Difference between revisions of "2002 AMC 12P Problems/Problem 6"
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== Solution 1== | == Solution 1== | ||
Let the amount of soccer players last year be <math>x</math>, the number of male players last year to be <math>m</math>, and the number of females players last year to be <math>f.</math> We want to find <math>\frac{1.2f}{1.1x},</math> since that's the fraction of female players now. From the problem, we are given | Let the amount of soccer players last year be <math>x</math>, the number of male players last year to be <math>m</math>, and the number of females players last year to be <math>f.</math> We want to find <math>\frac{1.2f}{1.1x},</math> since that's the fraction of female players now. From the problem, we are given | ||
| − | + | ||
| − | + | \begin{align*} | |
| + | x&=m+f \\ | ||
| + | 1.1x&=1.05m+1.2f \\ | ||
| + | \end{align*} | ||
| + | |||
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}} | {{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}} | ||
Latest revision as of 09:10, 15 July 2024
Problem
Participation in the local soccer league this year is 
higher than last year. The number of males increased by 
and the number of females increased by 
. What fraction of the soccer league is now female?
Solution 1
Let the amount of soccer players last year be 
, the number of male players last year to be 
, and the number of females players last year to be 
We want to find 
since that's the fraction of female players now. From the problem, we are given
\begin{align*} x&=m+f \\ 1.1x&=1.05m+1.2f \\ \end{align*}
Eliminating and solving for 
gives us our answer of 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
