Difference between revisions of "2016 AMC 8 Problems/Problem 12"
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− | ==Problem== | + | ==Problem 12== |
− | Jefferson Middle School has the same number of boys and girls. | + | 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> |
Revision as of 21:30, 22 March 2021
Problem 12
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?
Solutions
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. is a number that works. There will be girls and boys. So, there will be = girls on the trip and = boys on the trip. The total number of children on the trip is , so the fraction of girls on the trip is or
Solution 2
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
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AJHSME/AMC 8 Problems and Solutions |
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