Difference between revisions of "2016 AMC 8 Problems/Problem 12"
Reaganchoi (talk | contribs) m |
|||
| Line 9: | Line 9: | ||
<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> | ||
| − | {{AMC8 box|year=2016|num-b= | + | {{AMC8 box|year=2016|num-b=11|num-a=13}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:45, 23 November 2016
12. 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?
Solution
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 
| 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. 
