1999 AMC 8 Problems/Problem 12

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 $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{\text{(B)}\ 27\%}$

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\%}$

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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1999 AMC 8 Problems/Problem 12

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