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2016 AMC 8 Problems/Problem 12: Difference between revisions

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


<math>\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}</math>
<math>\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}</math>


==Solution==
==Solution 1==
 
Let there be <math>b</math> boys and <math>g</math> girls in the school. We see <math>g=b</math>, which means <math>\frac{3}{4}b+\frac{2}{3}b=\frac{17}{12}b</math> kids went on the trip and <math>\frac{3}{4}b</math> kids are girls. So, the answer is <math>\frac{\frac{3}{4}b}{\frac{17}{12}b}=\frac{9}{17}</math>, which is <math>\boxed{\textbf{(B)} \frac{9}{17}}</math>.
 
~CHECKMATE2021
 
==Solution 2==
 
Using WLOG (Without loss of generativity), Let there be <math>12</math> boys and <math>12</math> girls in the school. Now we can do <math>\frac{3}{4}\times{12}</math> + <math>\frac{2}{3}\times{12}</math> to get the total number of students going to the field trip to be <math>17</math>. Since we already know the number of girls to be <math>9</math>. We have our answer to be <math>\frac{9}{17}</math>. So, the answer is <math>\boxed{\textbf{(B)} \frac{9}{17}}</math>.
 
~algebraic_algorithmic
 
==Video Solution (CREATIVE THINKING!!!)==
https://youtu.be/Y4N4L_HcnKY
 
~Education, the Study of Everything
 
==Video Solution==
https://youtu.be/MnqS_-dUMV8
 
~savannahsolver


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.
==See Also==
<math>120</math> is a number that works. There will be <math>60</math> girls and <math>60</math> boys. So, there will be
{{AMC8 box|year=2016|num-b=11|num-a=13}}
<math>60\cdot\frac{3}{4}</math> = <math>45</math> girls on the trip and <math>60\cdot\frac{2}{3}</math> = <math>40</math> boys on the trip.
The total number of children on the trip is <math>85</math>, so the fraction of girls on the trip is <math>\frac{45}{85}</math> or <math>\boxed{(B) \frac{9}{17}}</math>
{{AMC8 box|year=2016|num-b=1|num-a=3}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Algebra Problems]]

Latest revision as of 17:06, 25 June 2025

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 $\frac{3}{4}\times{12}$ + $\frac{2}{3}\times{12}$ 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

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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