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Difference between revisions of "2002 AMC 12P Problems/Problem 6"

(→‎Solution)
(→‎Solution 2)
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<cmath>1.1x=1.05m+1.2f.</cmath>
 
<cmath>1.1x=1.05m+1.2f.</cmath>
 
Eliminating <math>m</math> and solving for <math>\frac{1.2f}{1.1x}</math> gives us our answer of <math>\boxed{\textbf{(B) } \frac {4}{11}}.</math>
 
Eliminating <math>m</math> and solving for <math>\frac{1.2f}{1.1x}</math> gives us our answer of <math>\boxed{\textbf{(B) } \frac {4}{11}}.</math>
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== Solution 2==
 
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WLOG, let the amount of soccer players last year be <math>100</math>, the number of male players last year to be <math>50</math>, and the number of females players last year to be <math>50.</math> That means that this year, the number of players is $110, the number of male players is
 
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=5|num-a=7}}
 
{{AMC12 box|year=2002|ab=P|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:20, 30 December 2023

Problem

Participation in the local soccer league this year is $10\%$ higher than last year. The number of males increased by $5\%$ and the number of females increased by $20\%$. What fraction of the soccer league is now female?

$\text{(A) }\frac{1}{3} \qquad \text{(B) }\frac{4}{11} \qquad \text{(C) }\frac{2}{5} \qquad \text{(D) }\frac{4}{9} \qquad \text{(E) }\frac{1}{2}$

Solution 1

Let the amount of soccer players last year be $x$, the number of male players last year to be $m$, and the number of females players last year to be $f.$ We want to find $\frac{1.2f}{1.1x},$ since that's the fraction of female players now. From the problem, we are given \[x=m+f\] \[1.1x=1.05m+1.2f.\] Eliminating $m$ and solving for $\frac{1.2f}{1.1x}$ gives us our answer of $\boxed{\textbf{(B) } \frac {4}{11}}.$

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AMC 12 Problems and Solutions

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