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

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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>
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2002|ab=P|num-b=12|num-a=14}}
 
{{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 17:14, 14 July 2024

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 10P (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 AMC 10 Problems and Solutions
2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 AMC 12 Problems and Solutions

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