Difference between revisions of "1999 AMC 8 Problems/Problem 12"
(Created page with "== Problem == The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is <math>11/4</math>. To the nearest whole percent, what per...") |
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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{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> |
==See also== | ==See also== | ||
{{AMC8 box|year=1999|num-b=11|num-a=13}} | {{AMC8 box|year=1999|num-b=11|num-a=13}} | ||
Revision as of 12:43, 23 December 2012
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 
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 | ||
