Difference between revisions of "2016 AMC 8 Problems/Problem 12"
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− | Jefferson Middle School has the same number of boys and 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 | ||
− | + | ==See Also== | |
− | + | {{AMC8 box|year=2016|num-b=11|num-a=13}} | |
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− | {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. of the girls and of the boys went on a field trip. What fraction of the students on the field trip were girls?
Solution 1
Let there be boys and girls in the school. We see , which means kids went on the trip and kids are girls. So, the answer is , which is .
~CHECKMATE2021
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.