Difference between revisions of "1999 AMC 8 Problems/Problem 12"
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==Solution== | ==Solution== | ||
The ratio means that for every <math>11</math> games won, <math>4</math> are lost, so the team has won <math>11x</math> games, lost <math>4x</math> games, and played <math>15x</math> games for some positive integer <math>x</math>. The percentage of games lost is just <math>\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{\text{(B)}\ 27\%}</math> | The ratio means that for every <math>11</math> games won, <math>4</math> are lost, so the team has won <math>11x</math> games, lost <math>4x</math> games, and played <math>15x</math> games for some positive integer <math>x</math>. The percentage of games lost is just <math>\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{\text{(B)}\ 27\%}</math> | ||
− | Solution 2 | + | ----Solution 2------ |
We see that the number of games won to the number of games lost is 4/15. Dividing gets the result 26 2/3%, which rounds up to 27%. | We see that the number of games won to the number of games lost is 4/15. Dividing gets the result 26 2/3%, which rounds up to 27%. | ||
Revision as of 11:33, 2 April 2018
Problem
The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is . To the nearest whole percent, what percent of its games did the team lose?
Solution
The ratio means that for every games won, are lost, so the team has won games, lost games, and played games for some positive integer . The percentage of games lost is just
Solution 2------
We see that the number of games won to the number of games lost is 4/15. Dividing gets the result 26 2/3%, which rounds up to 27%.
See also
1999 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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