Difference between revisions of "2016 AMC 8 Problems/Problem 12"
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~CHECKMATE2021 | ~CHECKMATE2021 | ||
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+ | Using WLOG (Without loss of generativity), Let there be <math>12</math> boys and <math>12</math> girls in the school. Now we can do <math>\frac{3}{4}12</math> + <math>\frac{2}{3}12</math> to get the total number of students going to the field trip to be <math>17</math>. Since we already know the number of girls to be <math>9</math>. We have our answer to be <math>\frac{9}{17}</math>. So, the answer is <math>\boxed{\textbf{(B)} \frac{9}{17}}</math>. | ||
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+ | ~algebraic_algorithmic | ||
==Video Solution (CREATIVE THINKING!!!)== | ==Video Solution (CREATIVE THINKING!!!)== |
Latest revision as of 20:11, 11 January 2025
Contents
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
Solution 2
Using WLOG (Without loss of generativity), Let there be boys and
girls in the school. Now we can do
+
to get the total number of students going to the field trip to be
. Since we already know the number of girls to be
. We have our answer to be
. So, the answer is
.
~algebraic_algorithmic
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.