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

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

~CHECKMATE2021

Solution 2

Using WLOG (Without loss of generativity), Let there be $12$ boys and $12$ girls in the school. Now we can do $\frac{3}{4}12$ + $\frac{2}{3}12$ to get the total number of students going to the field trip to be $17$ . Since we already know the number of girls to be $9$ . We have our answer to be $\frac{9}{17}$ . So, the answer is $\boxed{\textbf{(B)} \frac{9}{17}}$ .

~algebraic_algorithmic

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/Y4N4L_HcnKY

~Education, the Study of Everything

Video Solution

https://youtu.be/MnqS_-dUMV8

~savannahsolver

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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