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

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<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>
  
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==Solution==
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==Solution 1==
  
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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.
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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.
 
<math>120</math> is a number that works. There will be <math>60</math> girls and <math>60</math> boys. So, there will be  
 
<math>120</math> is a number that works. There will be <math>60</math> girls and <math>60</math> boys. So, there will be  
 
<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.  
 
<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>
 
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>
 
   
 
   
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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>
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==Solution 2==
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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}}
 
{{AMC8 box|year=2016|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:24, 27 November 2016

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

Solution 2

Let their be $b$ boys and $g$ girls in the school. We see $g=b$, which mean $\frac{3}{4}b+\frac{2}{3}b=\frac{17}{12}b$ 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 $\boxed{(B) \frac{9}{17}}$

2016 AMC 8 (Problems • Answer Key • Resources)
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Problem 11 		Followed by
Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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