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------
 
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%. Therefore,{\text{(B)}\27\%}$
 
  
 
==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 18:21, 11 September 2020

Problem

The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$. To the nearest whole percent, what percent of its games did the team lose?

$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$

Solution

The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$. The percentage of games lost is just $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{\text{(B)}\ 27\%}$

See also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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