Difference between revisions of "2002 AMC 12P Problems/Problem 6"
(→Problem) |
(→Solution 1) |
||
(3 intermediate revisions by the same user not shown) | |||
Line 14: | Line 14: | ||
</math> | </math> | ||
− | == Solution == | + | == 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 | |
+ | |||
+ | \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> | ||
== 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}} |
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
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.