2011 AMC 8 Problems/Problem 14

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?

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

Video Solution

https://youtu.be/rQUwNC0gqdg?t=697

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 $\frac{120+100}{270+180} = \frac{220}{450} = \boxed{\textbf{(C)}\ \frac{22}{45}}$

Solution 2 (Guess and Check)

Since we know there are $270$ kids at Colfax Middle School and we know the ratio of boys to girls is $5:4$ , we can use guess and check to find the number of girls at the dance (and vice versa). $5:4=150:120$ and $4:5=80:100$ . Now that the ratios sum to $270$ and $180$ , we know that in Colfax there are $120$ girls and in Winthrop, there are $100$ girls. $120+100=220$ total girls and $270+180=450$ total people. $\frac{220}{450}=\boxed{\textbf{(C)} \frac{22}{45}}$

Video Solution 1 by SpreadTheMathLove

2011 AMC 8 (Problems • Answer Key • Resources)
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.

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