2016 AMC 8 Problems/Problem 12

Problem

Jefferson Middle School has the same number of boys and girls. $\frac{3}{4}$ of the girls and $\frac{2}{3}$ of the boys went on a field trip. What fraction of the students on the field trip were girls?

$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}$

Solution 1

Set the number of children to a number that is divisible by two, four, and three. In this question, the number of children in the school is not a specific number because there are no actual numbers in the question, only ratios.This way, we can calculate the answer without dealing with decimals. The pdf version says,"What fraction of the students were girls?", which is not the question.

$120$ is a number that works. There will be $60$ girls and $60$ boys. So, there will be $60\cdot\frac{3}{4}$ = $45$ girls on the trip and $60\cdot\frac{2}{3}$ = $40$ boys on the trip. The total number of children on the trip is $85$ , so the fraction of girls on the trip is $\frac{45}{85}$ or $\boxed{\textbf{(B)} \frac{9}{17}}$ .

Solution 2

Let there be $b$ boys and $g$ girls in the school. We see $g=b$ , which means $\frac{3}{4}b+\frac{2}{3}b=\frac{17}{12}b$ kids went on the trip and $\frac{3}{4}b$ kids are girls. So, the answer is $\frac{\frac{3}{4}b}{\frac{17}{12}b}=\frac{9}{17}$ , which is $\boxed{\textbf{(B)} \frac{9}{17}}$ .

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2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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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