2016 AMC 8 Problems/Problem 12

Jefferson Middle School has the same number of boys and girls. Three-fourths of the girls and two-thirds of the boys went on a field trip. What fraction of the students 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

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. $120$ 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 $\boxed{(B) \frac{9}{17}}$

Solution 2: Let their be $b$ boys and $g$ girls in the school. We see $g=b$ , which mean $\frac{3}{4}b+\frac{2}{3}b=\frac{17}{12}b$ 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 $\boxed{(B) \frac{9}{17}}$ {AMC8 box|year=2016|num-b=11|num-a=13}} The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2016 AMC 8 Problems/Problem 12

Solution