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 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 $\boxed{\textbf{(B)} \frac{9}{17}}$

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 $\boxed{\textbf{(B)} \frac{9}{17}}$

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.

Page

Toolbox

Search

2016 AMC 8 Problems/Problem 12

Solution 1

Solution 2