Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 8 Problems/Problem 12"

(→‎Solution 1)
(Categorized problem)
 
(7 intermediate revisions by 4 users not shown)
Line 8: Line 8:
 
==Solution 1==
 
==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.
+
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>.
  
−
<math>120</math> is a number that works. There will be <math>60</math> girls and <math>60</math> boys. So, there will be
+
~CHECKMATE2021
−
<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{\textbf{(B)} \frac{9}{17}}</math>.
 
  
 
==Solution 2==
 
==Solution 2==
  
−
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>.
+
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!!!)==
 
==Video Solution (CREATIVE THINKING!!!)==
Line 31: Line 31:
 
{{AMC8 box|year=2016|num-b=11|num-a=13}}
 
{{AMC8 box|year=2016|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 18:06, 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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_12&oldid=252325"