Difference between revisions of "2016 AMC 8 Problems/Problem 12"
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<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{\textbf{(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{\textbf{(B)} \frac{9}{17}}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 05:35, 17 January 2023
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
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
Video Solution
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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 |
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