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

(Created page with "== Problem == How many positive integers <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer? <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B...")
 
(→‎Solution 1)
 
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== Problem ==
 
== Problem ==
−
How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
+
Participation in the local soccer league this year is <math>10\%</math> higher than last year. The number of males increased by <math>5\%</math> and the number of females increased by <math>20\%</math>. What fraction of the soccer league is now female?
  
−
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math>
+
<math>
 +
\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}
 +
</math>
  
−
== Solution ==
+
== Solution 1==
−
If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+
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 ==
−
{{AMC12 box|year=2000|num-b=6|num-a=8}}
+
{{AMC10 box|year=2002|ab=P|num-b=12|num-a=14}}
 +
{{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 $10\%$ 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

\begin{align*} x&=m+f \\ 1.1x&=1.05m+1.2f \\ \end{align*}

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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