Difference between revisions of "2016 AMC 8 Problems/Problem 12"
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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> | 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> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/MnqS_-dUMV8 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{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 10:30, 21 March 2022
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
~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.