Difference between revisions of "2011 AMC 8 Problems/Problem 14"

Line 10: Line 10:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=13|num-a=15}}
 
{{AMC8 box|year=2011|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Revision as of 00:59, 5 July 2013

Problem

There are $270$ students at Colfax Middle School, where the ratio of boys to girls is $5 : 4$. There are $180$ students at Winthrop Middle School, where the ratio of boys to girls is $4 : 5$. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?

$\textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45}$

Solution

At Colfax Middle School, there are $\frac49 \times 270 = 120$ girls. At Winthrop Middle School, there are $\frac59 \times 180 = 100$ girls. The ratio of girls to the total number of students is $\frac{120+100}{270+180} = \frac{220}{450} = \boxed{\textbf{(C)}\ \frac{22}{45}}$

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png