I’m addicted to Free Cell, a game on our computer.
At the moment I’ve won 3300 games, which is 60% of the games I've played.
I’d like to be able to say that I had won two-thirds of the games that I had played.
How many games would I have to win in a row to get to the two-thirds winning mark?
To solve this problem the students must read carefully the information they are given, and decide how to use this to find an unknown amount.
An algebraic approach is desirable, but it may not be intuitively used by students. To successfully solve this problem, students need to have had experience with finding fractions and percentages of whole numbers.
The algebraic solution given will provide a foundation for a valuable discussion with students.
I’m addicted to Free Cell, a game on our computer. At the moment I’ve won 3300 games, which is 60% of the games I've played. I’d like to be able to say that I had won two-thirds of the games that I had played. How many games would I have to win in a row to get to the two-thirds winning mark?
The information given is:
I have won 60% of my games. This number is 3300 so I can find the number of games that I have played. This is because 3300/games played = 60/100. So games played = (3300 x 100)/60 = 5500.
I’m going to play some more games and I’m going to win them all. So 2/3 = (3300 + more games) / (5500 + more games). In the table below, I’ll let the fraction on the right be F. This is now set up for a ‘guess and improve’ strategy and the use of a table.
More games played | F | Compared to 2/3 |
100 | 0.6071 | too small – increase games played |
1000 | 0.6615 | too small – increase games played |
2000 | 0.7067 | too large – decrease games played |
1500 | 0.6857 | too large – decrease games played |
1300 | 0.6765 | too large – decrease games played |
1200 | 0.6716 | too large – decrease games played |
1100 | 0.6667 | Bang on! |
So if I play another 1100 games and win them all, then I shall have won two-thirds of all the games I have played.
We already know that 2/3 = (3300 + more games) / (5500 + more games). So let m = more games and we then have 2/3 = (330 + m)/ (5500 + m). Multiplying both sides by 3 and 5500 + m gives 2(5500 + m) = 3(3300 + m),
so 11000 + 2m = 9900 + 3m,
or m = 1100.
Printed from https://nzmaths.co.nz/resource/free-cell at 6:11am on the 27th April 2024