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

(Solution)
(Solution 2)
 
(22 intermediate revisions by 16 users not shown)
Line 1: Line 1:
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
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/Y4N4L_HcnKY
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/MnqS_-dUMV8
 +
 
 +
~savannahsolver
  
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.
+
==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>
 
 
Solution 2: Let their be <math>b</math> boys and <math>g</math> girls in the school. We see <math>g=b</math>, which mean <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{(B) \frac{9}{17}}</math>
 
{AMC8 box|year=2016|num-b=11|num-a=13}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:32, 21 January 2024

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

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 (ProblemsAnswer KeyResources)
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. AMC logo.png