Difference between revisions of "1999 AMC 8 Problems/Problem 12"
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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> | ||
| + | ==Alternate Solution== | ||
| + | The Won/Lost ratio is 11/4 so, for some number <math>N</math>, the team won <math>11N</math> games and lost <math>4N</math> games. Thus, the team played <math>15N</math> games and the fraction of games lost is <math>\dfrac{4N}{15N}=\dfrac{4}{15}\approx 0.27=\boxed{27\%}.</math> | ||
| + | -Clara Garza | ||
| − | + | ==Video Solution== | |
| − | + | https://youtu.be/hF4QIyb7R_4 Soo, DRMS, NM | |
| − | |||
| − | |||
==See also== | ==See also== | ||
{{AMC8 box|year=1999|num-b=11|num-a=13}} | {{AMC8 box|year=1999|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:56, 9 June 2022
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 
Alternate Solution
The Won/Lost ratio is 11/4 so, for some number 
, the team won 
games and lost 
games. Thus, the team played 
games and the fraction of games lost is 
-Clara Garza
Video Solution
https://youtu.be/hF4QIyb7R_4 Soo, DRMS, NM
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
