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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>
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==Alternate Solution==
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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>
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-Clara Garza
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 +
==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 $11/4$. 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 $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{\text{(B)}\ 27\%}$

Alternate Solution

The Won/Lost ratio is 11/4 so, for some number $N$, the team won $11N$ games and lost $4N$ games. Thus, the team played $15N$ games and the fraction of games lost is $\dfrac{4N}{15N}=\dfrac{4}{15}\approx 0.27=\boxed{27\%}.$ -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

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